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3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
545
|
C
|
Woodcutters
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy"
] | null | null |
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
|
Print a single number — the maximum number of trees that you can cut down by the given rules.
|
[
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] |
[
"3\n",
"4\n"
] |
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
| 1,750
|
[
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4",
"output": "10"
},
{
"input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1",
"output": "2"
},
{
"input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8",
"output": "5"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1",
"output": "9"
},
{
"input": "3\n1 1\n1000 1000\n1000000000 1000000000",
"output": "3"
},
{
"input": "2\n1 999999999\n1000000000 1000000000",
"output": "2"
},
{
"input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000",
"output": "2"
},
{
"input": "2\n100000000 1000000000\n1000000000 1000000000",
"output": "2"
}
] | 1,685,292,862
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll n;
vector<vector<ll>> trees;
ll dfs(ll i, ll l, ll r) {
// cout << i << " " << l << " " << r << '\n';
ll ans = 0;
if (i+1 == n)
return 0;
i += 1;
ll x, h;
x = trees[i][0];
h = trees[i][1];
// cout << x << " " << h << '\n';
if (i == 0) {
ans = max(dfs(i,x-h,x)+1, max(dfs(i,x,x+h)+1, dfs(i,x,x)));
} else if (x < l || x > r) {
if ((x-h) > r) {
ans = dfs(i,min(l,x-h),max(x,r))+1;
}
if (x>r) {
ans = max(ans, dfs(i,min(l,x-h), max(x+h,r)) + 1);
}
ans = max(ans, dfs(i, min(l,x), max(x,r)));
}
return ans;
}
ll power(ll a, ll b) {
ll result = 1;
while (b) {
if (b&1) result *= a;
a *= a;
b /= 2;
}
return result;
}
int main() {
cin >> n;
ll r, h;
for (ll i = 0; i < n; ++i) {
cin >> r >> h;
vector<ll> b = {r, h};
trees.emplace_back(b);
}
// for (vector<ll> b : trees) {
// cout << b[0] << " " << b[1] << '\n';
// }
cout << dfs(-1, power(10,10), -1 * power(10,10)) << '\n';
}
|
Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
|
```python
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll n;
vector<vector<ll>> trees;
ll dfs(ll i, ll l, ll r) {
// cout << i << " " << l << " " << r << '\n';
ll ans = 0;
if (i+1 == n)
return 0;
i += 1;
ll x, h;
x = trees[i][0];
h = trees[i][1];
// cout << x << " " << h << '\n';
if (i == 0) {
ans = max(dfs(i,x-h,x)+1, max(dfs(i,x,x+h)+1, dfs(i,x,x)));
} else if (x < l || x > r) {
if ((x-h) > r) {
ans = dfs(i,min(l,x-h),max(x,r))+1;
}
if (x>r) {
ans = max(ans, dfs(i,min(l,x-h), max(x+h,r)) + 1);
}
ans = max(ans, dfs(i, min(l,x), max(x,r)));
}
return ans;
}
ll power(ll a, ll b) {
ll result = 1;
while (b) {
if (b&1) result *= a;
a *= a;
b /= 2;
}
return result;
}
int main() {
cin >> n;
ll r, h;
for (ll i = 0; i < n; ++i) {
cin >> r >> h;
vector<ll> b = {r, h};
trees.emplace_back(b);
}
// for (vector<ll> b : trees) {
// cout << b[0] << " " << b[1] << '\n';
// }
cout << dfs(-1, power(10,10), -1 * power(10,10)) << '\n';
}
```
| -1
|
|
99
|
B
|
Help Chef Gerasim
|
PROGRAMMING
| 1,300
|
[
"implementation",
"sortings"
] |
B. Help Chef Gerasim
|
0
|
256
|
In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup to another. Now Gerasim wants to check his hypothesis. The good thing is that chef Gerasim always pour the same number of milliliters of juice to all cups in the royal kitchen. Having thoroughly measured the juice in each cup, Gerasim asked you to write a program that will determine from which cup juice was poured to which one; otherwise, the program should determine that this time the pages set the table diligently.
To simplify your task we shall consider the cups to be bottomless so that the juice never overfills a cup and pours out, however much it can be. Besides, by some strange reason in a far away kingdom one can only pour to a cup or from one cup to another an integer number of milliliters of juice.
|
The first line contains integer *n* — the number of cups on the royal table (1<=≤<=*n*<=≤<=1000). Next *n* lines contain volumes of juice in each cup — non-negative integers, not exceeding 104.
|
If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "*v* ml. from cup #*a* to cup #*b*." (without the quotes), where *v* represents the volume of poured juice, *a* represents the number of the cup from which the juice was poured (the cups are numbered with consecutive positive integers starting from one in the order in which the cups are described in the input data), *b* represents the number of the cup into which the juice was poured. Finally, if the given juice's volumes cannot be obtained using no more than one pouring (for example, the pages poured the juice from one cup to another more than once or the royal kitchen maids poured the juice into the cups incorrectly), print "Unrecoverable configuration." (without the quotes).
|
[
"5\n270\n250\n250\n230\n250\n",
"5\n250\n250\n250\n250\n250\n",
"5\n270\n250\n249\n230\n250\n"
] |
[
"20 ml. from cup #4 to cup #1.\n",
"Exemplary pages.\n",
"Unrecoverable configuration.\n"
] |
none
| 1,000
|
[
{
"input": "5\n270\n250\n250\n230\n250",
"output": "20 ml. from cup #4 to cup #1."
},
{
"input": "5\n250\n250\n250\n250\n250",
"output": "Exemplary pages."
},
{
"input": "5\n270\n250\n249\n230\n250",
"output": "Unrecoverable configuration."
},
{
"input": "4\n200\n190\n210\n200",
"output": "10 ml. from cup #2 to cup #3."
},
{
"input": "4\n1\n2\n3\n4",
"output": "Unrecoverable configuration."
},
{
"input": "1\n0",
"output": "Exemplary pages."
},
{
"input": "2\n0\n0",
"output": "Exemplary pages."
},
{
"input": "2\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "2\n0\n2",
"output": "1 ml. from cup #1 to cup #2."
},
{
"input": "2\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "2\n1\n1",
"output": "Exemplary pages."
},
{
"input": "2\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "2\n2\n0",
"output": "1 ml. from cup #2 to cup #1."
},
{
"input": "2\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "2\n2\n2",
"output": "Exemplary pages."
},
{
"input": "3\n0\n0\n0",
"output": "Exemplary pages."
},
{
"input": "3\n0\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n0\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n0\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n0\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n0\n1\n2",
"output": "1 ml. from cup #1 to cup #3."
},
{
"input": "3\n0\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n0\n2\n1",
"output": "1 ml. from cup #1 to cup #2."
},
{
"input": "3\n0\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n0\n2",
"output": "1 ml. from cup #2 to cup #3."
},
{
"input": "3\n1\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n1\n1",
"output": "Exemplary pages."
},
{
"input": "3\n1\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n2\n0",
"output": "1 ml. from cup #3 to cup #2."
},
{
"input": "3\n1\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n1\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n0\n1",
"output": "1 ml. from cup #2 to cup #1."
},
{
"input": "3\n2\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n1\n0",
"output": "1 ml. from cup #3 to cup #1."
},
{
"input": "3\n2\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "3\n2\n2\n2",
"output": "Exemplary pages."
},
{
"input": "4\n0\n0\n0\n0",
"output": "Exemplary pages."
},
{
"input": "4\n0\n0\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n0\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n1\n2",
"output": "1 ml. from cup #1 to cup #4."
},
{
"input": "4\n0\n1\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n1\n2\n1",
"output": "1 ml. from cup #1 to cup #3."
},
{
"input": "4\n0\n1\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n1\n1",
"output": "1 ml. from cup #1 to cup #2."
},
{
"input": "4\n0\n2\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n0\n2\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n1\n2",
"output": "1 ml. from cup #2 to cup #4."
},
{
"input": "4\n1\n0\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n0\n2\n1",
"output": "1 ml. from cup #2 to cup #3."
},
{
"input": "4\n1\n0\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n0\n2",
"output": "1 ml. from cup #3 to cup #4."
},
{
"input": "4\n1\n1\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n1\n1",
"output": "Exemplary pages."
},
{
"input": "4\n1\n1\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n2\n0",
"output": "1 ml. from cup #4 to cup #3."
},
{
"input": "4\n1\n1\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n1\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n0\n1",
"output": "1 ml. from cup #3 to cup #2."
},
{
"input": "4\n1\n2\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n1\n0",
"output": "1 ml. from cup #4 to cup #2."
},
{
"input": "4\n1\n2\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n1\n2\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n1\n1",
"output": "1 ml. from cup #2 to cup #1."
},
{
"input": "4\n2\n0\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n0\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n0\n1",
"output": "1 ml. from cup #3 to cup #1."
},
{
"input": "4\n2\n1\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n1\n0",
"output": "1 ml. from cup #4 to cup #1."
},
{
"input": "4\n2\n1\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n1\n2\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n0\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n0\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n0\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n1\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n1\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n1\n2",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n2\n0",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n2\n1",
"output": "Unrecoverable configuration."
},
{
"input": "4\n2\n2\n2\n2",
"output": "Exemplary pages."
},
{
"input": "27\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n2626\n5599\n5599\n5599\n5599\n5599\n8572\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599\n5599",
"output": "2973 ml. from cup #11 to cup #17."
},
{
"input": "98\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n1204\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n702\n197\n702\n702\n702\n702\n702\n702\n702\n702\n702",
"output": "Unrecoverable configuration."
},
{
"input": "54\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6859\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n5389\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124\n6124",
"output": "735 ml. from cup #37 to cup #29."
},
{
"input": "50\n9636\n9678\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9636\n9596\n9636\n9636\n9636\n9636\n9636\n9636",
"output": "Unrecoverable configuration."
},
{
"input": "19\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n5001\n82\n5001\n9919\n5001\n5001",
"output": "Unrecoverable configuration."
},
{
"input": "74\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n3918\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n5330\n6742\n5330\n5330",
"output": "1412 ml. from cup #36 to cup #72."
}
] | 1,693,454,296
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 93
| 2,048,000
|
n = int(input())
arr = []
for _ in range(n):
arr.append(int(input()))
juice_amount = sum(arr) / len(arr)
arr_dict = {}
for element in arr:
arr_dict[element] = arr_dict.get(element, 0) + 1
if(len(arr_dict) == 1):
print("Exemplary pages.")
exit()
if(arr_dict[juice_amount] != len(arr) - 2):
print("Unrecoverable configuration.")
exit()
amount_poured = 0
sender = -1
receiver = -1
for i in range(len(arr)):
if(arr[i] > juice_amount):
amount_poured = arr[i] - juice_amount
receiver = i+1
if(arr[i] < juice_amount):
sender = i+1
print("%d ml. from cup #%d to cup #%d."%(amount_poured, sender, receiver))
|
Title: Help Chef Gerasim
Time Limit: 0 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup to another. Now Gerasim wants to check his hypothesis. The good thing is that chef Gerasim always pour the same number of milliliters of juice to all cups in the royal kitchen. Having thoroughly measured the juice in each cup, Gerasim asked you to write a program that will determine from which cup juice was poured to which one; otherwise, the program should determine that this time the pages set the table diligently.
To simplify your task we shall consider the cups to be bottomless so that the juice never overfills a cup and pours out, however much it can be. Besides, by some strange reason in a far away kingdom one can only pour to a cup or from one cup to another an integer number of milliliters of juice.
Input Specification:
The first line contains integer *n* — the number of cups on the royal table (1<=≤<=*n*<=≤<=1000). Next *n* lines contain volumes of juice in each cup — non-negative integers, not exceeding 104.
Output Specification:
If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "*v* ml. from cup #*a* to cup #*b*." (without the quotes), where *v* represents the volume of poured juice, *a* represents the number of the cup from which the juice was poured (the cups are numbered with consecutive positive integers starting from one in the order in which the cups are described in the input data), *b* represents the number of the cup into which the juice was poured. Finally, if the given juice's volumes cannot be obtained using no more than one pouring (for example, the pages poured the juice from one cup to another more than once or the royal kitchen maids poured the juice into the cups incorrectly), print "Unrecoverable configuration." (without the quotes).
Demo Input:
['5\n270\n250\n250\n230\n250\n', '5\n250\n250\n250\n250\n250\n', '5\n270\n250\n249\n230\n250\n']
Demo Output:
['20 ml. from cup #4 to cup #1.\n', 'Exemplary pages.\n', 'Unrecoverable configuration.\n']
Note:
none
|
```python
n = int(input())
arr = []
for _ in range(n):
arr.append(int(input()))
juice_amount = sum(arr) / len(arr)
arr_dict = {}
for element in arr:
arr_dict[element] = arr_dict.get(element, 0) + 1
if(len(arr_dict) == 1):
print("Exemplary pages.")
exit()
if(arr_dict[juice_amount] != len(arr) - 2):
print("Unrecoverable configuration.")
exit()
amount_poured = 0
sender = -1
receiver = -1
for i in range(len(arr)):
if(arr[i] > juice_amount):
amount_poured = arr[i] - juice_amount
receiver = i+1
if(arr[i] < juice_amount):
sender = i+1
print("%d ml. from cup #%d to cup #%d."%(amount_poured, sender, receiver))
```
| -1
|
766
|
B
|
Mahmoud and a Triangle
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null |
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
|
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
|
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
|
[
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] |
[
"YES\n",
"NO\n"
] |
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
| 1,000
|
[
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},
{
"input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761",
"output": "YES"
},
{
"input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974",
"output": "YES"
},
{
"input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291",
"output": "NO"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "YES"
},
{
"input": "4\n1 1000000000 1000000000 1000000000",
"output": "YES"
},
{
"input": "3\n1 1000000000 1",
"output": "NO"
},
{
"input": "5\n1 2 3 5 2",
"output": "YES"
},
{
"input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431",
"output": "NO"
},
{
"input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159",
"output": "YES"
},
{
"input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023",
"output": "YES"
},
{
"input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015",
"output": "YES"
},
{
"input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14",
"output": "YES"
},
{
"input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368",
"output": "NO"
},
{
"input": "3\n1 1000000000 999999999",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000",
"output": "NO"
},
{
"input": "5\n2 3 4 10 20",
"output": "YES"
},
{
"input": "6\n18 23 40 80 160 161",
"output": "YES"
},
{
"input": "4\n5 6 7 888",
"output": "YES"
},
{
"input": "9\n1 1 2 2 4 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 900 4 500 1500 5",
"output": "YES"
},
{
"input": "3\n2 2 3",
"output": "YES"
},
{
"input": "7\n1 2 100 200 250 1000000 2000000",
"output": "YES"
},
{
"input": "8\n2 3 5 5 5 6 6 13",
"output": "YES"
},
{
"input": "3\n2 3 4",
"output": "YES"
},
{
"input": "6\n1 1 1 4 5 100",
"output": "YES"
},
{
"input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377",
"output": "YES"
},
{
"input": "4\n2 3 4 8",
"output": "YES"
},
{
"input": "3\n5 6 7",
"output": "YES"
},
{
"input": "5\n1 4 5 6 1000000",
"output": "YES"
},
{
"input": "4\n5 6 7 20",
"output": "YES"
},
{
"input": "6\n1 1 1 1 1 65",
"output": "YES"
},
{
"input": "4\n3 4 5 100",
"output": "YES"
},
{
"input": "3\n2 4 5",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 10 1000",
"output": "YES"
},
{
"input": "4\n1 1 2 3",
"output": "NO"
},
{
"input": "11\n1 2 5 6 7 8 9 17 18 19 100",
"output": "YES"
},
{
"input": "4\n5 16 20 200",
"output": "YES"
},
{
"input": "5\n17 6 3 3 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "6\n1 1 1 2 3 5",
"output": "YES"
},
{
"input": "4\n2 4 6 6",
"output": "YES"
},
{
"input": "9\n1 2 4 4 4 4 7 8 20",
"output": "YES"
},
{
"input": "9\n1 1 2 5 5 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 600 4 1700 6000 5",
"output": "YES"
},
{
"input": "5\n5761 20966 27841 28800 29399",
"output": "YES"
},
{
"input": "9\n1 2 3 6 7 10 11 12 24",
"output": "YES"
},
{
"input": "4\n1 2 1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 4",
"output": "YES"
}
] | 1,651,803,289
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 87
| 156
| 16,691,200
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
# print(a)
f = 0
for i in range(2, n):
if a[i]<a[i-1]+a[i-2]:
f = 1
break
if f:
print("YES")
else:
print("NO")
|
Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
|
```python
n = int(input())
a = list(map(int, input().split()))
a.sort()
# print(a)
f = 0
for i in range(2, n):
if a[i]<a[i-1]+a[i-2]:
f = 1
break
if f:
print("YES")
else:
print("NO")
```
| 3
|
|
16
|
A
|
Flag
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Flag
|
2
|
64
|
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
|
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
|
Output YES, if the flag meets the new ISO standard, and NO otherwise.
|
[
"3 3\n000\n111\n222\n",
"3 3\n000\n000\n111\n",
"3 3\n000\n111\n002\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "3 3\n000\n111\n222",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "NO"
},
{
"input": "3 3\n000\n111\n002",
"output": "NO"
},
{
"input": "10 10\n2222222222\n5555555555\n0000000000\n4444444444\n1111111111\n3333333393\n3333333333\n5555555555\n0000000000\n8888888888",
"output": "NO"
},
{
"input": "10 13\n4442444444444\n8888888888888\n6666666666666\n0000000000000\n3333333333333\n4444444444444\n7777777777777\n8388888888888\n1111111111111\n5555555555555",
"output": "NO"
},
{
"input": "10 8\n33333333\n44444444\n11111115\n81888888\n44444444\n11111111\n66666666\n33330333\n33333333\n33333333",
"output": "NO"
},
{
"input": "5 5\n88888\n44444\n66666\n55555\n88888",
"output": "YES"
},
{
"input": "20 19\n1111111111111111111\n5555555555555555555\n0000000000000000000\n3333333333333333333\n1111111111111111111\n2222222222222222222\n4444444444444444444\n5555555555555555555\n0000000000000000000\n4444444444444444444\n0000000000000000000\n5555555555555555555\n7777777777777777777\n9999999999999999999\n2222222222222222222\n4444444444444444444\n1111111111111111111\n6666666666666666666\n7777777777777777777\n2222222222222222222",
"output": "YES"
},
{
"input": "1 100\n8888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888",
"output": "YES"
},
{
"input": "100 1\n5\n7\n9\n4\n7\n2\n5\n1\n6\n7\n2\n7\n6\n8\n7\n4\n0\n2\n9\n8\n9\n1\n6\n4\n3\n4\n7\n1\n9\n3\n0\n8\n3\n1\n7\n5\n3\n9\n5\n1\n3\n5\n8\n1\n9\n3\n9\n0\n6\n0\n7\n6\n5\n2\n8\n3\n7\n6\n5\n1\n8\n3\n6\n9\n6\n0\n5\n8\n5\n2\n9\n1\n0\n1\n8\n3\n2\n1\n0\n3\n9\n0\n5\n1\n0\n4\n9\n3\n0\n4\n8\n4\n8\n6\n3\n0\n4\n6\n8\n4",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "1 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111181111111111111111111111",
"output": "NO"
},
{
"input": "100 1\n3\n6\n4\n3\n0\n2\n8\n7\n3\n2\n1\n7\n1\n3\n2\n3\n6\n9\n0\n8\n5\n9\n7\n9\n2\n1\n4\n5\n1\n9\n2\n5\n1\n4\n6\n4\n9\n1\n0\n2\n1\n4\n7\n1\n4\n8\n0\n9\n2\n1\n6\n2\n8\n6\n9\n5\n8\n6\n4\n5\n9\n2\n7\n4\n1\n5\n8\n0\n9\n5\n4\n6\n5\n0\n6\n3\n6\n9\n7\n2\n0\n9\n7\n3\n2\n4\n9\n4\n7\n1\n2\n3\n1\n7\n9\n1\n9\n0\n4\n0",
"output": "YES"
}
] | 1,648,672,612
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
#Ejercicio 16A codeforce
#este ejercicio las personas ingresan numeros por pantalla y el programa solo dice si esta bien o no
listColors = []
matriz = []
equivocacion = 0
filas = input(" ")
f = 0
while f<int(filas[0]):
colors = input("")
listColors.append(colors)
for x in range(len(colors)):
if colors[x] != colors [x-1]:
equivocacion = 1
f+=1
for y in range(len(listColors)-1):
if listColors[y] == listColors[y+1]:
equivocacion = 1
if equivocacion == 1:
print("No")
else:
print ("Si")
|
Title: Flag
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
Input Specification:
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output Specification:
Output YES, if the flag meets the new ISO standard, and NO otherwise.
Demo Input:
['3 3\n000\n111\n222\n', '3 3\n000\n000\n111\n', '3 3\n000\n111\n002\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
#Ejercicio 16A codeforce
#este ejercicio las personas ingresan numeros por pantalla y el programa solo dice si esta bien o no
listColors = []
matriz = []
equivocacion = 0
filas = input(" ")
f = 0
while f<int(filas[0]):
colors = input("")
listColors.append(colors)
for x in range(len(colors)):
if colors[x] != colors [x-1]:
equivocacion = 1
f+=1
for y in range(len(listColors)-1):
if listColors[y] == listColors[y+1]:
equivocacion = 1
if equivocacion == 1:
print("No")
else:
print ("Si")
```
| 0
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,622,976,393
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 434
| 2,150,400
|
a=input()
s=0
for i in a:
if i==upper(i):
s+=1
if s>len(a)//2:
print(upper(a))
else:
print(lower(a))
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
a=input()
s=0
for i in a:
if i==upper(i):
s+=1
if s>len(a)//2:
print(upper(a))
else:
print(lower(a))
```
| -1
|
22
|
A
|
Second Order Statistics
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Second Order Statistics
|
2
|
256
|
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
|
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
|
[
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] |
[
"1\n",
"2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,595,045,254
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 10
| 218
| 6,656,000
|
"""
Author: VISHVESH BHAVSAR :)
"""
n=int(input())
ar=sorted(set([int(x) for x in input().split()]))
if n==1:
print("NO")
else:
#print(ar)
print(ar[1])
|
Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none
|
```python
"""
Author: VISHVESH BHAVSAR :)
"""
n=int(input())
ar=sorted(set([int(x) for x in input().split()]))
if n==1:
print("NO")
else:
#print(ar)
print(ar[1])
```
| -1
|
768
|
B
|
Code For 1
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"dfs and similar",
"divide and conquer"
] | null | null |
Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.
Initially Sam has a list with a single element *n*. Then he has to perform certain operations on this list. In each operation Sam must remove any element *x*, such that *x*<=><=1, from the list and insert at the same position , , sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.
Now the masters want the total number of 1s in the range *l* to *r* (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?
|
The first line contains three integers *n*, *l*, *r* (0<=≤<=*n*<=<<=250, 0<=≤<=*r*<=-<=*l*<=≤<=105, *r*<=≥<=1, *l*<=≥<=1) – initial element and the range *l* to *r*.
It is guaranteed that *r* is not greater than the length of the final list.
|
Output the total number of 1s in the range *l* to *r* in the final sequence.
|
[
"7 2 5\n",
"10 3 10\n"
] |
[
"4\n",
"5\n"
] |
Consider first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/288fbb682a6fa1934a47b763d6851f9d32a06150.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.
For the second example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/52e9bc51ef858cacc27fc274c7ba9419d5c1ded9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.
| 1,000
|
[
{
"input": "7 2 5",
"output": "4"
},
{
"input": "10 3 10",
"output": "5"
},
{
"input": "56 18 40",
"output": "20"
},
{
"input": "203 40 124",
"output": "67"
},
{
"input": "903316762502 354723010040 354723105411",
"output": "78355"
},
{
"input": "33534354842198 32529564319236 32529564342569",
"output": "22239"
},
{
"input": "62518534961045 50734311240112 50734311287877",
"output": "42439"
},
{
"input": "95173251245550 106288351347530 106288351372022",
"output": "16565"
},
{
"input": "542 321 956",
"output": "336"
},
{
"input": "3621 237 2637",
"output": "2124"
},
{
"input": "9056 336 896",
"output": "311"
},
{
"input": "36007 368 24490",
"output": "13253"
},
{
"input": "244269 149154 244246",
"output": "88609"
},
{
"input": "880234 669493 757150",
"output": "73585"
},
{
"input": "3740160 1031384 1104236",
"output": "64965"
},
{
"input": "11586121 15337246 15397874",
"output": "41868"
},
{
"input": "38658997 35923164 35985664",
"output": "36004"
},
{
"input": "192308932 207804787 207866400",
"output": "44142"
},
{
"input": "950099012 175922161 176000556",
"output": "69369"
},
{
"input": "2787326787 3799676481 3799680514",
"output": "2618"
},
{
"input": "14417262581 8527979363 8528075536",
"output": "80707"
},
{
"input": "39889373539 7747197212 7747278363",
"output": "47105"
},
{
"input": "251772781087 70597428577 70597479816",
"output": "46933"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "3 2 3",
"output": "2"
},
{
"input": "14 7 12",
"output": "5"
},
{
"input": "1125899906842623 1 100001",
"output": "100001"
},
{
"input": "1125899906842623 1125899906742623 1125899906842623",
"output": "100001"
},
{
"input": "1000 1 1023",
"output": "1000"
},
{
"input": "281474976710656 17179869184 17179869186",
"output": "1"
},
{
"input": "2 2 2",
"output": "0"
},
{
"input": "3 2 2",
"output": "1"
}
] | 1,697,702,774
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
def count_ones(n, l, r):
def recursive_process(num):
if num == 1:
return [1]
else:
half = num // 2
remainder = num % 2
return recursive_process(half) + [remainder] + recursive_process(half)
sequence = recursive_process(n)
ones_count = 0
for i in range(l - 1, r):
ones_count += sequence[i]
return ones_count
|
Title: Code For 1
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.
Initially Sam has a list with a single element *n*. Then he has to perform certain operations on this list. In each operation Sam must remove any element *x*, such that *x*<=><=1, from the list and insert at the same position , , sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.
Now the masters want the total number of 1s in the range *l* to *r* (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?
Input Specification:
The first line contains three integers *n*, *l*, *r* (0<=≤<=*n*<=<<=250, 0<=≤<=*r*<=-<=*l*<=≤<=105, *r*<=≥<=1, *l*<=≥<=1) – initial element and the range *l* to *r*.
It is guaranteed that *r* is not greater than the length of the final list.
Output Specification:
Output the total number of 1s in the range *l* to *r* in the final sequence.
Demo Input:
['7 2 5\n', '10 3 10\n']
Demo Output:
['4\n', '5\n']
Note:
Consider first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/288fbb682a6fa1934a47b763d6851f9d32a06150.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.
For the second example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/52e9bc51ef858cacc27fc274c7ba9419d5c1ded9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.
|
```python
def count_ones(n, l, r):
def recursive_process(num):
if num == 1:
return [1]
else:
half = num // 2
remainder = num % 2
return recursive_process(half) + [remainder] + recursive_process(half)
sequence = recursive_process(n)
ones_count = 0
for i in range(l - 1, r):
ones_count += sequence[i]
return ones_count
```
| 0
|
|
908
|
A
|
New Year and Counting Cards
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
|
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
|
Print a single integer, the minimum number of cards you must turn over to verify your claim.
|
[
"ee\n",
"z\n",
"0ay1\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
| 500
|
[
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,564,425,740
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 109
| 0
|
n=list(input())
t=['a','e','i','o','u']
count=0
for i in range(len(n)):
if n[i] in t and n[i].isdigit()==False:
count+=1
if n[i].isdigit() and int(n[i])%2!=0:
count+=1
print(count)
|
Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
|
```python
n=list(input())
t=['a','e','i','o','u']
count=0
for i in range(len(n)):
if n[i] in t and n[i].isdigit()==False:
count+=1
if n[i].isdigit() and int(n[i])%2!=0:
count+=1
print(count)
```
| 3
|
|
994
|
A
|
Fingerprints
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
|
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
|
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
|
[
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] |
[
"7 1 2\n",
"1 0\n"
] |
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
| 500
|
[
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8",
"output": "1 2 3 4 5 6 7 8 9 0"
},
{
"input": "1 1\n4\n4",
"output": "4"
},
{
"input": "3 7\n6 3 4\n4 9 0 1 7 8 6",
"output": "6 4"
},
{
"input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0",
"output": "0"
},
{
"input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6",
"output": "6 0 3 8 1"
},
{
"input": "8 2\n7 2 9 6 1 0 3 4\n6 3",
"output": "6 3"
},
{
"input": "5 4\n7 0 1 4 9\n0 9 5 3",
"output": "0 9"
},
{
"input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6",
"output": "6"
},
{
"input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2",
"output": "2 3"
},
{
"input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2",
"output": "3 9 2 4"
},
{
"input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6",
"output": "1 2 3 8 6 5"
},
{
"input": "8 2\n7 4 8 9 2 5 6 1\n6 4",
"output": "4 6"
},
{
"input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3",
"output": "0 3"
},
{
"input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8",
"output": "2 8 1 3"
},
{
"input": "1 6\n3\n6 8 2 4 5 3",
"output": "3"
},
{
"input": "1 8\n0\n9 2 4 8 1 5 0 7",
"output": "0"
},
{
"input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4",
"output": "7 9 4 1 0"
},
{
"input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1",
"output": "0 1"
},
{
"input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2",
"output": "2 9 6 1 4"
},
{
"input": "6 3\n8 3 9 2 7 6\n5 4 3",
"output": "3"
},
{
"input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5",
"output": "8 3 9 6"
},
{
"input": "1 2\n1\n1 0",
"output": "1"
},
{
"input": "3 6\n1 2 3\n4 5 6 1 2 3",
"output": "1 2 3"
},
{
"input": "1 2\n2\n1 2",
"output": "2"
},
{
"input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9",
"output": "9"
}
] | 1,612,405,939
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 93
| 0
|
n,m=map(int,input().split())
seq=list(map(str,input().split()))
fp=list(map(str,input().split()))
checklist=[]
for number in seq:
if(number in fp):
checklist.append(number)
print(" ".join(checklist))
|
Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
|
```python
n,m=map(int,input().split())
seq=list(map(str,input().split()))
fp=list(map(str,input().split()))
checklist=[]
for number in seq:
if(number in fp):
checklist.append(number)
print(" ".join(checklist))
```
| 3
|
|
999
|
A
|
Mishka and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
|
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
|
Print one integer — the maximum number of problems Mishka can solve.
|
[
"8 4\n4 2 3 1 5 1 6 4\n",
"5 2\n3 1 2 1 3\n",
"5 100\n12 34 55 43 21\n"
] |
[
"5\n",
"0\n",
"5\n"
] |
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
| 0
|
[
{
"input": "8 4\n4 2 3 1 5 1 6 4",
"output": "5"
},
{
"input": "5 2\n3 1 2 1 3",
"output": "0"
},
{
"input": "5 100\n12 34 55 43 21",
"output": "5"
},
{
"input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62",
"output": "100"
},
{
"input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98",
"output": "98"
},
{
"input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2",
"output": "39"
},
{
"input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16",
"output": "99"
},
{
"input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1",
"output": "1"
},
{
"input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "3 2\n1 4 1",
"output": "2"
},
{
"input": "1 2\n100",
"output": "0"
},
{
"input": "7 4\n4 2 3 4 4 2 3",
"output": "7"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "1 2\n15",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "5 3\n3 4 3 2 1",
"output": "4"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "1 5\n1",
"output": "1"
},
{
"input": "6 6\n7 1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n6 5 5 5 5",
"output": "4"
},
{
"input": "1 4\n2",
"output": "1"
},
{
"input": "9 4\n1 2 1 2 4 2 1 2 1",
"output": "9"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n5",
"output": "1"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4",
"output": "71"
},
{
"input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90",
"output": "63"
},
{
"input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8",
"output": "61"
},
{
"input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22",
"output": "72"
},
{
"input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2",
"output": "52"
},
{
"input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90",
"output": "60"
},
{
"input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10",
"output": "56"
},
{
"input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90",
"output": "61"
},
{
"input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24",
"output": "99"
},
{
"input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50",
"output": "99"
},
{
"input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48",
"output": "99"
},
{
"input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57",
"output": "99"
},
{
"input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93",
"output": "99"
},
{
"input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33",
"output": "3"
},
{
"input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51",
"output": "3"
},
{
"input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49",
"output": "6"
},
{
"input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8",
"output": "66"
},
{
"input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95",
"output": "0"
},
{
"input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66",
"output": "0"
},
{
"input": "2 8\n8 8",
"output": "2"
},
{
"input": "1 6\n3",
"output": "1"
},
{
"input": "1 5\n5",
"output": "1"
}
] | 1,694,583,597
| 297
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
n, k=map(int, input().split())
l=list(map(int, input().split()))
c=0
for i in l:
if i<=k:
c+=1
else:
break
if c<n:
for i in l[::-1]:
if i<=k:
c+=1
else:
break
print(c)
|
Title: Mishka and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Output Specification:
Print one integer — the maximum number of problems Mishka can solve.
Demo Input:
['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n']
Demo Output:
['5\n', '0\n', '5\n']
Note:
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
|
```python
n, k=map(int, input().split())
l=list(map(int, input().split()))
c=0
for i in l:
if i<=k:
c+=1
else:
break
if c<n:
for i in l[::-1]:
if i<=k:
c+=1
else:
break
print(c)
```
| 3
|
|
863
|
A
|
Quasi-palindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
|
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
|
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
|
[
"131\n",
"320\n",
"2010200\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 0
|
[
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
},
{
"input": "102000",
"output": "NO"
},
{
"input": "210000000",
"output": "NO"
},
{
"input": "213443120",
"output": "YES"
},
{
"input": "99",
"output": "YES"
},
{
"input": "22002",
"output": "NO"
},
{
"input": "1010",
"output": "YES"
},
{
"input": "1201",
"output": "NO"
},
{
"input": "6460046",
"output": "NO"
},
{
"input": "503435",
"output": "NO"
},
{
"input": "21002",
"output": "NO"
},
{
"input": "101001",
"output": "NO"
},
{
"input": "200102",
"output": "NO"
},
{
"input": "20010002",
"output": "NO"
},
{
"input": "33003",
"output": "NO"
},
{
"input": "100101",
"output": "NO"
},
{
"input": "1021",
"output": "NO"
},
{
"input": "1101",
"output": "NO"
},
{
"input": "10101100",
"output": "NO"
},
{
"input": "101",
"output": "YES"
},
{
"input": "1011",
"output": "NO"
},
{
"input": "11010",
"output": "NO"
},
{
"input": "10110",
"output": "NO"
},
{
"input": "110000",
"output": "YES"
},
{
"input": "2011",
"output": "NO"
},
{
"input": "10020001",
"output": "NO"
},
{
"input": "12505021",
"output": "NO"
},
{
"input": "12310",
"output": "NO"
},
{
"input": "100501",
"output": "NO"
},
{
"input": "11001",
"output": "NO"
},
{
"input": "20020002",
"output": "NO"
},
{
"input": "202002",
"output": "NO"
},
{
"input": "1001",
"output": "YES"
},
{
"input": "1020021",
"output": "NO"
},
{
"input": "60660",
"output": "NO"
},
{
"input": "98809",
"output": "NO"
},
{
"input": "11000000",
"output": "YES"
},
{
"input": "807008",
"output": "NO"
}
] | 1,584,293,651
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
num=int(input())
temp=num
rev=0
while temp!=0:
rev=(rev*10)+(temp%10)
temp=temp//10
if num==rev:
print("YES")
else:
print("NO")
|
Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
num=int(input())
temp=num
rev=0
while temp!=0:
rev=(rev*10)+(temp%10)
temp=temp//10
if num==rev:
print("YES")
else:
print("NO")
```
| 0
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,698,257,307
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
n = input()
distinct_characters = set()
for char in n:
if char.isalnum(): # You can adjust this condition to match your requirements
distinct_characters.add(char)
count = len(distinct_characters)
print(count)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
n = input()
distinct_characters = set()
for char in n:
if char.isalnum(): # You can adjust this condition to match your requirements
distinct_characters.add(char)
count = len(distinct_characters)
print(count)
```
| 3
|
|
845
|
D
|
Driving Test
|
PROGRAMMING
| 1,800
|
[
"data structures",
"dp",
"greedy"
] | null | null |
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
- speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type); - overtake is allowed: this sign means that after some car meets it, it can overtake any other car; - no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign); - no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number); 1. Polycarp's car overtakes the other car; 1. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer); 1. Polycarp's car goes past the "overtake is allowed" sign; 1. Polycarp's car goes past the "no speed limit"; 1. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
|
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=2·105) — number of events.
Each of the next *n* lines starts with integer *t* (1<=≤<=*t*<=≤<=6) — the type of the event.
An integer *s* (1<=≤<=*s*<=≤<=300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
|
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
|
[
"11\n1 100\n3 70\n4\n2\n3 120\n5\n3 120\n6\n1 150\n4\n3 300\n",
"5\n1 100\n3 200\n2\n4\n5\n",
"7\n1 20\n2\n6\n4\n6\n6\n2\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
| 0
|
[
{
"input": "11\n1 100\n3 70\n4\n2\n3 120\n5\n3 120\n6\n1 150\n4\n3 300",
"output": "2"
},
{
"input": "5\n1 100\n3 200\n2\n4\n5",
"output": "0"
},
{
"input": "7\n1 20\n2\n6\n4\n6\n6\n2",
"output": "2"
},
{
"input": "1\n1 100",
"output": "0"
},
{
"input": "2\n1 100\n2",
"output": "0"
},
{
"input": "10\n1 37\n6\n5\n2\n5\n6\n5\n2\n4\n2",
"output": "2"
},
{
"input": "10\n1 5\n4\n5\n4\n1 21\n6\n1 10\n1 1\n2\n3 1",
"output": "1"
}
] | 1,622,725,702
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 608
| 1,536,000
|
n = int(input())
speed_limits = []
obgon = 0
current_speed = 0
count = 0
for i in range(n):
event = list(input().split())
if (len(event) == 2):
type, x = map(int, event)
if type == 1: # 1 изменил скорость на х
current_speed = x
while len(speed_limits) != 0 and speed_limits[-1] < current_speed:
count += 1
speed_limits.pop()
elif type == 3: # 3 новый знак ограничение скорости х
speed_limits.append(x)
if speed_limits[-1] < current_speed:
count += 1
speed_limits.pop()
else:
type = int(event[0])
if type == 2: # 2 совершил обгон
if obgon == 0:
continue
else:
count += obgon
obgon = 0
elif type == 4: # 4 новый знак обгон разрешен
obgon = 0
elif type == 5: # 5 новый знак отмена ограничения скорости
speed_limits.clear()
elif type == 6: # 6 новый знак обгон запрещен
obgon += 1
print(count)
|
Title: Driving Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has just attempted to pass the driving test. He ran over the straight road with the signs of four types.
- speed limit: this sign comes with a positive integer number — maximal speed of the car after the sign (cancel the action of the previous sign of this type); - overtake is allowed: this sign means that after some car meets it, it can overtake any other car; - no speed limit: this sign cancels speed limit if any (car can move with arbitrary speed after this sign); - no overtake allowed: some car can't overtake any other car after this sign.
Polycarp goes past the signs consequentially, each new sign cancels the action of all the previous signs of it's kind (speed limit/overtake). It is possible that two or more "no overtake allowed" signs go one after another with zero "overtake is allowed" signs between them. It works with "no speed limit" and "overtake is allowed" signs as well.
In the beginning of the ride overtake is allowed and there is no speed limit.
You are given the sequence of events in chronological order — events which happened to Polycarp during the ride. There are events of following types:
1. Polycarp changes the speed of his car to specified (this event comes with a positive integer number); 1. Polycarp's car overtakes the other car; 1. Polycarp's car goes past the "speed limit" sign (this sign comes with a positive integer); 1. Polycarp's car goes past the "overtake is allowed" sign; 1. Polycarp's car goes past the "no speed limit"; 1. Polycarp's car goes past the "no overtake allowed";
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
After the exam Polycarp can justify his rule violations by telling the driving instructor that he just didn't notice some of the signs. What is the minimal number of signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view?
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=2·105) — number of events.
Each of the next *n* lines starts with integer *t* (1<=≤<=*t*<=≤<=6) — the type of the event.
An integer *s* (1<=≤<=*s*<=≤<=300) follows in the query of the first and the third type (if it is the query of first type, then it's new speed of Polycarp's car, if it is the query of third type, then it's new speed limit).
It is guaranteed that the first event in chronological order is the event of type 1 (Polycarp changed the speed of his car to specified).
Output Specification:
Print the minimal number of road signs Polycarp should say he didn't notice, so that he would make no rule violations from his point of view.
Demo Input:
['11\n1 100\n3 70\n4\n2\n3 120\n5\n3 120\n6\n1 150\n4\n3 300\n', '5\n1 100\n3 200\n2\n4\n5\n', '7\n1 20\n2\n6\n4\n6\n6\n2\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first example Polycarp should say he didn't notice the "speed limit" sign with the limit of 70 and the second "speed limit" sign with the limit of 120.
In the second example Polycarp didn't make any rule violation.
In the third example Polycarp should say he didn't notice both "no overtake allowed" that came after "overtake is allowed" sign.
|
```python
n = int(input())
speed_limits = []
obgon = 0
current_speed = 0
count = 0
for i in range(n):
event = list(input().split())
if (len(event) == 2):
type, x = map(int, event)
if type == 1: # 1 изменил скорость на х
current_speed = x
while len(speed_limits) != 0 and speed_limits[-1] < current_speed:
count += 1
speed_limits.pop()
elif type == 3: # 3 новый знак ограничение скорости х
speed_limits.append(x)
if speed_limits[-1] < current_speed:
count += 1
speed_limits.pop()
else:
type = int(event[0])
if type == 2: # 2 совершил обгон
if obgon == 0:
continue
else:
count += obgon
obgon = 0
elif type == 4: # 4 новый знак обгон разрешен
obgon = 0
elif type == 5: # 5 новый знак отмена ограничения скорости
speed_limits.clear()
elif type == 6: # 6 новый знак обгон запрещен
obgon += 1
print(count)
```
| 3
|
|
807
|
A
|
Is it rated?
|
PROGRAMMING
| 900
|
[
"implementation",
"sortings"
] | null | null |
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
|
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
|
[
"6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n",
"4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n",
"5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n"
] |
[
"rated\n",
"unrated\n",
"maybe\n"
] |
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
| 500
|
[
{
"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884",
"output": "rated"
},
{
"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n1 1\n1 1",
"output": "maybe"
},
{
"input": "2\n4126 4126\n4126 4126",
"output": "maybe"
},
{
"input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423",
"output": "rated"
},
{
"input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110",
"output": "unrated"
},
{
"input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143",
"output": "maybe"
},
{
"input": "2\n3936 3936\n2967 2967",
"output": "maybe"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 1\n1 2",
"output": "rated"
},
{
"input": "2\n2967 2967\n3936 3936",
"output": "unrated"
},
{
"input": "3\n1200 1200\n1200 1200\n1300 1300",
"output": "unrated"
},
{
"input": "3\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "3\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n3 2\n3 2",
"output": "rated"
},
{
"input": "3\n5 5\n4 4\n3 4",
"output": "rated"
},
{
"input": "3\n200 200\n200 200\n300 300",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n3 3",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699",
"output": "maybe"
},
{
"input": "2\n10 10\n8 8",
"output": "maybe"
},
{
"input": "3\n1500 1500\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n100 100\n100 100\n70 70\n80 80",
"output": "unrated"
},
{
"input": "2\n1 2\n2 1",
"output": "rated"
},
{
"input": "3\n5 5\n4 3\n3 3",
"output": "rated"
},
{
"input": "3\n1600 1650\n1500 1550\n1400 1450",
"output": "rated"
},
{
"input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700",
"output": "unrated"
},
{
"input": "2\n1600 1600\n1400 1400",
"output": "maybe"
},
{
"input": "2\n3 1\n9 8",
"output": "rated"
},
{
"input": "2\n2 1\n1 1",
"output": "rated"
},
{
"input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670",
"output": "unrated"
},
{
"input": "2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n10 11\n5 4",
"output": "rated"
},
{
"input": "2\n15 14\n13 12",
"output": "rated"
},
{
"input": "2\n2 1\n2 2",
"output": "rated"
},
{
"input": "3\n2670 2670\n3670 3670\n4106 4106",
"output": "unrated"
},
{
"input": "3\n4 5\n3 3\n2 2",
"output": "rated"
},
{
"input": "2\n10 9\n10 10",
"output": "rated"
},
{
"input": "3\n1011 1011\n1011 999\n2200 2100",
"output": "rated"
},
{
"input": "2\n3 3\n5 5",
"output": "unrated"
},
{
"input": "2\n1500 1500\n3000 2000",
"output": "rated"
},
{
"input": "2\n5 6\n5 5",
"output": "rated"
},
{
"input": "3\n2000 2000\n1500 1501\n500 500",
"output": "rated"
},
{
"input": "2\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n1 2\n1 1",
"output": "rated"
},
{
"input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n15 14\n14 13",
"output": "rated"
},
{
"input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900",
"output": "unrated"
},
{
"input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884",
"output": "rated"
},
{
"input": "2\n100 99\n100 100",
"output": "rated"
},
{
"input": "4\n2 2\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n100 101\n100 100\n100 100",
"output": "rated"
},
{
"input": "4\n1000 1001\n900 900\n950 950\n890 890",
"output": "rated"
},
{
"input": "2\n2 3\n1 1",
"output": "rated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n3 2\n2 2",
"output": "rated"
},
{
"input": "2\n3 2\n3 3",
"output": "rated"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n3 2\n3 3\n3 3",
"output": "rated"
},
{
"input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "3\n1000 1000\n500 500\n400 300",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000",
"output": "unrated"
},
{
"input": "2\n1 1\n2 3",
"output": "rated"
},
{
"input": "2\n6 2\n6 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246",
"output": "unrated"
},
{
"input": "2\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699",
"output": "maybe"
},
{
"input": "2\n20 30\n10 5",
"output": "rated"
},
{
"input": "3\n1 1\n2 2\n1 1",
"output": "unrated"
},
{
"input": "2\n1 2\n3 3",
"output": "rated"
},
{
"input": "5\n5 5\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 2\n2 1",
"output": "rated"
},
{
"input": "2\n100 100\n90 89",
"output": "rated"
},
{
"input": "2\n1000 900\n2000 2000",
"output": "rated"
},
{
"input": "2\n50 10\n10 50",
"output": "rated"
},
{
"input": "2\n200 200\n100 100",
"output": "maybe"
},
{
"input": "3\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "3\n1000 1000\n300 300\n100 100",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n3 3\n4 4",
"output": "unrated"
},
{
"input": "2\n5 3\n6 3",
"output": "rated"
},
{
"input": "2\n1200 1100\n1200 1000",
"output": "rated"
},
{
"input": "2\n5 5\n4 4",
"output": "maybe"
},
{
"input": "2\n5 5\n3 3",
"output": "maybe"
},
{
"input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100",
"output": "unrated"
},
{
"input": "5\n10 10\n9 9\n8 8\n7 7\n6 6",
"output": "maybe"
},
{
"input": "3\n1000 1000\n300 300\n10 10",
"output": "maybe"
},
{
"input": "5\n6 6\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n3 3\n1 1",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n1000 1000\n700 700",
"output": "maybe"
},
{
"input": "2\n4 3\n5 3",
"output": "rated"
},
{
"input": "2\n1000 1000\n1100 1100",
"output": "unrated"
},
{
"input": "4\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "3\n1 1\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n1 2\n1 3",
"output": "rated"
},
{
"input": "2\n3 3\n1 2",
"output": "rated"
},
{
"input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "5\n1 1\n2 2\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "2\n10 10\n1 2",
"output": "rated"
},
{
"input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900",
"output": "unrated"
},
{
"input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699",
"output": "unrated"
},
{
"input": "2\n100 100\n110 110",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n3 3\n3 2\n4 4",
"output": "rated"
},
{
"input": "3\n5 2\n4 4\n3 3",
"output": "rated"
},
{
"input": "4\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n1 1\n3 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n3 4",
"output": "rated"
},
{
"input": "3\n1 2\n2 2\n3 3",
"output": "rated"
},
{
"input": "3\n1 2\n1 2\n1 2",
"output": "rated"
},
{
"input": "2\n2 1\n2 1",
"output": "rated"
},
{
"input": "2\n1 2\n3 4",
"output": "rated"
},
{
"input": "2\n3 2\n2 3",
"output": "rated"
},
{
"input": "3\n1500 1500\n1600 1600\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1 1\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n2 2",
"output": "unrated"
},
{
"input": "2\n10 12\n8 8",
"output": "rated"
},
{
"input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500",
"output": "unrated"
},
{
"input": "2\n1 2\n2 2",
"output": "rated"
},
{
"input": "3\n1500 1400\n1200 1200\n1100 1100",
"output": "rated"
},
{
"input": "2\n10 12\n10 10",
"output": "rated"
},
{
"input": "3\n1500 1500\n1400 1400\n1300 1300",
"output": "maybe"
},
{
"input": "3\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "3\n2 6\n3 5\n4 4",
"output": "rated"
},
{
"input": "2\n5 6\n4 6",
"output": "rated"
},
{
"input": "4\n10 10\n10 10\n7 7\n8 8",
"output": "unrated"
},
{
"input": "2\n4 4\n3 3",
"output": "maybe"
}
] | 1,629,530,420
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 150
| 124
| 21,708,800
|
n=int(input())
before,after=[],[]
for i in range(n):
l=[int(i) for i in input().split()]
before.append(l[0])
after.append(l[1])
for i in range(n):
if before[i]!=after[i]:
print('rated')
exit()
if after!=sorted(after,reverse=True):
print('unrated')
exit()
print('maybe')
|
Title: Is it rated?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output Specification:
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Demo Input:
['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n']
Demo Output:
['rated\n', 'unrated\n', 'maybe\n']
Note:
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
```python
n=int(input())
before,after=[],[]
for i in range(n):
l=[int(i) for i in input().split()]
before.append(l[0])
after.append(l[1])
for i in range(n):
if before[i]!=after[i]:
print('rated')
exit()
if after!=sorted(after,reverse=True):
print('unrated')
exit()
print('maybe')
```
| 3
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,689,008,474
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
a = int(input())
H=[]
A=[]
C=0
for i in range(a):
b,c = map(int,input().split())
H.append(b)
A.append(c)
for i in range(a):
for j in range(a):
if H[i] == A[j]:
C+=1
print(C)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
a = int(input())
H=[]
A=[]
C=0
for i in range(a):
b,c = map(int,input().split())
H.append(b)
A.append(c)
for i in range(a):
for j in range(a):
if H[i] == A[j]:
C+=1
print(C)
```
| 3
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,589,552,299
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 140
| 20,172,800
|
n=int(input())
while n>5:
n = n - 4
s=n%2
n=(n-s)/2
if n==1:
print('Sheldon')
if n==2:
print('Leonard')
if n==3:
print('Penny')
if n==4:
print('Rajesh')
if n==5:print('Howard')
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
n=int(input())
while n>5:
n = n - 4
s=n%2
n=(n-s)/2
if n==1:
print('Sheldon')
if n==2:
print('Leonard')
if n==3:
print('Penny')
if n==4:
print('Rajesh')
if n==5:print('Howard')
```
| 3.892425
|
484
|
A
|
Bits
|
PROGRAMMING
| 1,700
|
[
"bitmasks",
"constructive algorithms"
] | null | null |
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
|
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
|
For each query print the answer in a separate line.
|
[
"3\n1 2\n2 4\n1 10\n"
] |
[
"1\n3\n7\n"
] |
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
| 500
|
[
{
"input": "3\n1 2\n2 4\n1 10",
"output": "1\n3\n7"
},
{
"input": "55\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n2 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n3 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n4 4\n4 5\n4 6\n4 7\n4 8\n4 9\n4 10\n5 5\n5 6\n5 7\n5 8\n5 9\n5 10\n6 6\n6 7\n6 8\n6 9\n6 10\n7 7\n7 8\n7 9\n7 10\n8 8\n8 9\n8 10\n9 9\n9 10\n10 10",
"output": "1\n1\n3\n3\n3\n3\n7\n7\n7\n7\n2\n3\n3\n3\n3\n7\n7\n7\n7\n3\n3\n3\n3\n7\n7\n7\n7\n4\n5\n5\n7\n7\n7\n7\n5\n5\n7\n7\n7\n7\n6\n7\n7\n7\n7\n7\n7\n7\n7\n8\n9\n9\n9\n9\n10"
},
{
"input": "18\n1 10\n1 100\n1 1000\n1 10000\n1 100000\n1 1000000\n1 10000000\n1 100000000\n1 1000000000\n1 10000000000\n1 100000000000\n1 1000000000000\n1 10000000000000\n1 100000000000000\n1 1000000000000000\n1 10000000000000000\n1 100000000000000000\n1 1000000000000000000",
"output": "7\n63\n511\n8191\n65535\n524287\n8388607\n67108863\n536870911\n8589934591\n68719476735\n549755813887\n8796093022207\n70368744177663\n562949953421311\n9007199254740991\n72057594037927935\n576460752303423487"
},
{
"input": "3\n0 0\n1 3\n2 4",
"output": "0\n3\n3"
},
{
"input": "17\n0 0\n0 8\n1 8\n36 39\n3 4\n3 7\n2 17\n8 12\n9 12\n10 12\n10 15\n6 14\n8 15\n9 15\n15 15\n100000000000000000 1000000000000000000\n99999999999999999 1000000000000000000",
"output": "0\n7\n7\n39\n3\n7\n15\n11\n11\n11\n15\n7\n15\n15\n15\n576460752303423487\n576460752303423487"
}
] | 1,681,826,675
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 15
| 1,000
| 7,987,200
|
def check(num):
ans = ''.join(num)
if f(ans) > r:
return False
return True
def f(s):
ans = 0
k = 0
for i in range(len(s)-1, -1, -1):
if s[i] == '1':
ans += (2**k)
k += 1
return ans
for _ in range(int(input())):
l, r = map(int, input().split())
s1 = ['0']*60 + list(bin(l)[2:])
ss = ['0']*60 + list(bin(l)[2:])
for i in range(len(s1) - 1, -1, -1):
if s1[i] != '1':
ss[i] = '1'
if check(ss):
s1[i] = '1'
else:
break
res = ''.join(s1)
print(f(res))
|
Title: Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
Input Specification:
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
Output Specification:
For each query print the answer in a separate line.
Demo Input:
['3\n1 2\n2 4\n1 10\n']
Demo Output:
['1\n3\n7\n']
Note:
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
|
```python
def check(num):
ans = ''.join(num)
if f(ans) > r:
return False
return True
def f(s):
ans = 0
k = 0
for i in range(len(s)-1, -1, -1):
if s[i] == '1':
ans += (2**k)
k += 1
return ans
for _ in range(int(input())):
l, r = map(int, input().split())
s1 = ['0']*60 + list(bin(l)[2:])
ss = ['0']*60 + list(bin(l)[2:])
for i in range(len(s1) - 1, -1, -1):
if s1[i] != '1':
ss[i] = '1'
if check(ss):
s1[i] = '1'
else:
break
res = ''.join(s1)
print(f(res))
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,656,365,754
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
number=int(input())
l=[]
g=0
for i in range(number):
a,b,c=map(int,input().split())
if g==0:
l.extend((a, b, c))
else:
l[0]+=a
l[1]+=b
l[2]+=c
g+=1
if l[0]== 0 and l[1]==0 and l[2]==0:
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
number=int(input())
l=[]
g=0
for i in range(number):
a,b,c=map(int,input().split())
if g==0:
l.extend((a, b, c))
else:
l[0]+=a
l[1]+=b
l[2]+=c
g+=1
if l[0]== 0 and l[1]==0 and l[2]==0:
print("YES")
else:
print("NO")
```
| 3.977
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*?
|
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met.
|
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.
|
[
"4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n"
] |
[
"4\n10\n0\n-1\n"
] |
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
| 0
|
[
{
"input": "4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1",
"output": "4\n10\n0\n-1"
},
{
"input": "8\n0 1 0 1\n0 2 1 2\n0 3 1 1\n1 2 0 1\n1 2 1 1\n2 2 0 1\n3 3 1 2\n4 4 1 1",
"output": "0\n2\n-1\n-1\n-1\n-1\n3\n0"
},
{
"input": "5\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2",
"output": "999999998\n999999998\n999999998\n999999998\n999999998"
},
{
"input": "5\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "1\n999999999 1000000000 1 2",
"output": "999999998"
},
{
"input": "1\n50 50 1 1",
"output": "0"
},
{
"input": "1\n100000000 100000000 1 2",
"output": "100000000"
},
{
"input": "1\n3 999999990 1 1000000000",
"output": "2000000010"
},
{
"input": "5\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n1 1 1 1",
"output": "4\n10\n0\n-1\n0"
},
{
"input": "5\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000",
"output": "9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000"
},
{
"input": "1\n0 1000000000 999999999 1000000000",
"output": "999999999000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n2 1000000000 999999999 1000000000\n3 1000000000 999999999 1000000000\n4 1000000000 999999999 1000000000\n5 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999997000000000\n999999996000000000\n999999995000000000\n999999994000000000"
},
{
"input": "1\n1 1 1 1",
"output": "0"
},
{
"input": "5\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999",
"output": "499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "5\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999",
"output": "899999999100000001\n449999999550000002\n899999999100000001\n449999999550000002\n899999999100000001"
},
{
"input": "1\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000"
},
{
"input": "1\n7 7 1 1",
"output": "0"
},
{
"input": "5\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2",
"output": "1000000000\n1000000000\n1000000000\n1000000000\n1000000000"
},
{
"input": "1\n1000000000 1000000000 1 1",
"output": "0"
},
{
"input": "1\n1 1000000000 999999999 1000000000",
"output": "999999998000000000"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n999999999 1000000000 1 1000000000\n1 2 0 1\n0 1 0 1",
"output": "999999998000000000\n999999998000000000\n-1\n0"
},
{
"input": "1\n1 1000000000 1 2",
"output": "999999998"
},
{
"input": "5\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2",
"output": "982449705\n982449705\n982449705\n982449705\n982449705"
},
{
"input": "5\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009",
"output": "405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966"
},
{
"input": "1\n5 10 0 1",
"output": "-1"
},
{
"input": "1\n2 2 1 1",
"output": "0"
},
{
"input": "5\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000",
"output": "999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001"
},
{
"input": "1\n0 5 0 1",
"output": "0"
},
{
"input": "5\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999",
"output": "9997999990001\n9997999990001\n9997999990001\n9997999990001\n9997999990001"
},
{
"input": "5\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000",
"output": "333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000"
},
{
"input": "1\n999999998 999999999 1 10",
"output": "8999999981"
},
{
"input": "5\n1 1 1 1\n2 2 1 1\n100 100 1 1\n1000000000 1000000000 1 1\n1000000000 1000000000 1 1",
"output": "0\n0\n0\n0\n0"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
}
] | 1,593,435,851
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 124
| 7,065,600
|
from functools import reduce
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def extended_gcd(a, b):
"""returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
s, old_s = 0, 1
r, old_r = b, a
while r:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
return old_r, old_s, (old_r - old_s * a) // b if b else 0
gcdm = lambda *args: reduce(gcd, args, 0)
lcm = lambda a, b: a * b // gcd(a, b)
lcmm = lambda *args: reduce(lcm, args, 1)
t = int(input())
for _ in range(t):
x, y, p, q = map(int, input().split())
g, a, b = extended_gcd(q, -p)
a *= p * y - q * x
b *= p * y - q * x
if x == 0 and y == 0:
print(q)
continue
if q == p and x == y:
print(0)
continue
if q == p and x != y:
print(-1)
continue
k = min((a * g) // p, g * (b - a) // (q - p))
a = a + k * ((-p) // g)
b = b - k * (q // g)
print(-b)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*?
Input Specification:
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met.
Output Specification:
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.
Demo Input:
['4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n']
Demo Output:
['4\n10\n0\n-1\n']
Note:
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
|
```python
from functools import reduce
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def extended_gcd(a, b):
"""returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
s, old_s = 0, 1
r, old_r = b, a
while r:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
return old_r, old_s, (old_r - old_s * a) // b if b else 0
gcdm = lambda *args: reduce(gcd, args, 0)
lcm = lambda a, b: a * b // gcd(a, b)
lcmm = lambda *args: reduce(lcm, args, 1)
t = int(input())
for _ in range(t):
x, y, p, q = map(int, input().split())
g, a, b = extended_gcd(q, -p)
a *= p * y - q * x
b *= p * y - q * x
if x == 0 and y == 0:
print(q)
continue
if q == p and x == y:
print(0)
continue
if q == p and x != y:
print(-1)
continue
k = min((a * g) // p, g * (b - a) // (q - p))
a = a + k * ((-p) // g)
b = b - k * (q // g)
print(-b)
```
| -1
|
|
811
|
A
|
Vladik and Courtesy
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
|
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
|
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
|
[
"1 1\n",
"7 6\n"
] |
[
"Valera\n",
"Vladik\n"
] |
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
"output": "Valera"
},
{
"input": "999963734 999994456",
"output": "Vladik"
},
{
"input": "1000000000 1000000000",
"output": "Vladik"
},
{
"input": "946 879",
"output": "Valera"
},
{
"input": "10819 45238",
"output": "Vladik"
},
{
"input": "101357 236928",
"output": "Vladik"
},
{
"input": "1033090 7376359",
"output": "Vladik"
},
{
"input": "9754309 9525494",
"output": "Valera"
},
{
"input": "90706344 99960537",
"output": "Vladik"
},
{
"input": "965161805 908862070",
"output": "Valera"
},
{
"input": "9 11",
"output": "Valera"
},
{
"input": "3 2",
"output": "Vladik"
},
{
"input": "6 6",
"output": "Vladik"
},
{
"input": "4 4",
"output": "Valera"
},
{
"input": "5 5",
"output": "Valera"
},
{
"input": "5 4",
"output": "Valera"
},
{
"input": "12345680 1",
"output": "Valera"
},
{
"input": "9 10",
"output": "Valera"
},
{
"input": "678 76687",
"output": "Vladik"
},
{
"input": "1 678",
"output": "Vladik"
},
{
"input": "45 1678",
"output": "Vladik"
},
{
"input": "3 3",
"output": "Vladik"
},
{
"input": "10 11",
"output": "Valera"
},
{
"input": "2 1",
"output": "Valera"
},
{
"input": "1 2",
"output": "Vladik"
},
{
"input": "2 2",
"output": "Vladik"
},
{
"input": "4 5",
"output": "Valera"
},
{
"input": "9 6",
"output": "Valera"
},
{
"input": "1 5",
"output": "Vladik"
},
{
"input": "7 8",
"output": "Vladik"
},
{
"input": "1000000000 999982505",
"output": "Valera"
},
{
"input": "12 12",
"output": "Vladik"
},
{
"input": "1000 950",
"output": "Valera"
},
{
"input": "10 9",
"output": "Valera"
},
{
"input": "100 9",
"output": "Valera"
},
{
"input": "1000 996",
"output": "Vladik"
},
{
"input": "9 5",
"output": "Valera"
}
] | 1,504,801,085
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 42
| 93
| 409,600
|
a, b = [int(x) for x in input().split()]
n = 0
s = 0
m = 0
i = 1
while s + i <= a:
s += i
i += 2
n += 1
s = 0
i = 2
while s + i <= b:
s += i
i += 2
m += 1
if n > m:
print("Valera")
else:
print("Vladik")
|
Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
a, b = [int(x) for x in input().split()]
n = 0
s = 0
m = 0
i = 1
while s + i <= a:
s += i
i += 2
n += 1
s = 0
i = 2
while s + i <= b:
s += i
i += 2
m += 1
if n > m:
print("Valera")
else:
print("Vladik")
```
| 3
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,630,595,603
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 6,963,200
|
r = {}
for _ in range(int(input())):
n,t = map(str,input().split())
t = int(t)
if(n in r):
r[n] += int(t)
else:
r[n] = int(t)
r = dict(sorted(r.items(), key=lambda x: x[1], reverse=True))
print(list(r.keys())[0])
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
r = {}
for _ in range(int(input())):
n,t = map(str,input().split())
t = int(t)
if(n in r):
r[n] += int(t)
else:
r[n] = int(t)
r = dict(sorted(r.items(), key=lambda x: x[1], reverse=True))
print(list(r.keys())[0])
```
| 0
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Vasya's telephone contains *n* photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo *n*. Similarly, by swiping right from the last photo you reach photo 1. It takes *a* seconds to swipe from photo to adjacent.
For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation and can't be rotated. It takes *b* second to change orientation of the photo.
Vasya has *T* seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends 1 second to notice all details in it. If photo is in the wrong orientation, he spends *b* seconds on rotating it before watching it. If Vasya has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.
Help Vasya find the maximum number of photos he is able to watch during *T* seconds.
|
The first line of the input contains 4 integers *n*,<=*a*,<=*b*,<=*T* (1<=≤<=*n*<=≤<=5·105, 1<=≤<=*a*,<=*b*<=≤<=1000, 1<=≤<=*T*<=≤<=109) — the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.
Second line of the input contains a string of length *n* containing symbols 'w' and 'h'.
If the *i*-th position of a string contains 'w', then the photo *i* should be seen in the horizontal orientation.
If the *i*-th position of a string contains 'h', then the photo *i* should be seen in vertical orientation.
|
Output the only integer, the maximum number of photos Vasya is able to watch during those *T* seconds.
|
[
"4 2 3 10\nwwhw\n",
"5 2 4 13\nhhwhh\n",
"5 2 4 1000\nhhwhh\n",
"3 1 100 10\nwhw\n"
] |
[
"2\n",
"4\n",
"5\n",
"0\n"
] |
In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.
Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.
| 0
|
[
{
"input": "4 2 3 10\nwwhw",
"output": "2"
},
{
"input": "5 2 4 13\nhhwhh",
"output": "4"
},
{
"input": "5 2 4 1000\nhhwhh",
"output": "5"
},
{
"input": "3 1 100 10\nwhw",
"output": "0"
},
{
"input": "10 2 3 32\nhhwwhwhwwh",
"output": "7"
},
{
"input": "1 2 3 3\nw",
"output": "0"
},
{
"input": "100 20 100 10202\nwwwwhhwhhwhhwhhhhhwwwhhhwwwhwwhwhhwwhhwwwhwwhwwwhwhwhwwhhhwhwhhwhwwhhwhwhwwwhwwwwhwhwwwwhwhhhwhwhwww",
"output": "100"
},
{
"input": "20 10 10 1\nhwhwhwhwhwhwhwhwhhhw",
"output": "1"
},
{
"input": "12 10 10 1\nwhwhwhwhwhwh",
"output": "0"
},
{
"input": "2 5 5 1000000000\nwh",
"output": "2"
},
{
"input": "16 1 1000 2100\nhhhwwwhhhwhhhwww",
"output": "5"
},
{
"input": "5 2 4 13\nhhhwh",
"output": "4"
},
{
"input": "7 1 1000 13\nhhhhwhh",
"output": "6"
},
{
"input": "10 1 1000 10\nhhhhhhwwhh",
"output": "5"
},
{
"input": "7 1 100 8\nhhhwwwh",
"output": "4"
},
{
"input": "5 2 4 12\nhhhwh",
"output": "4"
}
] | 1,587,575,069
| 7,169
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main(){
int n, a, b, T;
cin>>n>>a>>b>>T;
string orent;
cin>>orent;
char oren[orent.length()+1];
strcpy(oren, orent.c_str());
int weights[n];
for(int i = 0; i < n; i++){
if(oren[i] == 'w'){
weights[i] = 1+b;
}else{
weights[i] = 1;
}
}
int jugad=0, score=0;
for(int i=0; i<n; i++){
jugad+=(a+weights[i]);
}
if(jugad<=T){
score++;
}
int i = 0;
int rp=0, lp=0;
int ans = 0;
while(T>=0 && score==0){
T -= weights[i];
weights[i]=0;
int ln=-1,rn=-1;
int gl=0, gr=0;
//left
int l = (i+n-1)%n;
int t = T;
while(t>=0 && l!= rp){
if(weights[l]!=0){
ln++;
}else{
gl++;
}
t-=a;
t-=weights[l];
l--;
}
//right
int r = (i+1)%n;
t = T;
while(t>=0 && r!= lp){
if(weights[r] != 0){
rn++;
}else{
gr++;
}
t-=a;
t-=weights[r];
r++;
}
if(ln>rn){
i=(i+n-1)%n;
T-=a;
lp=i;
}else if(ln<rn){
i = (i+1)%n;
T-=a;
rp=i;
}else{
if(gl<=gr){
i=(i+n-1)%n;
T-=a;
lp=i;
}else{
i = (i+1)%n;
T-=a;
rp=i;
}
}
if(weights[i]!=0){
ans++;
}
if(lp=rp){
break;
}
}
if(score!=0){
ans=n;
}
cout<<ans<<endl;
}
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's telephone contains *n* photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo *n*. Similarly, by swiping right from the last photo you reach photo 1. It takes *a* seconds to swipe from photo to adjacent.
For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation and can't be rotated. It takes *b* second to change orientation of the photo.
Vasya has *T* seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends 1 second to notice all details in it. If photo is in the wrong orientation, he spends *b* seconds on rotating it before watching it. If Vasya has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.
Help Vasya find the maximum number of photos he is able to watch during *T* seconds.
Input Specification:
The first line of the input contains 4 integers *n*,<=*a*,<=*b*,<=*T* (1<=≤<=*n*<=≤<=5·105, 1<=≤<=*a*,<=*b*<=≤<=1000, 1<=≤<=*T*<=≤<=109) — the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.
Second line of the input contains a string of length *n* containing symbols 'w' and 'h'.
If the *i*-th position of a string contains 'w', then the photo *i* should be seen in the horizontal orientation.
If the *i*-th position of a string contains 'h', then the photo *i* should be seen in vertical orientation.
Output Specification:
Output the only integer, the maximum number of photos Vasya is able to watch during those *T* seconds.
Demo Input:
['4 2 3 10\nwwhw\n', '5 2 4 13\nhhwhh\n', '5 2 4 1000\nhhwhh\n', '3 1 100 10\nwhw\n']
Demo Output:
['2\n', '4\n', '5\n', '0\n']
Note:
In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.
Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.
|
```python
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main(){
int n, a, b, T;
cin>>n>>a>>b>>T;
string orent;
cin>>orent;
char oren[orent.length()+1];
strcpy(oren, orent.c_str());
int weights[n];
for(int i = 0; i < n; i++){
if(oren[i] == 'w'){
weights[i] = 1+b;
}else{
weights[i] = 1;
}
}
int jugad=0, score=0;
for(int i=0; i<n; i++){
jugad+=(a+weights[i]);
}
if(jugad<=T){
score++;
}
int i = 0;
int rp=0, lp=0;
int ans = 0;
while(T>=0 && score==0){
T -= weights[i];
weights[i]=0;
int ln=-1,rn=-1;
int gl=0, gr=0;
//left
int l = (i+n-1)%n;
int t = T;
while(t>=0 && l!= rp){
if(weights[l]!=0){
ln++;
}else{
gl++;
}
t-=a;
t-=weights[l];
l--;
}
//right
int r = (i+1)%n;
t = T;
while(t>=0 && r!= lp){
if(weights[r] != 0){
rn++;
}else{
gr++;
}
t-=a;
t-=weights[r];
r++;
}
if(ln>rn){
i=(i+n-1)%n;
T-=a;
lp=i;
}else if(ln<rn){
i = (i+1)%n;
T-=a;
rp=i;
}else{
if(gl<=gr){
i=(i+n-1)%n;
T-=a;
lp=i;
}else{
i = (i+1)%n;
T-=a;
rp=i;
}
}
if(weights[i]!=0){
ans++;
}
if(lp=rp){
break;
}
}
if(score!=0){
ans=n;
}
cout<<ans<<endl;
}
```
| -1
|
|
913
|
B
|
Christmas Spruce
|
PROGRAMMING
| 1,200
|
[
"implementation",
"trees"
] | null | null |
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).
|
The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children.
|
Print "Yes" if the tree is a spruce and "No" otherwise.
|
[
"4\n1\n1\n1\n",
"7\n1\n1\n1\n2\n2\n2\n",
"8\n1\n1\n1\n1\n3\n3\n3\n"
] |
[
"Yes\n",
"No\n",
"Yes\n"
] |
The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 750
|
[
{
"input": "4\n1\n1\n1",
"output": "Yes"
},
{
"input": "7\n1\n1\n1\n2\n2\n2",
"output": "No"
},
{
"input": "8\n1\n1\n1\n1\n3\n3\n3",
"output": "Yes"
},
{
"input": "3\n1\n1",
"output": "No"
},
{
"input": "13\n1\n2\n2\n2\n1\n6\n6\n6\n1\n10\n10\n10",
"output": "No"
},
{
"input": "7\n1\n2\n2\n1\n1\n1",
"output": "No"
},
{
"input": "7\n1\n1\n1\n1\n2\n2",
"output": "No"
},
{
"input": "8\n1\n1\n1\n1\n5\n5\n5",
"output": "Yes"
},
{
"input": "9\n1\n1\n1\n1\n2\n6\n6\n6",
"output": "No"
},
{
"input": "12\n1\n1\n1\n2\n5\n5\n5\n5\n1\n2\n2",
"output": "No"
},
{
"input": "20\n1\n1\n1\n1\n2\n2\n2\n3\n3\n3\n4\n4\n4\n5\n5\n5\n1\n1\n1",
"output": "Yes"
},
{
"input": "7\n1\n1\n1\n3\n3\n3",
"output": "No"
}
] | 1,646,828,420
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
if __name__ == '__main__':
_n = int(input())
mas = []
flag = True
for i in range(_n-1):
mas.append(int(input()))
help_mas = set(mas)
res = [0]*(_n+1)
for i in range(len(mas)):
if i+2 not in help_mas:
res[mas[i]] += 1
for i in res[1:]:
if not(i == 3 or i == 0):
flag = False
if flag:
print('Yes')
else:
print('No')
|
Title: Christmas Spruce
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).
Input Specification:
The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children.
Output Specification:
Print "Yes" if the tree is a spruce and "No" otherwise.
Demo Input:
['4\n1\n1\n1\n', '7\n1\n1\n1\n2\n2\n2\n', '8\n1\n1\n1\n1\n3\n3\n3\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
if __name__ == '__main__':
_n = int(input())
mas = []
flag = True
for i in range(_n-1):
mas.append(int(input()))
help_mas = set(mas)
res = [0]*(_n+1)
for i in range(len(mas)):
if i+2 not in help_mas:
res[mas[i]] += 1
for i in res[1:]:
if not(i == 3 or i == 0):
flag = False
if flag:
print('Yes')
else:
print('No')
```
| 0
|
|
203
|
A
|
Two Problems
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
A boy Valera registered on site Codeforces as Valera, and wrote his first Codeforces Round #300. He boasted to a friend Arkady about winning as much as *x* points for his first contest. But Arkady did not believe his friend's words and decided to check whether Valera could have shown such a result.
He knows that the contest number 300 was unusual because there were only two problems. The contest lasted for *t* minutes, the minutes are numbered starting from zero. The first problem had the initial cost of *a* points, and every minute its cost reduced by *d**a* points. The second problem had the initial cost of *b* points, and every minute this cost reduced by *d**b* points. Thus, as soon as the zero minute of the contest is over, the first problem will cost *a*<=-<=*d**a* points, and the second problem will cost *b*<=-<=*d**b* points. It is guaranteed that at any moment of the contest each problem has a non-negative cost.
Arkady asks you to find out whether Valera could have got exactly *x* points for this contest. You should assume that Valera could have solved any number of the offered problems. You should also assume that for each problem Valera made no more than one attempt, besides, he could have submitted both problems at the same minute of the contest, starting with minute 0 and ending with minute number *t*<=-<=1. Please note that Valera can't submit a solution exactly *t* minutes after the start of the contest or later.
|
The single line of the input contains six integers *x*,<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b* (0<=≤<=*x*<=≤<=600; 1<=≤<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b*<=≤<=300) — Valera's result, the contest's duration, the initial cost of the first problem, the initial cost of the second problem, the number of points that the first and the second problem lose per minute, correspondingly.
It is guaranteed that at each minute of the contest each problem has a non-negative cost, that is, *a*<=-<=*i*·*d**a*<=≥<=0 and *b*<=-<=*i*·*d**b*<=≥<=0 for all 0<=≤<=*i*<=≤<=*t*<=-<=1.
|
If Valera could have earned exactly *x* points at a contest, print "YES", otherwise print "NO" (without the quotes).
|
[
"30 5 20 20 3 5\n",
"10 4 100 5 5 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample Valera could have acted like this: he could have submitted the first problem at minute 0 and the second problem — at minute 2. Then the first problem brings him 20 points and the second problem brings him 10 points, that in total gives the required 30 points.
| 500
|
[
{
"input": "30 5 20 20 3 5",
"output": "YES"
},
{
"input": "10 4 100 5 5 1",
"output": "NO"
},
{
"input": "0 7 30 50 3 4",
"output": "YES"
},
{
"input": "50 10 30 20 1 2",
"output": "YES"
},
{
"input": "40 1 40 5 11 2",
"output": "YES"
},
{
"input": "35 8 20 20 1 2",
"output": "YES"
},
{
"input": "10 2 27 4 11 1",
"output": "NO"
},
{
"input": "64 12 258 141 10 7",
"output": "YES"
},
{
"input": "5 3 11 100 2 4",
"output": "NO"
},
{
"input": "5 4 11 80 2 4",
"output": "YES"
},
{
"input": "28 3 16 20 3 10",
"output": "NO"
},
{
"input": "6 2 11 1 11 1",
"output": "NO"
},
{
"input": "15 5 230 213 32 25",
"output": "NO"
},
{
"input": "223 92 123 118 1 1",
"output": "YES"
},
{
"input": "375 6 133 267 19 36",
"output": "NO"
},
{
"input": "80 5 39 40 1 1",
"output": "NO"
},
{
"input": "543 4 31 69 6 5",
"output": "NO"
},
{
"input": "38 100 99 245 1 1",
"output": "YES"
},
{
"input": "3 1 20 15 17 5",
"output": "NO"
},
{
"input": "360 5 215 4 52 1",
"output": "NO"
},
{
"input": "363 2 280 239 5 231",
"output": "NO"
},
{
"input": "46 7 18 6 3 1",
"output": "NO"
},
{
"input": "5 3 135 12 21 6",
"output": "NO"
},
{
"input": "15 5 230 213 32 25",
"output": "NO"
},
{
"input": "2 5 29 36 5 6",
"output": "NO"
},
{
"input": "59 4 113 45 25 12",
"output": "YES"
},
{
"input": "74 72 104 71 1 1",
"output": "YES"
},
{
"input": "16 24 26 23 1 1",
"output": "YES"
},
{
"input": "11 1 10 1 10 1",
"output": "YES"
},
{
"input": "13 3 11 14 5 1",
"output": "YES"
},
{
"input": "1 2 1 2 1 2",
"output": "YES"
},
{
"input": "145 26 25 150 1 4",
"output": "YES"
},
{
"input": "59 18 50 17 1 1",
"output": "YES"
},
{
"input": "230 125 175 124 1 1",
"output": "YES"
},
{
"input": "142 1 66 76 18 39",
"output": "YES"
},
{
"input": "0 3 5 6 1 2",
"output": "YES"
},
{
"input": "6 46 95 45 1 1",
"output": "YES"
},
{
"input": "16 73 92 72 1 1",
"output": "YES"
},
{
"input": "3 18 272 17 6 1",
"output": "YES"
},
{
"input": "1 21 178 20 1 1",
"output": "YES"
},
{
"input": "0 15 86 84 5 2",
"output": "YES"
},
{
"input": "0 35 208 98 6 2",
"output": "YES"
},
{
"input": "0 11 67 82 6 4",
"output": "YES"
},
{
"input": "30 9 18 83 1 7",
"output": "NO"
},
{
"input": "18 12 11 54 1 3",
"output": "NO"
},
{
"input": "41 77 96 145 1 1",
"output": "YES"
},
{
"input": "27 45 44 169 1 2",
"output": "YES"
},
{
"input": "50 5 30 60 3 3",
"output": "NO"
},
{
"input": "49 2 50 20 2 1",
"output": "NO"
},
{
"input": "49 2 50 20 2 1",
"output": "NO"
},
{
"input": "17 10 10 20 1 2",
"output": "YES"
}
] | 1,583,080,574
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 248
| 307,200
|
x, t, a, b, da, db = map(int, input().split())
A = [0] + [a-i*da for i in range(t)]
B = {0} | {b-i*db for i in range(t)}
print('YES' if any(x-a in B for a in A) else 'NO')
|
Title: Two Problems
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A boy Valera registered on site Codeforces as Valera, and wrote his first Codeforces Round #300. He boasted to a friend Arkady about winning as much as *x* points for his first contest. But Arkady did not believe his friend's words and decided to check whether Valera could have shown such a result.
He knows that the contest number 300 was unusual because there were only two problems. The contest lasted for *t* minutes, the minutes are numbered starting from zero. The first problem had the initial cost of *a* points, and every minute its cost reduced by *d**a* points. The second problem had the initial cost of *b* points, and every minute this cost reduced by *d**b* points. Thus, as soon as the zero minute of the contest is over, the first problem will cost *a*<=-<=*d**a* points, and the second problem will cost *b*<=-<=*d**b* points. It is guaranteed that at any moment of the contest each problem has a non-negative cost.
Arkady asks you to find out whether Valera could have got exactly *x* points for this contest. You should assume that Valera could have solved any number of the offered problems. You should also assume that for each problem Valera made no more than one attempt, besides, he could have submitted both problems at the same minute of the contest, starting with minute 0 and ending with minute number *t*<=-<=1. Please note that Valera can't submit a solution exactly *t* minutes after the start of the contest or later.
Input Specification:
The single line of the input contains six integers *x*,<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b* (0<=≤<=*x*<=≤<=600; 1<=≤<=*t*,<=*a*,<=*b*,<=*d**a*,<=*d**b*<=≤<=300) — Valera's result, the contest's duration, the initial cost of the first problem, the initial cost of the second problem, the number of points that the first and the second problem lose per minute, correspondingly.
It is guaranteed that at each minute of the contest each problem has a non-negative cost, that is, *a*<=-<=*i*·*d**a*<=≥<=0 and *b*<=-<=*i*·*d**b*<=≥<=0 for all 0<=≤<=*i*<=≤<=*t*<=-<=1.
Output Specification:
If Valera could have earned exactly *x* points at a contest, print "YES", otherwise print "NO" (without the quotes).
Demo Input:
['30 5 20 20 3 5\n', '10 4 100 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Valera could have acted like this: he could have submitted the first problem at minute 0 and the second problem — at minute 2. Then the first problem brings him 20 points and the second problem brings him 10 points, that in total gives the required 30 points.
|
```python
x, t, a, b, da, db = map(int, input().split())
A = [0] + [a-i*da for i in range(t)]
B = {0} | {b-i*db for i in range(t)}
print('YES' if any(x-a in B for a in A) else 'NO')
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,637,035,823
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 9
| 62
| 0
|
s = input()
t = input()
ans = 'YES'
for i in range(len(s)):
if s[i] != t[len(s)-i-1]:
ans = 'NO'
break
print(ans)
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = input()
t = input()
ans = 'YES'
for i in range(len(s)):
if s[i] != t[len(s)-i-1]:
ans = 'NO'
break
print(ans)
```
| -1
|
463
|
A
|
Caisa and Sugar
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.
Unfortunately, he has just *s* dollars for sugar. But that's not a reason to be sad, because there are *n* types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).
Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.
|
The first line contains two space-separated integers *n*,<=*s* (1<=≤<=*n*,<=*s*<=≤<=100).
The *i*-th of the next *n* lines contains two integers *x**i*, *y**i* (1<=≤<=*x**i*<=≤<=100; 0<=≤<=*y**i*<=<<=100), where *x**i* represents the number of dollars and *y**i* the number of cents needed in order to buy the *i*-th type of sugar.
|
Print a single integer representing the maximum number of sweets he can buy, or -1 if he can't buy any type of sugar.
|
[
"5 10\n3 90\n12 0\n9 70\n5 50\n7 0\n",
"5 5\n10 10\n20 20\n30 30\n40 40\n50 50\n"
] |
[
"50\n",
"-1\n"
] |
In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take 50 sweets as a change.
| 500
|
[
{
"input": "5 10\n3 90\n12 0\n9 70\n5 50\n7 0",
"output": "50"
},
{
"input": "5 5\n10 10\n20 20\n30 30\n40 40\n50 50",
"output": "-1"
},
{
"input": "1 2\n1 0",
"output": "0"
},
{
"input": "2 10\n20 99\n30 99",
"output": "-1"
},
{
"input": "15 21\n16 51\n33 44\n32 92\n19 91\n67 81\n94 94\n40 67\n77 38\n42 22\n48 19\n9 35\n90 36\n57 80\n58 97\n13 15",
"output": "85"
},
{
"input": "1 2\n5 0",
"output": "-1"
},
{
"input": "37 10\n57 61\n78 79\n59 60\n48 17\n39 22\n11 87\n19 55\n56 78\n25 12\n15 55\n13 37\n49 38\n37 57\n19 37\n74 18\n81 27\n8 82\n43 9\n79 17\n92 41\n29 69\n77 1\n33 98\n11 31\n40 56\n46 76\n16 51\n10 94\n32 3\n23 27\n8 57\n40 54\n16 66\n36 33\n28 45\n67 26\n81 55",
"output": "43"
},
{
"input": "46 27\n53 88\n89 56\n98 5\n48 35\n88 19\n41 22\n71 75\n75 88\n57 51\n42 63\n100 91\n58 78\n99 75\n56 3\n49 70\n59 81\n38 0\n59 98\n35 86\n6 5\n80 21\n97 45\n86 2\n54 22\n38 42\n34 94\n25 37\n65 42\n46 74\n46 19\n53 93\n47 55\n55 89\n76 56\n62 88\n62 32\n78 41\n44 9\n32 14\n9 48\n43 68\n58 93\n37 18\n94 21\n41 81\n28 38",
"output": "95"
},
{
"input": "69 9\n39 15\n70 44\n90 89\n84 58\n83 97\n24 28\n90 18\n7 96\n43 86\n99 50\n77 12\n12 34\n100 19\n21 86\n55 30\n99 29\n26 92\n82 67\n78 14\n64 16\n98 12\n31 39\n36 86\n73 36\n87 29\n42 23\n58 15\n35 52\n91 56\n39 6\n88 45\n38 36\n57 12\n79 71\n40 33\n98 42\n34 36\n23 79\n69 68\n50 47\n23 6\n22 90\n96 79\n88 55\n63 85\n83 25\n15 1\n15 72\n96 70\n56 44\n87 34\n24 5\n85 71\n92 97\n97 52\n60 40\n84 18\n50 32\n89 78\n16 36\n15 2\n56 98\n23 28\n51 69\n15 17\n52 57\n63 52\n25 26\n25 47",
"output": "4"
},
{
"input": "44 77\n95 38\n51 48\n89 76\n18 40\n97 15\n6 57\n30 44\n84 8\n93 60\n75 96\n14 61\n72 97\n17 71\n11 85\n29 2\n93 86\n88 74\n52 94\n88 75\n43 20\n79 83\n28 99\n8 7\n20 62\n59 51\n65 94\n44 23\n95 36\n74 11\n66 57\n44 1\n29 57\n53 82\n11 79\n100 91\n75 4\n5 29\n90 53\n16 11\n16 8\n55 73\n98 79\n68 98\n10 43",
"output": "99"
},
{
"input": "62 37\n33 3\n7 38\n3 4\n80 48\n27 5\n38 4\n1 89\n81 25\n11 42\n63 19\n68 99\n17 90\n43 33\n51 62\n24 8\n64 87\n78 56\n61 21\n70 79\n38 91\n97 96\n38 74\n100 96\n39 70\n90 62\n62 53\n96 8\n42 30\n93 38\n99 44\n32 67\n24 61\n84 8\n41 35\n28 4\n53 77\n52 23\n72 42\n13 3\n66 61\n15 1\n10 78\n61 10\n1 3\n100 52\n21 98\n12 30\n91 95\n32 46\n4 30\n60 10\n86 55\n98 63\n67 56\n70 9\n63 83\n62 84\n47 24\n80 22\n27 98\n49 78\n61 98",
"output": "99"
},
{
"input": "6 84\n95 74\n34 76\n12 46\n89 34\n77 28\n26 56",
"output": "72"
},
{
"input": "1 10\n10 10",
"output": "-1"
},
{
"input": "7 7\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0",
"output": "0"
},
{
"input": "1 9\n9 0",
"output": "0"
},
{
"input": "1 1\n1 0",
"output": "0"
},
{
"input": "1 9\n9 1",
"output": "-1"
},
{
"input": "1 10\n10 0",
"output": "0"
},
{
"input": "1 2\n2 1",
"output": "-1"
},
{
"input": "2 2\n1 50\n2 0",
"output": "50"
},
{
"input": "2 3\n2 10\n3 0",
"output": "90"
},
{
"input": "2 2\n1 10\n2 0",
"output": "90"
},
{
"input": "2 10\n9 10\n1 0",
"output": "90"
},
{
"input": "3 7\n1 56\n5 49\n7 0",
"output": "51"
},
{
"input": "2 1\n1 50\n1 0",
"output": "0"
},
{
"input": "2 10\n10 50\n10 0",
"output": "0"
},
{
"input": "2 10\n9 20\n10 20",
"output": "80"
},
{
"input": "3 4\n1 0\n2 0\n4 1",
"output": "0"
},
{
"input": "2 2\n2 0\n3 1",
"output": "0"
}
] | 1,638,411,261
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 30
| 0
|
n, s = map(int, input().split())
s *= 100
maxim = 0
for _ in range(n):
x, y = map(int, input().split())
cost = x * 100 + y
maxim = max(s % cost % 100, maxim)
print(-1 if maxim == 0 else maxim)
|
Title: Caisa and Sugar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.
Unfortunately, he has just *s* dollars for sugar. But that's not a reason to be sad, because there are *n* types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).
Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.
Input Specification:
The first line contains two space-separated integers *n*,<=*s* (1<=≤<=*n*,<=*s*<=≤<=100).
The *i*-th of the next *n* lines contains two integers *x**i*, *y**i* (1<=≤<=*x**i*<=≤<=100; 0<=≤<=*y**i*<=<<=100), where *x**i* represents the number of dollars and *y**i* the number of cents needed in order to buy the *i*-th type of sugar.
Output Specification:
Print a single integer representing the maximum number of sweets he can buy, or -1 if he can't buy any type of sugar.
Demo Input:
['5 10\n3 90\n12 0\n9 70\n5 50\n7 0\n', '5 5\n10 10\n20 20\n30 30\n40 40\n50 50\n']
Demo Output:
['50\n', '-1\n']
Note:
In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take 50 sweets as a change.
|
```python
n, s = map(int, input().split())
s *= 100
maxim = 0
for _ in range(n):
x, y = map(int, input().split())
cost = x * 100 + y
maxim = max(s % cost % 100, maxim)
print(-1 if maxim == 0 else maxim)
```
| 0
|
|
672
|
B
|
Different is Good
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation",
"strings"
] | null | null |
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
|
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
|
[
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
| 1,000
|
[
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"output": "12"
},
{
"input": "25\ntcqpchnqskqjacruoaqilgebu",
"output": "7"
},
{
"input": "13\naebaecedabbee",
"output": "8"
},
{
"input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "10\nbababbdaee",
"output": "6"
},
{
"input": "11\ndbadcdbdbca",
"output": "7"
},
{
"input": "12\nacceaabddaaa",
"output": "7"
},
{
"input": "13\nabddfbfaeecfa",
"output": "7"
},
{
"input": "14\neeceecacdbcbbb",
"output": "9"
},
{
"input": "15\ndcbceaaggabaheb",
"output": "8"
},
{
"input": "16\nhgiegfbadgcicbhd",
"output": "7"
},
{
"input": "17\nabhfibbdddfghgfdi",
"output": "10"
},
{
"input": "26\nbbbbbabbaababaaabaaababbaa",
"output": "24"
},
{
"input": "26\nahnxdnbfbcrirerssyzydihuee",
"output": "11"
},
{
"input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc",
"output": "8"
},
{
"input": "26\nrvxmulriorilidecqwmfaemifj",
"output": "10"
},
{
"input": "26\naowpmreooavnmamogdoopuisge",
"output": "12"
},
{
"input": "26\ninimevtuefhvuefirdehmmfudh",
"output": "15"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "27\nqdcfjtblgglnilgassirrjekcjt",
"output": "-1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyza",
"output": "-1"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "0"
},
{
"input": "5\nzzzzz",
"output": "4"
},
{
"input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa",
"output": "-1"
},
{
"input": "1\nq",
"output": "0"
},
{
"input": "27\nqwertyuioplkjhgfdsazxcvbnmm",
"output": "-1"
},
{
"input": "9\nxxxyyyzzz",
"output": "6"
},
{
"input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk",
"output": "-1"
},
{
"input": "27\nqwertyuiopasdfghjklzxcvbnmm",
"output": "-1"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxya",
"output": "1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyzz",
"output": "-1"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "26\nqwertyuioplkjhgfdsazxcvbnm",
"output": "0"
},
{
"input": "10\nzzzzzzzzzz",
"output": "9"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "30\nabcdefghtyabcdefghtyabcdefghty",
"output": "-1"
}
] | 1,658,899,314
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
n=int(input())
s=set(input())
print(abs(n-s))
|
Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
|
```python
n=int(input())
s=set(input())
print(abs(n-s))
```
| -1
|
|
931
|
A
|
Friends Meeting
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
|
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
|
Print the minimum possible total tiredness if the friends meet in the same point.
|
[
"3\n4\n",
"101\n99\n",
"5\n10\n"
] |
[
"1\n",
"2\n",
"9\n"
] |
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
| 500
|
[
{
"input": "3\n4",
"output": "1"
},
{
"input": "101\n99",
"output": "2"
},
{
"input": "5\n10",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n1000",
"output": "250000"
},
{
"input": "999\n1000",
"output": "1"
},
{
"input": "1000\n999",
"output": "1"
},
{
"input": "1000\n1",
"output": "250000"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n999",
"output": "249001"
},
{
"input": "2\n998",
"output": "248502"
},
{
"input": "999\n2",
"output": "249001"
},
{
"input": "998\n2",
"output": "248502"
},
{
"input": "2\n1000",
"output": "249500"
},
{
"input": "1000\n2",
"output": "249500"
},
{
"input": "1\n999",
"output": "249500"
},
{
"input": "999\n1",
"output": "249500"
},
{
"input": "188\n762",
"output": "82656"
},
{
"input": "596\n777",
"output": "8281"
},
{
"input": "773\n70",
"output": "123904"
},
{
"input": "825\n729",
"output": "2352"
},
{
"input": "944\n348",
"output": "89102"
},
{
"input": "352\n445",
"output": "2209"
},
{
"input": "529\n656",
"output": "4096"
},
{
"input": "19\n315",
"output": "22052"
},
{
"input": "138\n370",
"output": "13572"
},
{
"input": "546\n593",
"output": "576"
},
{
"input": "285\n242",
"output": "484"
},
{
"input": "773\n901",
"output": "4160"
},
{
"input": "892\n520",
"output": "34782"
},
{
"input": "864\n179",
"output": "117649"
},
{
"input": "479\n470",
"output": "25"
},
{
"input": "967\n487",
"output": "57840"
},
{
"input": "648\n106",
"output": "73712"
},
{
"input": "58\n765",
"output": "125316"
},
{
"input": "235\n56",
"output": "8100"
},
{
"input": "285\n153",
"output": "4422"
},
{
"input": "943\n13",
"output": "216690"
},
{
"input": "675\n541",
"output": "4556"
},
{
"input": "4\n912",
"output": "206570"
}
] | 1,520,178,716
| 1,016
|
Python 3
|
OK
|
TESTS
| 40
| 62
| 5,632,000
|
a = int(input())
b = int(input())
health = 0
i = 1
if a > b :
x = (a - b )// 2 + b
while a != b:
if a != x:
a -= 1
health += i
if b != x:
b += 1
health += i
i += 1
else:
x = (b - a ) // 2 + a
while a != b:
if a != x:
a += 1
health += i
if b != x:
health += i
b -= 1
i += 1
print(health)
|
Title: Friends Meeting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
Output Specification:
Print the minimum possible total tiredness if the friends meet in the same point.
Demo Input:
['3\n4\n', '101\n99\n', '5\n10\n']
Demo Output:
['1\n', '2\n', '9\n']
Note:
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.
|
```python
a = int(input())
b = int(input())
health = 0
i = 1
if a > b :
x = (a - b )// 2 + b
while a != b:
if a != x:
a -= 1
health += i
if b != x:
b += 1
health += i
i += 1
else:
x = (b - a ) // 2 + a
while a != b:
if a != x:
a += 1
health += i
if b != x:
health += i
b -= 1
i += 1
print(health)
```
| 3
|
|
994
|
A
|
Fingerprints
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
|
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
|
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
|
[
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] |
[
"7 1 2\n",
"1 0\n"
] |
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
| 500
|
[
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8",
"output": "1 2 3 4 5 6 7 8 9 0"
},
{
"input": "1 1\n4\n4",
"output": "4"
},
{
"input": "3 7\n6 3 4\n4 9 0 1 7 8 6",
"output": "6 4"
},
{
"input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0",
"output": "0"
},
{
"input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6",
"output": "6 0 3 8 1"
},
{
"input": "8 2\n7 2 9 6 1 0 3 4\n6 3",
"output": "6 3"
},
{
"input": "5 4\n7 0 1 4 9\n0 9 5 3",
"output": "0 9"
},
{
"input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6",
"output": "6"
},
{
"input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2",
"output": "2 3"
},
{
"input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2",
"output": "3 9 2 4"
},
{
"input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6",
"output": "1 2 3 8 6 5"
},
{
"input": "8 2\n7 4 8 9 2 5 6 1\n6 4",
"output": "4 6"
},
{
"input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3",
"output": "0 3"
},
{
"input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8",
"output": "2 8 1 3"
},
{
"input": "1 6\n3\n6 8 2 4 5 3",
"output": "3"
},
{
"input": "1 8\n0\n9 2 4 8 1 5 0 7",
"output": "0"
},
{
"input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4",
"output": "7 9 4 1 0"
},
{
"input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1",
"output": "0 1"
},
{
"input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2",
"output": "2 9 6 1 4"
},
{
"input": "6 3\n8 3 9 2 7 6\n5 4 3",
"output": "3"
},
{
"input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5",
"output": "8 3 9 6"
},
{
"input": "1 2\n1\n1 0",
"output": "1"
},
{
"input": "3 6\n1 2 3\n4 5 6 1 2 3",
"output": "1 2 3"
},
{
"input": "1 2\n2\n1 2",
"output": "2"
},
{
"input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9",
"output": "9"
}
] | 1,566,424,165
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 124
| 0
|
n, m = map(int, input().split())
xs = map(int, input().split())
ys = map(int, input().split())
flags = [False] * 10
for y in ys:
flags[y] = True
res = (str(x) for x in xs if flags[x])
print(" ".join(res))
|
Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
|
```python
n, m = map(int, input().split())
xs = map(int, input().split())
ys = map(int, input().split())
flags = [False] * 10
for y in ys:
flags[y] = True
res = (str(x) for x in xs if flags[x])
print(" ".join(res))
```
| 3
|
|
22
|
B
|
Bargaining Table
|
PROGRAMMING
| 1,500
|
[
"brute force",
"dp"
] |
B. Bargaining Table
|
2
|
256
|
Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular room *n*<=×<=*m* meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.
|
The first line contains 2 space-separated numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=25) — the office room dimensions. Then there follow *n* lines with *m* characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.
|
Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.
|
[
"3 3\n000\n010\n000\n",
"5 4\n1100\n0000\n0000\n0000\n0000\n"
] |
[
"8\n",
"16\n"
] |
none
| 0
|
[
{
"input": "3 3\n000\n010\n000",
"output": "8"
},
{
"input": "5 4\n1100\n0000\n0000\n0000\n0000",
"output": "16"
},
{
"input": "3 3\n000\n110\n000",
"output": "8"
},
{
"input": "4 2\n00\n10\n11\n00",
"output": "6"
},
{
"input": "3 5\n00001\n00000\n10100",
"output": "12"
},
{
"input": "1 1\n0",
"output": "4"
},
{
"input": "10 11\n11111111101\n01111111111\n11101111111\n01111110111\n11111111111\n11111111111\n11111111111\n11110111111\n11111111111\n11111111111",
"output": "4"
},
{
"input": "7 7\n1110111\n1111111\n1111101\n1111101\n1111111\n1100111\n1011111",
"output": "6"
},
{
"input": "10 10\n0110001011\n0101010111\n0010110100\n1010000110\n0111100011\n1010100100\n1010010000\n1011100011\n1110011000\n0010100101",
"output": "16"
},
{
"input": "20 20\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000\n00000000000000000000",
"output": "80"
},
{
"input": "25 1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n0\n1\n1\n1\n0\n1\n1\n1\n1",
"output": "4"
},
{
"input": "25 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "52"
},
{
"input": "1 25\n1111111111111111011010110",
"output": "4"
},
{
"input": "1 25\n0000000000000000000000000",
"output": "52"
},
{
"input": "3 4\n1111\n1101\n1011",
"output": "4"
}
] | 1,642,116,378
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 23
| 342
| 26,009,600
|
# Codeforces Beta Round #22 (Div. 2 Only)
# B. Bargaining Table
n, m = map(int, input().split())
office = []
for _ in range(n):
office += [list(input())]
# dp is a set of 4-tuples (x, y, l, w) which describes the coordinates of a
# table. (x, y, l, w) is in dp iff the rectangle from (x, y) to (x + l, y + w)
# is free
dp = set()
# First add 1x1 tables
for i in range(n):
for j in range(m):
if office[i][j] == '0':
dp.add((i, j, 1, 1))
# Then add 1xn tables
for x in range(n):
for y in range(m):
for length in range(1, m - y + 1):
if office[x][y + length - 1] == '0':
dp.add((x, y, 1, length))
else:
break
# Then add nx1 tables
for x in range(n):
for y in range(m):
for height in range(1, n - x + 1):
if office[x + height - 1][y] == '0':
dp.add((x, y, height, 1))
else:
break
for x in range(n):
for y in range(m):
if office[x][y] == '1':
continue
for length in range(2, m - y + 1):
for height in range(2, n - x + 1):
if (x, y, height - 1, length) in dp and \
(x, y, height, length - 1) in dp and \
office[x + height - 1][y + length - 1] == '0':
dp.add((x, y, height, length))
max_perimeter = 0
for x, y, l, w in dp:
max_perimeter = max(max_perimeter, 2 * (l + w))
print(max_perimeter)
|
Title: Bargaining Table
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular room *n*<=×<=*m* meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.
Input Specification:
The first line contains 2 space-separated numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=25) — the office room dimensions. Then there follow *n* lines with *m* characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.
Output Specification:
Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.
Demo Input:
['3 3\n000\n010\n000\n', '5 4\n1100\n0000\n0000\n0000\n0000\n']
Demo Output:
['8\n', '16\n']
Note:
none
|
```python
# Codeforces Beta Round #22 (Div. 2 Only)
# B. Bargaining Table
n, m = map(int, input().split())
office = []
for _ in range(n):
office += [list(input())]
# dp is a set of 4-tuples (x, y, l, w) which describes the coordinates of a
# table. (x, y, l, w) is in dp iff the rectangle from (x, y) to (x + l, y + w)
# is free
dp = set()
# First add 1x1 tables
for i in range(n):
for j in range(m):
if office[i][j] == '0':
dp.add((i, j, 1, 1))
# Then add 1xn tables
for x in range(n):
for y in range(m):
for length in range(1, m - y + 1):
if office[x][y + length - 1] == '0':
dp.add((x, y, 1, length))
else:
break
# Then add nx1 tables
for x in range(n):
for y in range(m):
for height in range(1, n - x + 1):
if office[x + height - 1][y] == '0':
dp.add((x, y, height, 1))
else:
break
for x in range(n):
for y in range(m):
if office[x][y] == '1':
continue
for length in range(2, m - y + 1):
for height in range(2, n - x + 1):
if (x, y, height - 1, length) in dp and \
(x, y, height, length - 1) in dp and \
office[x + height - 1][y + length - 1] == '0':
dp.add((x, y, height, length))
max_perimeter = 0
for x, y, l, w in dp:
max_perimeter = max(max_perimeter, 2 * (l + w))
print(max_perimeter)
```
| 3.866053
|
884
|
B
|
Japanese Crosswords Strike Back
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
|
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
|
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
|
[
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,601,464,983
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 66
| 186
| 10,137,600
|
n,x=map(int,input().split())
l=list(map(int,input().split()))
if sum(l)>x:
print("NO")
elif sum(l)+n-1==x or (n==1 and (l[0]==x)):
print("YES")
else:
print("NO")
|
Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n,x=map(int,input().split())
l=list(map(int,input().split()))
if sum(l)>x:
print("NO")
elif sum(l)+n-1==x or (n==1 and (l[0]==x)):
print("YES")
else:
print("NO")
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,663,858,075
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 92
| 0
|
kelime=list(input())
word=list(input())
kelime.reverse()
if kelime==word:
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
kelime=list(input())
word=list(input())
kelime.reverse()
if kelime==word:
print("YES")
else:
print("NO")
```
| 3.977
|
954
|
A
|
Diagonal Walking
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
|
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
|
Print the minimum possible length of the sequence of moves after all replacements are done.
|
[
"5\nRUURU\n",
"17\nUUURRRRRUUURURUUU\n"
] |
[
"3\n",
"13\n"
] |
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
| 0
|
[
{
"input": "5\nRUURU",
"output": "3"
},
{
"input": "17\nUUURRRRRUUURURUUU",
"output": "13"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU",
"output": "100"
},
{
"input": "100\nRRURRUUUURURRRURRRRURRRRRRURRUURRRUUURUURURRURUURUURRUURUURRURURUUUUURUUUUUURRUUURRRURRURRRUURRUUUUR",
"output": "67"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUURUUUUUUUUUUUUUUUUUUUUU",
"output": "99"
},
{
"input": "3\nRUR",
"output": "2"
},
{
"input": "1\nR",
"output": "1"
},
{
"input": "5\nRURUU",
"output": "3"
},
{
"input": "1\nU",
"output": "1"
},
{
"input": "2\nUR",
"output": "1"
},
{
"input": "23\nUUUUUUUUUUUUUUUUUUUUUUU",
"output": "23"
}
] | 1,637,575,333
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
n=int(input())
s=input()
count=0
for i in range(n-1):
if (s[i]=='U' and s[i+1]=='R')or(s[i]=='R' and s[i+1]=='U'):
count+=1
elif (s[i]=='U' and s[i+1]=='U')or(s[i]=='R' and s[i+1]=='R'):
count+=1
print(count)
|
Title: Diagonal Walking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
Output Specification:
Print the minimum possible length of the sequence of moves after all replacements are done.
Demo Input:
['5\nRUURU\n', '17\nUUURRRRRUUURURUUU\n']
Demo Output:
['3\n', '13\n']
Note:
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
|
```python
n=int(input())
s=input()
count=0
for i in range(n-1):
if (s[i]=='U' and s[i+1]=='R')or(s[i]=='R' and s[i+1]=='U'):
count+=1
elif (s[i]=='U' and s[i+1]=='U')or(s[i]=='R' and s[i+1]=='R'):
count+=1
print(count)
```
| 0
|
|
237
|
A
|
Free Cash
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
|
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
|
[
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] |
[
"2\n",
"1\n"
] |
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
| 500
|
[
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,663,917,806
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 9
| 2,000
| 8,806,400
|
dic={}
for i in range(int(input())):
s=input()
if s in list(dic.keys()):
dic[s]+=1
else:
dic[s]=1
print(max(dic.values()))
|
Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
|
```python
dic={}
for i in range(int(input())):
s=input()
if s in list(dic.keys()):
dic[s]+=1
else:
dic[s]=1
print(max(dic.values()))
```
| 0
|
|
489
|
C
|
Given Length and Sum of Digits...
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
|
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
|
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
|
[
"2 15\n",
"3 0\n"
] |
[
"69 96\n",
"-1 -1\n"
] |
none
| 1,500
|
[
{
"input": "2 15",
"output": "69 96"
},
{
"input": "3 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "100 100",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000099999999999 9999999999910000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1 900",
"output": "-1 -1"
},
{
"input": "1 9",
"output": "9 9"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "1 2",
"output": "2 2"
},
{
"input": "1 8",
"output": "8 8"
},
{
"input": "1 10",
"output": "-1 -1"
},
{
"input": "1 11",
"output": "-1 -1"
},
{
"input": "2 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "2 2",
"output": "11 20"
},
{
"input": "2 8",
"output": "17 80"
},
{
"input": "2 10",
"output": "19 91"
},
{
"input": "2 11",
"output": "29 92"
},
{
"input": "2 16",
"output": "79 97"
},
{
"input": "2 17",
"output": "89 98"
},
{
"input": "2 18",
"output": "99 99"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "2 20",
"output": "-1 -1"
},
{
"input": "2 900",
"output": "-1 -1"
},
{
"input": "3 1",
"output": "100 100"
},
{
"input": "3 2",
"output": "101 200"
},
{
"input": "3 3",
"output": "102 300"
},
{
"input": "3 9",
"output": "108 900"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "3 20",
"output": "299 992"
},
{
"input": "3 21",
"output": "399 993"
},
{
"input": "3 26",
"output": "899 998"
},
{
"input": "3 27",
"output": "999 999"
},
{
"input": "3 28",
"output": "-1 -1"
},
{
"input": "3 100",
"output": "-1 -1"
},
{
"input": "100 0",
"output": "-1 -1"
},
{
"input": "100 1",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 2",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 9",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 10",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009 9100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 11",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000019 9200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 296",
"output": "1000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999999999999 9999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 297",
"output": "1000000000000000000000000000000000000000000000000000000000000000000899999999999999999999999999999999 9999999999999999999999999999999990000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 298",
"output": "1000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999 9999999999999999999999999999999991000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 299",
"output": "1000000000000000000000000000000000000000000000000000000000000000001999999999999999999999999999999999 9999999999999999999999999999999992000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 300",
"output": "1000000000000000000000000000000000000000000000000000000000000000002999999999999999999999999999999999 9999999999999999999999999999999993000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 301",
"output": "1000000000000000000000000000000000000000000000000000000000000000003999999999999999999999999999999999 9999999999999999999999999999999994000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 895",
"output": "4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994"
},
{
"input": "100 896",
"output": "5999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995"
},
{
"input": "100 897",
"output": "6999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996"
},
{
"input": "100 898",
"output": "7999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997"
},
{
"input": "100 899",
"output": "8999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998"
},
{
"input": "100 900",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 900",
"output": "-1 -1"
},
{
"input": "99 891",
"output": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 892",
"output": "-1 -1"
},
{
"input": "96 364",
"output": "100000000000000000000000000000000000000000000000000000039999999999999999999999999999999999999999 999999999999999999999999999999999999999940000000000000000000000000000000000000000000000000000000"
},
{
"input": "97 206",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000079999999999999999999999 9999999999999999999999800000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "98 250",
"output": "10000000000000000000000000000000000000000000000000000000000000000000006999999999999999999999999999 99999999999999999999999999970000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "99 96",
"output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000059999999999 999999999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 215",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999 9999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "5 18",
"output": "10089 99000"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "3 3",
"output": "102 300"
}
] | 1,695,055,025
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 31
| 0
|
def len_and_sum(m, s):
if 9 * m < s or s < 1:
return '-1 -1'
s1 = s
maxz, minz = '', ''
for i in range(m):
if s >= 9:
maxz += '9'
s -= 9
else:
maxz += str(s)
s = 0
minz = maxz[::-1]
if minz[0] == '0':
for i in range(m):
if minz[i] != '0':
minz = minz[:i] + str(int(minz[i]) - 1) + minz[i + 1:]
minz = '1' + minz[1:]
break
return minz + ' ' + maxz
def main():
m, s = map(int, input().split())
print(len_and_sum(m, s))
if __name__ == "__main__":
main()
|
Title: Given Length and Sum of Digits...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
Input Specification:
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
Output Specification:
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
Demo Input:
['2 15\n', '3 0\n']
Demo Output:
['69 96\n', '-1 -1\n']
Note:
none
|
```python
def len_and_sum(m, s):
if 9 * m < s or s < 1:
return '-1 -1'
s1 = s
maxz, minz = '', ''
for i in range(m):
if s >= 9:
maxz += '9'
s -= 9
else:
maxz += str(s)
s = 0
minz = maxz[::-1]
if minz[0] == '0':
for i in range(m):
if minz[i] != '0':
minz = minz[:i] + str(int(minz[i]) - 1) + minz[i + 1:]
minz = '1' + minz[1:]
break
return minz + ' ' + maxz
def main():
m, s = map(int, input().split())
print(len_and_sum(m, s))
if __name__ == "__main__":
main()
```
| 0
|
|
272
|
B
|
Dima and Sequence
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Dima got into number sequences. Now he's got sequence *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers. Also, Dima has got a function *f*(*x*), which can be defined with the following recurrence:
- *f*(0)<==<=0; - *f*(2·*x*)<==<=*f*(*x*); - *f*(2·*x*<=+<=1)<==<=*f*(*x*)<=+<=1.
Dima wonders, how many pairs of indexes (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*) are there, such that *f*(*a**i*)<==<=*f*(*a**j*). Help him, count the number of such pairs.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
The numbers in the lines are separated by single spaces.
|
In a single line print the answer to the problem.
Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"3\n1 2 4\n",
"3\n5 3 1\n"
] |
[
"3\n",
"1\n"
] |
In the first sample any pair (*i*, *j*) will do, so the answer is 3.
In the second sample only pair (1, 2) will do.
| 1,000
|
[
{
"input": "3\n1 2 4",
"output": "3"
},
{
"input": "3\n5 3 1",
"output": "1"
},
{
"input": "2\n469264357 996569493",
"output": "0"
},
{
"input": "6\n396640239 62005863 473635171 329666981 510631133 207643327",
"output": "2"
},
{
"input": "8\n851991424 32517099 310793856 776130403 342626527 58796623 49544509 517126753",
"output": "2"
},
{
"input": "7\n481003311 553247971 728349004 258700257 916143165 398096105 412826266",
"output": "2"
},
{
"input": "4\n363034183 741262741 657823174 453546052",
"output": "1"
},
{
"input": "8\n7 1 2 7 6 8 6 5",
"output": "7"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n7 1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
}
] | 1,559,322,009
| 2,889
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 0
|
n = int(input())
def norm(a):
if a % 2 == 0:
while a % 2 == 0:
a //= 2
if a > 1:
while (a - 1) % 4 == 0:
a = (a - 1) // 2 + 1
return a
m = sorted([norm(int(x)) for x in input().split()])
s = 0
for i in range(len(m)-1):
if m[i] == m[i+1]:
s += 1
print(s)
|
Title: Dima and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima got into number sequences. Now he's got sequence *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers. Also, Dima has got a function *f*(*x*), which can be defined with the following recurrence:
- *f*(0)<==<=0; - *f*(2·*x*)<==<=*f*(*x*); - *f*(2·*x*<=+<=1)<==<=*f*(*x*)<=+<=1.
Dima wonders, how many pairs of indexes (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*) are there, such that *f*(*a**i*)<==<=*f*(*a**j*). Help him, count the number of such pairs.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print the answer to the problem.
Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3\n1 2 4\n', '3\n5 3 1\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample any pair (*i*, *j*) will do, so the answer is 3.
In the second sample only pair (1, 2) will do.
|
```python
n = int(input())
def norm(a):
if a % 2 == 0:
while a % 2 == 0:
a //= 2
if a > 1:
while (a - 1) % 4 == 0:
a = (a - 1) // 2 + 1
return a
m = sorted([norm(int(x)) for x in input().split()])
s = 0
for i in range(len(m)-1):
if m[i] == m[i+1]:
s += 1
print(s)
```
| 0
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,605,791,998
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 0
|
s = input()
t = input()
if len(s) != len(t):
print('Error')
else:
tmp = list(s)
n = len(tmp)
for i in range(n):
tmp[i], tmp[n-i-1] = tmp[n-i-1], tmp[i]
word = ''.join(tmp)
if word == t:
print('YES')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = input()
t = input()
if len(s) != len(t):
print('Error')
else:
tmp = list(s)
n = len(tmp)
for i in range(n):
tmp[i], tmp[n-i-1] = tmp[n-i-1], tmp[i]
word = ''.join(tmp)
if word == t:
print('YES')
else:
print('NO')
```
| 0
|
659
|
E
|
New Reform
|
PROGRAMMING
| 1,600
|
[
"data structures",
"dfs and similar",
"dsu",
"graphs",
"greedy"
] | null | null |
Berland has *n* cities connected by *m* bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guaranteed that you can get from any city to any other one, using only the existing roads.
The President of Berland decided to make changes to the road system and instructed the Ministry of Transport to make this reform. Now, each road should be unidirectional (only lead from one city to another).
In order not to cause great resentment among residents, the reform needs to be conducted so that there can be as few separate cities as possible. A city is considered separate, if no road leads into it, while it is allowed to have roads leading from this city.
Help the Ministry of Transport to find the minimum possible number of separate cities after the reform.
|
The first line of the input contains two positive integers, *n* and *m* — the number of the cities and the number of roads in Berland (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000).
Next *m* lines contain the descriptions of the roads: the *i*-th road is determined by two distinct integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the numbers of the cities connected by the *i*-th road.
It is guaranteed that there is no more than one road between each pair of cities, but it is not guaranteed that from any city you can get to any other one, using only roads.
|
Print a single integer — the minimum number of separated cities after the reform.
|
[
"4 3\n2 1\n1 3\n4 3\n",
"5 5\n2 1\n1 3\n2 3\n2 5\n4 3\n",
"6 5\n1 2\n2 3\n4 5\n4 6\n5 6\n"
] |
[
"1\n",
"0\n",
"1\n"
] |
In the first sample the following road orientation is allowed: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/04481aced8a9d501ae5d785ab654c542ff5497a1.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d220a75a508edc0d540dbaec5e198345049b66f.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The second sample: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/030fc9181b578c2d906254d38dc56da5554323eb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6011a68d6de05246ad6ba8aa24a5c5c71cd450a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d220a75a508edc0d540dbaec5e198345049b66f.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The third sample: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7113d83514735a488d7b85262585381d26986195.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/98300067e46d036076e429b208db829296ebca9d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/146f3f3d208399fbab4d1d8c2800f07bf7b463e5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 1,500
|
[
{
"input": "4 3\n2 1\n1 3\n4 3",
"output": "1"
},
{
"input": "5 5\n2 1\n1 3\n2 3\n2 5\n4 3",
"output": "0"
},
{
"input": "6 5\n1 2\n2 3\n4 5\n4 6\n5 6",
"output": "1"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "0"
},
{
"input": "10 45\n3 5\n2 3\n4 8\n2 5\n6 8\n5 7\n2 1\n3 7\n5 10\n6 1\n9 4\n3 6\n9 10\n6 7\n1 7\n7 9\n6 9\n9 3\n4 2\n2 6\n5 6\n5 8\n3 4\n10 8\n7 8\n4 6\n9 1\n5 9\n7 4\n1 10\n9 2\n2 8\n6 10\n9 8\n1 5\n7 2\n10 3\n3 8\n4 10\n4 1\n10 7\n1 3\n1 8\n10 2\n4 5",
"output": "0"
},
{
"input": "20 20\n16 3\n15 14\n6 14\n13 19\n7 13\n3 13\n3 2\n17 11\n14 20\n19 10\n4 13\n3 8\n18 4\n12 7\n6 3\n11 13\n17 19\n5 14\n9 2\n11 1",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 2",
"output": "0"
}
] | 1,596,071,037
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 998
| 43,417,600
|
#cycle detection
def dfs(node,par):
seen[node] = 1
for each in adj[node]:
if seen[each] == 0:
if dfs(each,node) == True: return True
else:
if each != par: return True
return False
n,m = map(int,input().split())
adj = [list() for i in range(n+1)]
seen = [0] * (n+1)
for _ in range(m):
a,b = map(int,input().split())
adj[a].append(b)
adj[b].append(a)
ans = 0
for i in range(1,n):
if seen[i] == 0:
if not dfs(i,i):
ans += 1
print(ans)
|
Title: New Reform
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland has *n* cities connected by *m* bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guaranteed that you can get from any city to any other one, using only the existing roads.
The President of Berland decided to make changes to the road system and instructed the Ministry of Transport to make this reform. Now, each road should be unidirectional (only lead from one city to another).
In order not to cause great resentment among residents, the reform needs to be conducted so that there can be as few separate cities as possible. A city is considered separate, if no road leads into it, while it is allowed to have roads leading from this city.
Help the Ministry of Transport to find the minimum possible number of separate cities after the reform.
Input Specification:
The first line of the input contains two positive integers, *n* and *m* — the number of the cities and the number of roads in Berland (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000).
Next *m* lines contain the descriptions of the roads: the *i*-th road is determined by two distinct integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the numbers of the cities connected by the *i*-th road.
It is guaranteed that there is no more than one road between each pair of cities, but it is not guaranteed that from any city you can get to any other one, using only roads.
Output Specification:
Print a single integer — the minimum number of separated cities after the reform.
Demo Input:
['4 3\n2 1\n1 3\n4 3\n', '5 5\n2 1\n1 3\n2 3\n2 5\n4 3\n', '6 5\n1 2\n2 3\n4 5\n4 6\n5 6\n']
Demo Output:
['1\n', '0\n', '1\n']
Note:
In the first sample the following road orientation is allowed: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/04481aced8a9d501ae5d785ab654c542ff5497a1.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d220a75a508edc0d540dbaec5e198345049b66f.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The second sample: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/030fc9181b578c2d906254d38dc56da5554323eb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6011a68d6de05246ad6ba8aa24a5c5c71cd450a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d220a75a508edc0d540dbaec5e198345049b66f.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The third sample: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e5b18c46402af724bd3841d549d5d6f52fc16253.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/29f71c065c3536e88b54429c734103ad3604f68b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7113d83514735a488d7b85262585381d26986195.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/98300067e46d036076e429b208db829296ebca9d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/146f3f3d208399fbab4d1d8c2800f07bf7b463e5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
#cycle detection
def dfs(node,par):
seen[node] = 1
for each in adj[node]:
if seen[each] == 0:
if dfs(each,node) == True: return True
else:
if each != par: return True
return False
n,m = map(int,input().split())
adj = [list() for i in range(n+1)]
seen = [0] * (n+1)
for _ in range(m):
a,b = map(int,input().split())
adj[a].append(b)
adj[b].append(a)
ans = 0
for i in range(1,n):
if seen[i] == 0:
if not dfs(i,i):
ans += 1
print(ans)
```
| 3
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,566,668,030
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 466
| 1,126,400
|
from fractions import Fraction
y, w = [int(i) for i in input().split()]
greater = max(y, w)
greaters = 6 - greater + 1
result = Fraction(greaters, 6)
if result.denominator == 1:
print(str(result.numerator) + '/1')
else:
print(result)
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
from fractions import Fraction
y, w = [int(i) for i in input().split()]
greater = max(y, w)
greaters = 6 - greater + 1
result = Fraction(greaters, 6)
if result.denominator == 1:
print(str(result.numerator) + '/1')
else:
print(result)
```
| 3.758608
|
729
|
B
|
Spotlights
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation"
] | null | null |
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
|
Print one integer — the number of good positions for placing the spotlight.
|
[
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] |
[
"9\n",
"20\n"
] |
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
| 1,000
|
[
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 1\n1\n0\n0",
"output": "2"
},
{
"input": "5 7\n0 0 0 0 0 0 1\n0 0 0 0 0 0 1\n0 0 0 1 0 0 0\n0 0 0 0 0 0 0\n0 0 0 0 0 0 0",
"output": "25"
},
{
"input": "10 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "5 7\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1\n0 1 1 1 1 1 1\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1",
"output": "26"
},
{
"input": "10 20\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "121"
},
{
"input": "1 2\n0 1",
"output": "1"
},
{
"input": "1 2\n1 0",
"output": "1"
},
{
"input": "1 2\n1 1",
"output": "0"
},
{
"input": "2 1\n1\n0",
"output": "1"
},
{
"input": "2 1\n0\n1",
"output": "1"
},
{
"input": "2 1\n1\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "4 4\n1 1 1 1\n1 0 0 1\n1 0 0 1\n1 1 1 1",
"output": "16"
}
] | 1,479,637,393
| 4,693
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 6
| 1,000
| 5,939,200
|
a, b = map(int, input().split())
scene = [list(map(int, input().split())) for i in range(a)]
lines = [False] * a
columns = [False] * b
ans = 0
for i in range(a):
for j in range(b):
if scene[i][j] == 0:
if lines[i] == True:
ans += 1
else:
lines[i] = True
lines[i] = False
for j in range(b - 1, -1, -1):
if scene[i][j] == 0:
if lines[i] == True:
ans += 1
else:
lines[i] = True
for j in range(b):
for i in range(a):
if scene[i][j] == 0:
if columns[j] == True:
ans += 1
else:
columns[j] = True
columns[j] = False
for i in range(a - 1, -1, -1):
if scene[i][j] == 0:
if columns[j] == True:
ans += 1
else:
columns[j] = True
print(ans)
|
Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
|
```python
a, b = map(int, input().split())
scene = [list(map(int, input().split())) for i in range(a)]
lines = [False] * a
columns = [False] * b
ans = 0
for i in range(a):
for j in range(b):
if scene[i][j] == 0:
if lines[i] == True:
ans += 1
else:
lines[i] = True
lines[i] = False
for j in range(b - 1, -1, -1):
if scene[i][j] == 0:
if lines[i] == True:
ans += 1
else:
lines[i] = True
for j in range(b):
for i in range(a):
if scene[i][j] == 0:
if columns[j] == True:
ans += 1
else:
columns[j] = True
columns[j] = False
for i in range(a - 1, -1, -1):
if scene[i][j] == 0:
if columns[j] == True:
ans += 1
else:
columns[j] = True
print(ans)
```
| 0
|
|
280
|
C
|
Game on Tree
|
PROGRAMMING
| 2,200
|
[
"implementation",
"math",
"probabilities",
"trees"
] | null | null |
Momiji has got a rooted tree, consisting of *n* nodes. The tree nodes are numbered by integers from 1 to *n*. The root has number 1. Momiji decided to play a game on this tree.
The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by *v*) and removes all the subtree nodes with the root in node *v* from the tree. Node *v* gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.
Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of nodes in the tree. The next *n*<=-<=1 lines contain the tree edges. The *i*-th line contains integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the nodes that are connected by the *i*-th edge.
It is guaranteed that the given graph is a tree.
|
Print a single real number — the expectation of the number of steps in the described game.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"2\n1 2\n",
"3\n1 2\n1 3\n"
] |
[
"1.50000000000000000000\n",
"2.00000000000000000000\n"
] |
In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are:
In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are:
| 1,500
|
[
{
"input": "2\n1 2",
"output": "1.50000000000000000000"
},
{
"input": "3\n1 2\n1 3",
"output": "2.00000000000000000000"
},
{
"input": "10\n1 2\n2 3\n3 4\n1 5\n2 6\n6 7\n4 8\n6 9\n9 10",
"output": "3.81666666666666690000"
},
{
"input": "6\n1 3\n2 4\n5 6\n3 6\n5 4",
"output": "2.45000000000000020000"
}
] | 1,588,047,835
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 40
| 1,000
| 10,342,400
|
# encoding: utf-8
"""
有一棵树，每次随机选一个点，
将其及其子孙节点全部删除，问将树删空需要的操作次数期望
树根的概率必然为1
一级子节点的概率为树根排在其后的概率，1/2
二级子节点的概率为树根与一级均排在其后的概率，1/3
...
"""
from math import sqrt
from queue import Queue
import sys
sys.setrecursionlimit(10**6)
if __name__ == '__main__':
# sys.stdin = open('1.txt', 'r')
n = int(input())
edges = [[] for i in range(n+1)]
for i in range(1, n):
a, b = [int(i) for i in input().split()]
edges[a].append(b); edges[b].append(a)
d = [0 for i in range(n+1)]
q = Queue()
q.put(1)
d[1] = 1
ans = 1.0
while q.empty() is False:
u = q.get()
for v in edges[u]:
if d[v] == 0:
d[v] = d[u] + 1
q.put(v)
ans += 1 / d[v]
print(ans)
|
Title: Game on Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Momiji has got a rooted tree, consisting of *n* nodes. The tree nodes are numbered by integers from 1 to *n*. The root has number 1. Momiji decided to play a game on this tree.
The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by *v*) and removes all the subtree nodes with the root in node *v* from the tree. Node *v* gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.
Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of nodes in the tree. The next *n*<=-<=1 lines contain the tree edges. The *i*-th line contains integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the nodes that are connected by the *i*-th edge.
It is guaranteed that the given graph is a tree.
Output Specification:
Print a single real number — the expectation of the number of steps in the described game.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['2\n1 2\n', '3\n1 2\n1 3\n']
Demo Output:
['1.50000000000000000000\n', '2.00000000000000000000\n']
Note:
In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are:
In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are:
|
```python
# encoding: utf-8
"""
有一棵树，每次随机选一个点，
将其及其子孙节点全部删除，问将树删空需要的操作次数期望
树根的概率必然为1
一级子节点的概率为树根排在其后的概率，1/2
二级子节点的概率为树根与一级均排在其后的概率，1/3
...
"""
from math import sqrt
from queue import Queue
import sys
sys.setrecursionlimit(10**6)
if __name__ == '__main__':
# sys.stdin = open('1.txt', 'r')
n = int(input())
edges = [[] for i in range(n+1)]
for i in range(1, n):
a, b = [int(i) for i in input().split()]
edges[a].append(b); edges[b].append(a)
d = [0 for i in range(n+1)]
q = Queue()
q.put(1)
d[1] = 1
ans = 1.0
while q.empty() is False:
u = q.get()
for v in edges[u]:
if d[v] == 0:
d[v] = d[u] + 1
q.put(v)
ans += 1 / d[v]
print(ans)
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,597,124,293
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 108
| 6,656,000
|
def f(a,b):
l=len(a)
r=[' '] *l
for i in range(0,l):
if a[i]==b[i]:
r[i]='0'
else:
r[i]='1'
return "".join(r)
a=input()
b=input()
print(f(a,b))
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
def f(a,b):
l=len(a)
r=[' '] *l
for i in range(0,l):
if a[i]==b[i]:
r[i]='0'
else:
r[i]='1'
return "".join(r)
a=input()
b=input()
print(f(a,b))
```
| 3.960602
|
490
|
A
|
Team Olympiad
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
|
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
|
[
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] |
[
"2\n3 5 2\n6 7 4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,657,447,537
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 155
| 10,035,200
|
n = int(input())
for i in range(n):
t = list(map(int, input().split()))
cnt1, cnt2, cnt3 = t.count(1), t.count(2), t.count(3)
min_mun = min(cnt1, cnt2, cnt3)
if cnt1 == 0 or cnt2 == 0 or cnt3 == 0:
print(0)
else:
li = []
while cnt1 and cnt2 and cnt3:
one, two, three = t.pop(1) + 1, t.pop(2) + 1, t.pop(3) + 1
li.append([one, two, three])
print(min_num)
for i in range(min_num):
print(li[i].split(' '))
|
Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none
|
```python
n = int(input())
for i in range(n):
t = list(map(int, input().split()))
cnt1, cnt2, cnt3 = t.count(1), t.count(2), t.count(3)
min_mun = min(cnt1, cnt2, cnt3)
if cnt1 == 0 or cnt2 == 0 or cnt3 == 0:
print(0)
else:
li = []
while cnt1 and cnt2 and cnt3:
one, two, three = t.pop(1) + 1, t.pop(2) + 1, t.pop(3) + 1
li.append([one, two, three])
print(min_num)
for i in range(min_num):
print(li[i].split(' '))
```
| -1
|
|
659
|
A
|
Round House
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
|
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
|
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
|
[
"6 2 -5\n",
"5 1 3\n",
"3 2 7\n"
] |
[
"3\n",
"4\n",
"3\n"
] |
The first example is illustrated by the picture in the statements.
| 500
|
[
{
"input": "6 2 -5",
"output": "3"
},
{
"input": "5 1 3",
"output": "4"
},
{
"input": "3 2 7",
"output": "3"
},
{
"input": "1 1 0",
"output": "1"
},
{
"input": "1 1 -1",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "100 1 -1",
"output": "100"
},
{
"input": "100 54 100",
"output": "54"
},
{
"input": "100 37 -100",
"output": "37"
},
{
"input": "99 41 0",
"output": "41"
},
{
"input": "97 37 -92",
"output": "42"
},
{
"input": "99 38 59",
"output": "97"
},
{
"input": "35 34 1",
"output": "35"
},
{
"input": "48 1 -1",
"output": "48"
},
{
"input": "87 65 -76",
"output": "76"
},
{
"input": "76 26 29",
"output": "55"
},
{
"input": "100 65 0",
"output": "65"
},
{
"input": "2 1 100",
"output": "1"
},
{
"input": "3 2 -100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "1 1 -100",
"output": "1"
},
{
"input": "3 1 -100",
"output": "3"
},
{
"input": "4 3 -100",
"output": "3"
},
{
"input": "3 2 -12",
"output": "2"
},
{
"input": "2 2 -100",
"output": "2"
},
{
"input": "3 2 -90",
"output": "2"
},
{
"input": "6 2 -10",
"output": "4"
},
{
"input": "3 3 -100",
"output": "2"
},
{
"input": "5 2 4",
"output": "1"
},
{
"input": "6 4 5",
"output": "3"
},
{
"input": "3 2 -6",
"output": "2"
},
{
"input": "5 1 -99",
"output": "2"
},
{
"input": "6 2 5",
"output": "1"
},
{
"input": "10 1 -100",
"output": "1"
},
{
"input": "2 2 1",
"output": "1"
},
{
"input": "3 3 1",
"output": "1"
},
{
"input": "6 4 4",
"output": "2"
},
{
"input": "17 17 2",
"output": "2"
},
{
"input": "6 6 1",
"output": "1"
},
{
"input": "5 3 -2",
"output": "1"
},
{
"input": "6 2 -100",
"output": "4"
},
{
"input": "5 3 -100",
"output": "3"
},
{
"input": "5 4 3",
"output": "2"
},
{
"input": "3 2 2",
"output": "1"
},
{
"input": "5 5 2",
"output": "2"
},
{
"input": "3 2 5",
"output": "1"
},
{
"input": "5 5 -1",
"output": "4"
},
{
"input": "5 3 3",
"output": "1"
},
{
"input": "4 2 3",
"output": "1"
},
{
"input": "88 76 74",
"output": "62"
}
] | 1,621,245,934
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 18
| 93
| 0
|
n,a,w=map(int,input().split())
w1=abs(w)
w1=w1%n
if(w1==0):
print(a)
elif(w>0):
if(a+w1<=n):
print(a+w1)
else:
print(n-a-w1)
elif(w<0):
w1=n-w1
if(a+w1<=n):
print(a+w1)
else:
print(n-a-w1)
|
Title: Round House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
Input Specification:
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output Specification:
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
Demo Input:
['6 2 -5\n', '5 1 3\n', '3 2 7\n']
Demo Output:
['3\n', '4\n', '3\n']
Note:
The first example is illustrated by the picture in the statements.
|
```python
n,a,w=map(int,input().split())
w1=abs(w)
w1=w1%n
if(w1==0):
print(a)
elif(w>0):
if(a+w1<=n):
print(a+w1)
else:
print(n-a-w1)
elif(w<0):
w1=n-w1
if(a+w1<=n):
print(a+w1)
else:
print(n-a-w1)
```
| 0
|
|
902
|
A
|
Visiting a Friend
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
|
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
|
[
"3 5\n0 2\n2 4\n3 5\n",
"3 7\n0 4\n2 5\n6 7\n"
] |
[
"YES\n",
"NO\n"
] |
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports.
| 500
|
[
{
"input": "3 5\n0 2\n2 4\n3 5",
"output": "YES"
},
{
"input": "3 7\n0 4\n2 5\n6 7",
"output": "NO"
},
{
"input": "1 1\n0 0",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 8\n8 8\n9 9\n9 9\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 100\n71 73\n80 94\n81 92\n84 85\n85 100\n88 91\n91 95\n92 98\n92 98\n99 100\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 6\n1 6\n1 2\n1 3\n1 2\n1 3\n2 5\n2 4\n2 3\n2 4\n2 6\n2 2\n2 5\n2 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 9\n3 3\n3 7\n3 9\n3 3\n3 9\n4 6\n4 7\n4 5\n4 7\n5 8\n5 5\n5 9\n5 7\n5 5\n6 6\n6 9\n6 7\n6 8\n6 9\n6 8\n7 7\n7 8\n7 7\n7 8\n8 9\n8 8\n8 9\n8 8\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 10\n8 10\n9 9\n9 9\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 100\n1 38\n2 82\n5 35\n7 71\n8 53\n11 49\n15 27\n17 84\n17 75\n18 99\n18 43\n18 69\n21 89\n27 60\n27 29\n38 62\n38 77\n39 83\n40 66\n48 80\n48 100\n50 51\n50 61\n53 77\n53 63\n55 58\n56 68\n60 82\n62 95\n66 74\n67 83\n69 88\n69 81\n69 88\n69 98\n70 91\n70 76\n71 90\n72 99\n81 99\n85 87\n88 97\n88 93\n90 97\n90 97\n92 98\n98 99\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 10\n1 9\n1 6\n1 2\n1 3\n1 2\n2 6\n2 5\n2 4\n2 3\n2 10\n2 2\n2 6\n2 2\n3 10\n3 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 10\n3 5\n3 3\n3 7\n4 8\n4 8\n4 9\n4 6\n5 7\n5 10\n5 7\n5 8\n5 5\n6 8\n6 9\n6 10\n6 6\n6 9\n6 7\n7 8\n7 9\n7 10\n7 10\n8 8\n8 8\n8 9\n8 10\n9 10\n9 9\n9 10\n9 10\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 4\n1 5\n1 9\n1 1\n1 6\n1 6\n2 5\n2 7\n2 7\n2 7\n2 7\n3 4\n3 7\n3 9\n3 5\n3 3\n4 4\n4 6\n4 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n5 8\n5 9\n5 8\n6 8\n6 7\n6 8\n6 9\n6 9\n6 6\n6 9\n6 7\n7 7\n7 7\n7 7\n7 8\n7 7\n7 8\n7 8\n7 9\n8 8\n8 8\n8 8\n8 8\n8 8\n8 9\n8 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 40\n38 38\n40 40",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 92\n92 97\n96 99\n97 98\n97 99\n99 99\n100 100\n100 100\n100 100",
"output": "NO"
},
{
"input": "1 10\n0 10",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 40\n23 23\n23 28\n24 29\n25 38\n26 35\n27 37\n28 39\n28 33\n28 40\n28 33\n29 31\n29 33\n30 38\n30 36\n30 30\n30 38\n31 37\n31 35\n31 32\n31 36\n33 39\n33 40\n35 38\n36 38\n37 38\n37 40\n38 39\n38 40\n38 39\n39 39\n39 40\n40 40\n40 40\n40 40\n40 40",
"output": "YES"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 40\n11 31\n12 26\n13 25\n14 32\n17 19\n21 29\n22 36\n24 27\n25 39\n25 27\n27 32\n27 29\n27 39\n27 29\n28 38\n30 38\n32 40\n32 38\n33 33\n33 40\n34 35\n34 34\n34 38\n34 38\n35 37\n36 39\n36 39\n37 37\n38 40\n39 39\n40 40",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 22\n23 28\n23 39\n24 24\n25 27\n26 38\n27 39\n28 33\n28 39\n28 34\n28 33\n29 30\n29 35\n30 30\n30 38\n30 34\n30 31\n31 36\n31 31\n31 32\n31 38\n33 34\n33 34\n35 36\n36 38\n37 38\n37 39\n38 38\n38 38\n38 38\n39 39\n39 39\n40 40\n40 40\n40 40\n40 40",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n88 99",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 10\n7 10\n9 10",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 5\n4 8\n5 5\n5 7\n6 7\n6 6\n7 7\n7 7\n7 7\n7 8\n7 8\n8 8\n8 8\n8 9\n8 8\n8 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n32 37",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 10\n4 6\n5 9\n5 5\n6 7\n6 10\n7 8\n7 7\n7 7\n7 7\n7 10\n8 8\n8 8\n8 10\n8 8\n8 8\n9 10\n9 10\n9 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "1 1\n0 1",
"output": "YES"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 36\n38 38\n40 40",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 82\n71 94\n80 90\n81 88\n84 93\n85 89\n88 92\n91 97\n92 99\n92 97\n99 99\n100 100",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n79 100",
"output": "YES"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n37 40",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 10\n1 2\n1 5\n1 10\n1 8\n1 1\n2 8\n2 7\n2 5\n2 5\n2 7\n3 5\n3 7\n3 5\n3 4\n3 7\n4 7\n4 8\n4 6\n5 7\n5 10\n5 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n6 10\n6 9\n6 7\n6 10\n6 8\n6 7\n6 10\n6 10\n7 8\n7 9\n7 8\n7 8\n7 8\n7 8\n7 7\n7 7\n8 8\n8 8\n8 10\n8 9\n8 9\n8 9\n8 9\n9 9\n9 10\n9 9\n9 9\n9 9\n9 9\n9 10\n9 10\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 7\n1 69\n2 83\n5 67\n7 82\n8 31\n11 25\n15 44\n17 75\n17 27\n18 43\n18 69\n18 40\n21 66\n27 29\n27 64\n38 77\n38 90\n39 52\n40 60\n48 91\n48 98\n50 89\n50 63\n53 54\n53 95\n55 76\n56 59\n60 96\n62 86\n66 70\n67 77\n69 88\n69 98\n69 80\n69 95\n70 74\n70 77\n71 99\n72 73\n81 87\n85 99\n88 96\n88 91\n90 97\n90 99\n92 92\n98 99\n100 100",
"output": "NO"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 35\n11 23\n12 36\n13 31\n14 31\n17 17\n21 25\n22 33\n24 26\n25 32\n25 25\n27 39\n27 29\n27 34\n27 32\n28 34\n30 36\n32 37\n32 33\n33 35\n33 33\n34 38\n34 38\n34 36\n34 36\n35 36\n36 36\n36 39\n37 37\n38 39\n39 39\n40 40",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 9\n7 7\n9 9",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 100\n92 97\n96 96\n97 98\n97 100\n99 100\n100 100\n100 100\n100 100",
"output": "YES"
},
{
"input": "2 2\n0 2\n0 1",
"output": "YES"
},
{
"input": "2 100\n0 49\n50 100",
"output": "NO"
},
{
"input": "2 100\n0 50\n50 100",
"output": "YES"
},
{
"input": "10 100\n0 10\n10 20\n20 30\n30 40\n40 50\n50 60\n60 70\n70 80\n80 90\n90 100",
"output": "YES"
},
{
"input": "100 10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 10",
"output": "YES"
},
{
"input": "1 4\n1 4",
"output": "NO"
},
{
"input": "3 5\n0 2\n2 5\n3 4",
"output": "YES"
},
{
"input": "2 5\n0 5\n1 2",
"output": "YES"
},
{
"input": "5 5\n0 2\n0 2\n3 5\n3 5\n3 5",
"output": "NO"
},
{
"input": "4 10\n0 3\n2 10\n4 10\n5 7",
"output": "YES"
},
{
"input": "3 8\n0 0\n1 8\n2 8",
"output": "NO"
},
{
"input": "6 8\n0 3\n1 5\n2 3\n5 6\n6 7\n7 8",
"output": "YES"
}
] | 1,513,769,171
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
using namespace std;
const int maxn=105;
vector<int>g[105];
bool vis[105];
void add(int a,int b){
g[a].push_back(b);
g[b].push_back(a);
}
void dfs(int m){
for(int i=0;i<g[m].size();i++){
if(!vis[g[m][i]]){
vis[g[m][i]]=true;
dfs(g[m][i]);
}
}
}
int main()
{ int m,n;
int a,b;
bool flag1=false;
bool flag2=false;
scanf("%d%d",&m,&n);
for(int i=1;i<=m;i++){
scanf("%d%d",&a,&b);
for(int i=a+1;i<=b;i++){
add(a,i);
}
}
dfs(1);
if(vis[n])
puts("yes");
else
puts("no");
return 0;
}
|
Title: Visiting a Friend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
Output Specification:
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['3 5\n0 2\n2 4\n3 5\n', '3 7\n0 4\n2 5\n6 7\n']
Demo Output:
['YES\n', 'NO\n']
Note:
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports.
|
```python
#include <bits/stdc++.h>
using namespace std;
const int maxn=105;
vector<int>g[105];
bool vis[105];
void add(int a,int b){
g[a].push_back(b);
g[b].push_back(a);
}
void dfs(int m){
for(int i=0;i<g[m].size();i++){
if(!vis[g[m][i]]){
vis[g[m][i]]=true;
dfs(g[m][i]);
}
}
}
int main()
{ int m,n;
int a,b;
bool flag1=false;
bool flag2=false;
scanf("%d%d",&m,&n);
for(int i=1;i<=m;i++){
scanf("%d%d",&a,&b);
for(int i=a+1;i<=b;i++){
add(a,i);
}
}
dfs(1);
if(vis[n])
puts("yes");
else
puts("no");
return 0;
}
```
| -1
|
|
615
|
A
|
Bulbs
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
|
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
|
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
|
[
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
| 500
|
[
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},
{
"input": "1 5\n5 1 1 1 1 5",
"output": "NO"
},
{
"input": "2 5\n4 3 1 4 2\n4 2 3 4 5",
"output": "YES"
},
{
"input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1",
"output": "YES"
},
{
"input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6",
"output": "YES"
},
{
"input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 4\n3 1 2 3",
"output": "NO"
},
{
"input": "1 4\n3 2 3 4",
"output": "NO"
},
{
"input": "2 4\n3 2 3 4\n1 1",
"output": "YES"
},
{
"input": "2 4\n3 1 2 3\n1 4",
"output": "YES"
},
{
"input": "5 1\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "1 1\n0",
"output": "NO"
},
{
"input": "1 10\n10 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "1 100\n99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "NO"
},
{
"input": "1 3\n3 1 2 1",
"output": "NO"
},
{
"input": "1 100\n100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "YES"
}
] | 1,665,466,735
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
n, m = [int(t) for t in input().split()]
bulbs = [0] * m
for _ in range(n):
on = [int(t) - 1 for t in input().split()]
for i in on:
bulbs[i] = 1
if 0 in bulbs:
print('NO')
else:
print('YES')
|
Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
|
```python
n, m = [int(t) for t in input().split()]
bulbs = [0] * m
for _ in range(n):
on = [int(t) - 1 for t in input().split()]
for i in on:
bulbs[i] = 1
if 0 in bulbs:
print('NO')
else:
print('YES')
```
| 0
|
|
476
|
A
|
Dreamoon and Stairs
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
|
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
|
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
|
[
"10 2\n",
"3 5\n"
] |
[
"6\n",
"-1\n"
] |
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
| 500
|
[
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 3",
"output": "5001"
},
{
"input": "10000 10",
"output": "5000"
},
{
"input": "9999 3",
"output": "5001"
},
{
"input": "9999 2",
"output": "5000"
},
{
"input": "9999 10",
"output": "5000"
},
{
"input": "9999 9",
"output": "5004"
},
{
"input": "18 10",
"output": "10"
},
{
"input": "19 10",
"output": "10"
},
{
"input": "20 10",
"output": "10"
},
{
"input": "21 10",
"output": "20"
},
{
"input": "7688 5",
"output": "3845"
},
{
"input": "4608 5",
"output": "2305"
},
{
"input": "3979 2",
"output": "1990"
},
{
"input": "9985 6",
"output": "4998"
},
{
"input": "3230 8",
"output": "1616"
},
{
"input": "24 9",
"output": "18"
},
{
"input": "3275 8",
"output": "1640"
},
{
"input": "6240 7",
"output": "3122"
},
{
"input": "5227 4",
"output": "2616"
},
{
"input": "3832 6",
"output": "1920"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "6 3",
"output": "3"
},
{
"input": "10 5",
"output": "5"
},
{
"input": "3 2",
"output": "2"
},
{
"input": "6 4",
"output": "4"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "9 9",
"output": "9"
},
{
"input": "4 5",
"output": "-1"
}
] | 1,680,864,131
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 0
|
#Khushal Sindhav
#Indian Institute Of Technology, Jodhpur
# 2022
import sys
input=sys.stdin.readline
def exe():
no_of_two=n//2
no_of_one=n%2
tot=(no_of_one+no_of_two)
no=tot//m
if(tot%m!=0):
no+=1
a=no*m
req=a-tot
if(no_of_two>=req):
return req+tot
else:
return -1
else:
return no
return
n,m=map(int,input().split())
print(exe())
|
Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
|
```python
#Khushal Sindhav
#Indian Institute Of Technology, Jodhpur
# 2022
import sys
input=sys.stdin.readline
def exe():
no_of_two=n//2
no_of_one=n%2
tot=(no_of_one+no_of_two)
no=tot//m
if(tot%m!=0):
no+=1
a=no*m
req=a-tot
if(no_of_two>=req):
return req+tot
else:
return -1
else:
return no
return
n,m=map(int,input().split())
print(exe())
```
| 0
|
|
340
|
A
|
The Wall
|
PROGRAMMING
| 1,200
|
[
"math"
] | null | null |
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
|
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
|
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
|
[
"2 3 6 18\n"
] |
[
"3"
] |
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
| 500
|
[
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5",
"output": "0"
},
{
"input": "555 777 1 1000000",
"output": "257"
},
{
"input": "1000 1000 1 32323",
"output": "32"
},
{
"input": "45 125 93451125 100000000",
"output": "5821"
},
{
"input": "101 171 1 1000000000",
"output": "57900"
},
{
"input": "165 255 69696 1000000000",
"output": "356482"
},
{
"input": "555 777 666013 1000000000",
"output": "257229"
},
{
"input": "23 46 123321 900000000",
"output": "19562537"
},
{
"input": "321 123 15 1000000",
"output": "75"
},
{
"input": "819 1000 9532 152901000",
"output": "186"
},
{
"input": "819 1000 10000 1000000",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "1 2 2 1000003",
"output": "500001"
},
{
"input": "1 1 1 1000000000",
"output": "1000000000"
},
{
"input": "10 15 69 195610342",
"output": "6520342"
},
{
"input": "2 1 1 1000000000",
"output": "500000000"
},
{
"input": "1000 1000 1 20",
"output": "0"
},
{
"input": "1 1 1 2000000000",
"output": "2000000000"
},
{
"input": "1 2 1 2000000000",
"output": "1000000000"
},
{
"input": "2 1 1 2000000000",
"output": "1000000000"
},
{
"input": "2 3 1 1000000000",
"output": "166666666"
},
{
"input": "2 3 1 2000000000",
"output": "333333333"
},
{
"input": "3 7 1 1000000000",
"output": "47619047"
},
{
"input": "1 1 1000000000 2000000000",
"output": "1000000001"
},
{
"input": "2 2 1 2000000000",
"output": "1000000000"
},
{
"input": "1 1 2 2000000000",
"output": "1999999999"
},
{
"input": "3 2 1 2000000000",
"output": "333333333"
},
{
"input": "1 1 2000000000 2000000000",
"output": "1"
},
{
"input": "2 3 7 7",
"output": "0"
},
{
"input": "3 3 3 7",
"output": "2"
}
] | 1,478,790,868
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 124
| 0
|
import math
if __name__ == '__main__':
x, y, a, b = str(input()).split()
x, y, a, b = int(x), int(y), int(a), int(b)
n = x * y // math.gcd(x, y)
print(1 + b // n - (a + n - 1) // n)
|
Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
|
```python
import math
if __name__ == '__main__':
x, y, a, b = str(input()).split()
x, y, a, b = int(x), int(y), int(a), int(b)
n = x * y // math.gcd(x, y)
print(1 + b // n - (a + n - 1) // n)
```
| 3
|
|
13
|
A
|
Numbers
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Numbers
|
1
|
64
|
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
|
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
|
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
|
[
"5\n",
"3\n"
] |
[
"7/3\n",
"2/1\n"
] |
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
| 0
|
[
{
"input": "5",
"output": "7/3"
},
{
"input": "3",
"output": "2/1"
},
{
"input": "1000",
"output": "90132/499"
},
{
"input": "927",
"output": "155449/925"
},
{
"input": "260",
"output": "6265/129"
},
{
"input": "131",
"output": "3370/129"
},
{
"input": "386",
"output": "857/12"
},
{
"input": "277",
"output": "2864/55"
},
{
"input": "766",
"output": "53217/382"
},
{
"input": "28",
"output": "85/13"
},
{
"input": "406",
"output": "7560/101"
},
{
"input": "757",
"output": "103847/755"
},
{
"input": "6",
"output": "9/4"
},
{
"input": "239",
"output": "10885/237"
},
{
"input": "322",
"output": "2399/40"
},
{
"input": "98",
"output": "317/16"
},
{
"input": "208",
"output": "4063/103"
},
{
"input": "786",
"output": "55777/392"
},
{
"input": "879",
"output": "140290/877"
},
{
"input": "702",
"output": "89217/700"
},
{
"input": "948",
"output": "7369/43"
},
{
"input": "537",
"output": "52753/535"
},
{
"input": "984",
"output": "174589/982"
},
{
"input": "934",
"output": "157951/932"
},
{
"input": "726",
"output": "95491/724"
},
{
"input": "127",
"output": "3154/125"
},
{
"input": "504",
"output": "23086/251"
},
{
"input": "125",
"output": "3080/123"
},
{
"input": "604",
"output": "33178/301"
},
{
"input": "115",
"output": "2600/113"
},
{
"input": "27",
"output": "167/25"
},
{
"input": "687",
"output": "85854/685"
},
{
"input": "880",
"output": "69915/439"
},
{
"input": "173",
"output": "640/19"
},
{
"input": "264",
"output": "6438/131"
},
{
"input": "785",
"output": "111560/783"
},
{
"input": "399",
"output": "29399/397"
},
{
"input": "514",
"output": "6031/64"
},
{
"input": "381",
"output": "26717/379"
},
{
"input": "592",
"output": "63769/590"
},
{
"input": "417",
"output": "32002/415"
},
{
"input": "588",
"output": "62723/586"
},
{
"input": "852",
"output": "131069/850"
},
{
"input": "959",
"output": "5059/29"
},
{
"input": "841",
"output": "127737/839"
},
{
"input": "733",
"output": "97598/731"
},
{
"input": "692",
"output": "87017/690"
},
{
"input": "69",
"output": "983/67"
},
{
"input": "223",
"output": "556/13"
},
{
"input": "93",
"output": "246/13"
},
{
"input": "643",
"output": "75503/641"
},
{
"input": "119",
"output": "2833/117"
},
{
"input": "498",
"output": "1459/16"
},
{
"input": "155",
"output": "4637/153"
},
{
"input": "305",
"output": "17350/303"
},
{
"input": "454",
"output": "37893/452"
},
{
"input": "88",
"output": "1529/86"
},
{
"input": "850",
"output": "32645/212"
},
{
"input": "474",
"output": "20581/236"
},
{
"input": "309",
"output": "17731/307"
},
{
"input": "762",
"output": "105083/760"
},
{
"input": "591",
"output": "63761/589"
},
{
"input": "457",
"output": "38317/455"
},
{
"input": "141",
"output": "3832/139"
},
{
"input": "385",
"output": "27232/383"
},
{
"input": "387",
"output": "27628/385"
},
{
"input": "469",
"output": "40306/467"
},
{
"input": "624",
"output": "35285/311"
},
{
"input": "330",
"output": "487/8"
},
{
"input": "31",
"output": "222/29"
},
{
"input": "975",
"output": "171679/973"
},
{
"input": "584",
"output": "62183/582"
},
{
"input": "668",
"output": "81127/666"
},
{
"input": "331",
"output": "20297/329"
},
{
"input": "189",
"output": "6789/187"
},
{
"input": "251",
"output": "11939/249"
},
{
"input": "876",
"output": "69196/437"
},
{
"input": "615",
"output": "68987/613"
},
{
"input": "451",
"output": "37258/449"
},
{
"input": "499",
"output": "45727/497"
},
{
"input": "699",
"output": "89117/697"
},
{
"input": "619",
"output": "70019/617"
},
{
"input": "413",
"output": "10515/137"
},
{
"input": "197",
"output": "7399/195"
},
{
"input": "794",
"output": "14281/99"
},
{
"input": "659",
"output": "79403/657"
},
{
"input": "653",
"output": "77695/651"
},
{
"input": "23",
"output": "45/7"
},
{
"input": "430",
"output": "16985/214"
},
{
"input": "249",
"output": "11659/247"
},
{
"input": "837",
"output": "126869/835"
},
{
"input": "258",
"output": "12373/256"
},
{
"input": "995",
"output": "59665/331"
},
{
"input": "102",
"output": "504/25"
},
{
"input": "989",
"output": "177124/987"
},
{
"input": "376",
"output": "13008/187"
},
{
"input": "657",
"output": "15715/131"
},
{
"input": "746",
"output": "50509/372"
},
{
"input": "602",
"output": "13177/120"
}
] | 1,603,697,691
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 186
| 0
|
number = input()
number = int(number)
all_bases = [i for i in range(int(number)) if i > 1]
nums_list = []
for base in all_bases:
remainder = number
while True:
nums_list.append(remainder%base)
remainder = remainder//base
if remainder == remainder%base:
nums_list.append(remainder%base)
break
print(str(sum(nums_list))+str("/")+str(len(all_bases)))
|
Title: Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input Specification:
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
Output Specification:
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
Demo Input:
['5\n', '3\n']
Demo Output:
['7/3\n', '2/1\n']
Note:
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
|
```python
number = input()
number = int(number)
all_bases = [i for i in range(int(number)) if i > 1]
nums_list = []
for base in all_bases:
remainder = number
while True:
nums_list.append(remainder%base)
remainder = remainder//base
if remainder == remainder%base:
nums_list.append(remainder%base)
break
print(str(sum(nums_list))+str("/")+str(len(all_bases)))
```
| 0
|
143
|
B
|
Help Kingdom of Far Far Away 2
|
PROGRAMMING
| 1,200
|
[
"implementation",
"strings"
] | null | null |
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
- A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point). - To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678 - In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests). - When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets. - Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
|
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
- The number's notation only contains characters from the set {"0" – "9", "-", "."}. - The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits - A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0"). - The minus sign (if it is present) is unique and stands in the very beginning of the number's notation - If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign. - The input data contains no spaces. - The number's notation contains at least one decimal digit.
|
Print the number given in the input in the financial format by the rules described in the problem statement.
|
[
"2012\n",
"0.000\n",
"-0.00987654321\n",
"-12345678.9\n"
] |
[
"$2,012.00",
"$0.00",
"($0.00)",
"($12,345,678.90)"
] |
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
| 1,000
|
[
{
"input": "2012",
"output": "$2,012.00"
},
{
"input": "0.000",
"output": "$0.00"
},
{
"input": "-0.00987654321",
"output": "($0.00)"
},
{
"input": "-12345678.9",
"output": "($12,345,678.90)"
},
{
"input": "0.99999999999999999999",
"output": "$0.99"
},
{
"input": "-999999999.9999999999",
"output": "($999,999,999.99)"
},
{
"input": "4.30",
"output": "$4.30"
},
{
"input": "-3136",
"output": "($3,136.00)"
},
{
"input": "47.849",
"output": "$47.84"
},
{
"input": "0",
"output": "$0.00"
},
{
"input": "-1",
"output": "($1.00)"
},
{
"input": "5.3944",
"output": "$5.39"
},
{
"input": "-359789",
"output": "($359,789.00)"
},
{
"input": "-999999",
"output": "($999,999.00)"
},
{
"input": "50117.75",
"output": "$50,117.75"
},
{
"input": "-2717.859",
"output": "($2,717.85)"
},
{
"input": "446900763",
"output": "$446,900,763.00"
},
{
"input": "-92.04295",
"output": "($92.04)"
},
{
"input": "1000000000",
"output": "$1,000,000,000.00"
},
{
"input": "-4097961.5",
"output": "($4,097,961.50)"
},
{
"input": "-83348637.91",
"output": "($83,348,637.91)"
},
{
"input": "741968647.01",
"output": "$741,968,647.01"
},
{
"input": "8590210736.2",
"output": "$8,590,210,736.20"
},
{
"input": "-337322633.10",
"output": "($337,322,633.10)"
},
{
"input": "-9389724657.706",
"output": "($9,389,724,657.70)"
},
{
"input": "-337807291537795",
"output": "($337,807,291,537,795.00)"
},
{
"input": "-1000000000000000",
"output": "($1,000,000,000,000,000.00)"
},
{
"input": "1000000000000000000",
"output": "$1,000,000,000,000,000,000.00"
},
{
"input": "64852365412711705.4",
"output": "$64,852,365,412,711,705.40"
},
{
"input": "-14193044875680849641.0",
"output": "($14,193,044,875,680,849,641.00)"
},
{
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"output": "($9,087,207,850,675,188,568.44)"
},
{
"input": "-999999999999999999999999",
"output": "($999,999,999,999,999,999,999,999.00)"
},
{
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{
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"output": "($9,946,519,009,668,593,136,622,791,780,335,166,786,329.96)"
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"output": "($39,243,277,445,578,948,100,023,610,303,161,362.21)"
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{
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"output": "($999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999.00)"
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},
{
"input": "2089113443991831781611590658416581830404242017.85102926202385542583311855337073083712400492547136479",
"output": "$2,089,113,443,991,831,781,611,590,658,416,581,830,404,242,017.85"
},
{
"input": "-93446155923266881322196606839694485100712773936897171033382798807975023881552872455711005123932.747",
"output": "($93,446,155,923,266,881,322,196,606,839,694,485,100,712,773,936,897,171,033,382,798,807,975,023,881,552,872,455,711,005,123,932.74)"
},
{
"input": "960516596871944593730108478032758053821336372808735358607440437077013969634756697387966042842288.508",
"output": "$960,516,596,871,944,593,730,108,478,032,758,053,821,336,372,808,735,358,607,440,437,077,013,969,634,756,697,387,966,042,842,288.50"
},
{
"input": "7542946645993289345871768107036410651745989844030221776852993379463784193885567707317993804499615689",
"output": "$7,542,946,645,993,289,345,871,768,107,036,410,651,745,989,844,030,221,776,852,993,379,463,784,193,885,567,707,317,993,804,499,615,689.00"
},
{
"input": "-62833497045916718064314002220718776776624697240820362462669558147156815011509869423334004968891.075",
"output": "($62,833,497,045,916,718,064,314,002,220,718,776,776,624,697,240,820,362,462,669,558,147,156,815,011,509,869,423,334,004,968,891.07)"
},
{
"input": "369983878656471317107141313973936685655559201630341263457253892446495.822347697919107135036916507458",
"output": "$369,983,878,656,471,317,107,141,313,973,936,685,655,559,201,630,341,263,457,253,892,446,495.82"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "$1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.00"
},
{
"input": "-7200722479435658295856375503813639375609209638447823589904775057990210002452424572601761458228411.3",
"output": "($7,200,722,479,435,658,295,856,375,503,813,639,375,609,209,638,447,823,589,904,775,057,990,210,002,452,424,572,601,761,458,228,411.30)"
},
{
"input": "1.62929379626674077244098830537592273171157251593607257308766051098303017164327540412154291842807913",
"output": "$1.62"
},
{
"input": "9094697811219913240397316094992038813655777565859532452.35345453828434088557646454113264025096745262",
"output": "$9,094,697,811,219,913,240,397,316,094,992,038,813,655,777,565,859,532,452.35"
},
{
"input": "-241995182456075514870952227695034085165209475359259147742565065759917424411707290789641890279251.11",
"output": "($241,995,182,456,075,514,870,952,227,695,034,085,165,209,475,359,259,147,742,565,065,759,917,424,411,707,290,789,641,890,279,251.11)"
},
{
"input": "2567340036354357844391998756110821468858185018763415770617907336824217629234299240638243305079104961",
"output": "$2,567,340,036,354,357,844,391,998,756,110,821,468,858,185,018,763,415,770,617,907,336,824,217,629,234,299,240,638,243,305,079,104,961.00"
}
] | 1,660,823,556
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 124
| 2,867,200
|
import collections
import heapq
import sys
import math
import itertools
import bisect
from io import BytesIO, IOBase
import os
# sys.setrecursionlimit(10000)
def values(): return tuple(map(int, sys.stdin.readline().split()))
def inlst(): return [int(i) for i in sys.stdin.readline().split()]
def inp(): return int(sys.stdin.readline())
def inpstr(): return sys.stdin.readline().strip()
def words(): return [i for i in sys.stdin.readline().split()]
def chars(): return list(sys.stdin.readline().strip())
def solve():
c=inpstr()
n=len(c)
y=1
if c[0] == '-':
y =(float(c[1:]) * -1)
c=c[1:]
if '.' in c :
c = list(c)
ind=c.index('.')
c=c[:ind+3]
c="".join(c)
i=ind-3
c=list(c)
while i>0:
c.insert(i,',')
i-=3
c="".join(c)
if y<0:
m='('+'$'+c+')'
else:
m='$'+c
print(m)
'''
-12357444245.326585
'''
if __name__ == "__main__":
# sys.stdin = open('/home/mohamed/Documents/w.txt','r')
# for i in range(inp()):
solve()
# l=[1,2,3,4]
# l2=[3,5,8]
# print(set(l)&set(l2))
# print(inter)
|
Title: Help Kingdom of Far Far Away 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
- A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point). - To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678 - In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests). - When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets. - Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input Specification:
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
- The number's notation only contains characters from the set {"0" – "9", "-", "."}. - The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits - A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0"). - The minus sign (if it is present) is unique and stands in the very beginning of the number's notation - If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign. - The input data contains no spaces. - The number's notation contains at least one decimal digit.
Output Specification:
Print the number given in the input in the financial format by the rules described in the problem statement.
Demo Input:
['2012\n', '0.000\n', '-0.00987654321\n', '-12345678.9\n']
Demo Output:
['$2,012.00', '$0.00', '($0.00)', '($12,345,678.90)']
Note:
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
|
```python
import collections
import heapq
import sys
import math
import itertools
import bisect
from io import BytesIO, IOBase
import os
# sys.setrecursionlimit(10000)
def values(): return tuple(map(int, sys.stdin.readline().split()))
def inlst(): return [int(i) for i in sys.stdin.readline().split()]
def inp(): return int(sys.stdin.readline())
def inpstr(): return sys.stdin.readline().strip()
def words(): return [i for i in sys.stdin.readline().split()]
def chars(): return list(sys.stdin.readline().strip())
def solve():
c=inpstr()
n=len(c)
y=1
if c[0] == '-':
y =(float(c[1:]) * -1)
c=c[1:]
if '.' in c :
c = list(c)
ind=c.index('.')
c=c[:ind+3]
c="".join(c)
i=ind-3
c=list(c)
while i>0:
c.insert(i,',')
i-=3
c="".join(c)
if y<0:
m='('+'$'+c+')'
else:
m='$'+c
print(m)
'''
-12357444245.326585
'''
if __name__ == "__main__":
# sys.stdin = open('/home/mohamed/Documents/w.txt','r')
# for i in range(inp()):
solve()
# l=[1,2,3,4]
# l2=[3,5,8]
# print(set(l)&set(l2))
# print(inter)
```
| -1
|
|
299
|
A
|
Ksusha and Array
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory",
"sortings"
] | null | null |
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
|
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
|
[
"3\n2 2 4\n",
"5\n2 1 3 1 6\n",
"3\n2 3 5\n"
] |
[
"2\n",
"1\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "3\n2 2 4",
"output": "2"
},
{
"input": "5\n2 1 3 1 6",
"output": "1"
},
{
"input": "3\n2 3 5",
"output": "-1"
},
{
"input": "1\n331358794",
"output": "331358794"
},
{
"input": "5\n506904227 214303304 136194869 838256937 183952885",
"output": "-1"
},
{
"input": "2\n500000000 1000000000",
"output": "500000000"
},
{
"input": "2\n4 6",
"output": "-1"
},
{
"input": "5\n10 8 6 4 2",
"output": "2"
},
{
"input": "2\n6 10",
"output": "-1"
},
{
"input": "1\n1000000000",
"output": "1000000000"
},
{
"input": "2\n6 8",
"output": "-1"
},
{
"input": "5\n2 2 2 2 1000000000",
"output": "2"
},
{
"input": "2\n6 4",
"output": "-1"
}
] | 1,518,379,204
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 5,632,000
|
from sys import stdin
def read(): return map(int, stdin.readline().split())
read()
a = list(read())
x = max(a)
for y in a:
if x % y != 0:
x = -1
break
print(x)
|
Title: Ksusha and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksusha is a beginner coder. Today she starts studying arrays. She has array *a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* positive integers.
Her university teacher gave her a task. Find such number in the array, that all array elements are divisible by it. Help her and find the number!
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105), showing how many numbers the array has. The next line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the array elements.
Output Specification:
Print a single integer — the number from the array, such that all array elements are divisible by it. If such number doesn't exist, print -1.
If there are multiple answers, you are allowed to print any of them.
Demo Input:
['3\n2 2 4\n', '5\n2 1 3 1 6\n', '3\n2 3 5\n']
Demo Output:
['2\n', '1\n', '-1\n']
Note:
none
|
```python
from sys import stdin
def read(): return map(int, stdin.readline().split())
read()
a = list(read())
x = max(a)
for y in a:
if x % y != 0:
x = -1
break
print(x)
```
| 0
|
|
514
|
A
|
Chewbaсca and Number
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
|
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
|
[
"27\n",
"4545\n"
] |
[
"22\n",
"4444\n"
] |
none
| 500
|
[
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
"output": "21223442"
},
{
"input": "91730629",
"output": "91230320"
},
{
"input": "420062703497",
"output": "420032203402"
},
{
"input": "332711047202",
"output": "332211042202"
},
{
"input": "3395184971407775",
"output": "3304114021402224"
},
{
"input": "8464062628894325",
"output": "1434032321104324"
},
{
"input": "164324828731963982",
"output": "134324121231033012"
},
{
"input": "384979173822804784",
"output": "314020123122104214"
},
{
"input": "41312150450968417",
"output": "41312140440031412"
},
{
"input": "2156",
"output": "2143"
},
{
"input": "1932",
"output": "1032"
},
{
"input": "5902",
"output": "4002"
},
{
"input": "5728",
"output": "4221"
},
{
"input": "8537",
"output": "1432"
},
{
"input": "55403857",
"output": "44403142"
},
{
"input": "270739",
"output": "220230"
},
{
"input": "28746918",
"output": "21243011"
},
{
"input": "10279211",
"output": "10220211"
},
{
"input": "40289679",
"output": "40210320"
},
{
"input": "545203238506",
"output": "444203231403"
},
{
"input": "461117063340",
"output": "431112033340"
},
{
"input": "658492686568",
"output": "341402313431"
},
{
"input": "857373361868",
"output": "142323331131"
},
{
"input": "429325660016",
"output": "420324330013"
},
{
"input": "9894448650287940",
"output": "9104441340212040"
},
{
"input": "6354510839296263",
"output": "3344410130203233"
},
{
"input": "6873575462224593",
"output": "3123424432224403"
},
{
"input": "4237951492601449",
"output": "4232041402301440"
},
{
"input": "2680352384836991",
"output": "2310342314133001"
},
{
"input": "606187734191890310",
"output": "303112234101100310"
},
{
"input": "351499943576823355",
"output": "341400043423123344"
},
{
"input": "180593481782177068",
"output": "110403411212122031"
},
{
"input": "999999999999999999",
"output": "900000000000000000"
},
{
"input": "1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "9999",
"output": "9000"
},
{
"input": "99",
"output": "90"
},
{
"input": "9991",
"output": "9001"
}
] | 1,677,145,774
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 31
| 0
|
s=list(input())
# print(s)
if(len(s)==1 and s[0]=='9'):
print(9)
else:
for i in range(len(s)):
s[i]=str(min(int(s[i]),9-int(s[i])))
for i in range(len(s)):
print(s[i],end='')
|
Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none
|
```python
s=list(input())
# print(s)
if(len(s)==1 and s[0]=='9'):
print(9)
else:
for i in range(len(s)):
s[i]=str(min(int(s[i]),9-int(s[i])))
for i in range(len(s)):
print(s[i],end='')
```
| 0
|
|
460
|
A
|
Vasya and Socks
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
|
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
|
Print a single integer — the answer to the problem.
|
[
"2 2\n",
"9 3\n"
] |
[
"3\n",
"13\n"
] |
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
| 500
|
[
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"output": "19"
},
{
"input": "10 9",
"output": "11"
},
{
"input": "100 100",
"output": "101"
},
{
"input": "2 27",
"output": "2"
},
{
"input": "99 100",
"output": "99"
},
{
"input": "99 2",
"output": "197"
},
{
"input": "100 3",
"output": "149"
},
{
"input": "98 3",
"output": "146"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 2",
"output": "199"
},
{
"input": "62 4",
"output": "82"
},
{
"input": "99 10",
"output": "109"
},
{
"input": "100 5",
"output": "124"
},
{
"input": "80 80",
"output": "81"
},
{
"input": "95 16",
"output": "101"
},
{
"input": "75 16",
"output": "79"
},
{
"input": "99 74",
"output": "100"
},
{
"input": "20 21",
"output": "20"
},
{
"input": "52 96",
"output": "52"
},
{
"input": "24 5",
"output": "29"
}
] | 1,696,004,752
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 31
| 0
|
n, m = map(int, input().split())
days = 0
socks = n
for day in range(1, 101):
if socks < day:
break
if day % m == 0:
socks += 1
days += 1
print(days)
|
Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
|
```python
n, m = map(int, input().split())
days = 0
socks = n
for day in range(1, 101):
if socks < day:
break
if day % m == 0:
socks += 1
days += 1
print(days)
```
| 0
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
|
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,683,744,035
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 62
| 0
|
import sys
b, r = [int(x) for x in sys.stdin.readline().split()]
max = b if b >= r else r
min = b if b <= r else r
print(str(min) + ' ' + str((max - min) // 2))
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
import sys
b, r = [int(x) for x in sys.stdin.readline().split()]
max = b if b >= r else r
min = b if b <= r else r
print(str(min) + ' ' + str((max - min) // 2))
```
| 3
|
|
725
|
A
|
Jumping Ball
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
|
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
|
[
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] |
[
"2",
"5",
"0"
] |
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
| 500
|
[
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input": "1\n<",
"output": "1"
},
{
"input": "2\n<>",
"output": "2"
},
{
"input": "3\n<>>",
"output": "3"
},
{
"input": "3\n><>",
"output": "1"
},
{
"input": "2\n><",
"output": "0"
},
{
"input": "2\n>>",
"output": "2"
},
{
"input": "2\n<<",
"output": "2"
},
{
"input": "1\n>",
"output": "1"
},
{
"input": "3\n>><",
"output": "0"
},
{
"input": "3\n>>>",
"output": "3"
},
{
"input": "3\n<><",
"output": "1"
},
{
"input": "10\n<<<><<<>>>",
"output": "6"
},
{
"input": "20\n><><<><<<>>>>>>>>>>>",
"output": "11"
},
{
"input": "20\n<<<<<<<<<<><<<<>>>>>",
"output": "15"
},
{
"input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "49"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<",
"output": "100"
},
{
"input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "100"
},
{
"input": "12\n<<>><<>><<>>",
"output": "4"
},
{
"input": "6\n<<><>>",
"output": "4"
},
{
"input": "6\n><>>>>",
"output": "4"
},
{
"input": "8\n>>>><<<>",
"output": "1"
},
{
"input": "4\n<><>",
"output": "2"
},
{
"input": "4\n><><",
"output": "0"
},
{
"input": "7\n<<>>><>",
"output": "3"
},
{
"input": "10\n><><>>>>>>",
"output": "6"
},
{
"input": "5\n<><>>",
"output": "3"
},
{
"input": "12\n<><<<<>>>>>>",
"output": "7"
},
{
"input": "6\n<>><<>",
"output": "2"
},
{
"input": "6\n>>><>>",
"output": "2"
},
{
"input": "10\n><><>>>><>",
"output": "1"
},
{
"input": "5\n><>>>",
"output": "3"
},
{
"input": "5\n<<><>",
"output": "3"
},
{
"input": "5\n<><<<",
"output": "1"
},
{
"input": "4\n<><<",
"output": "1"
},
{
"input": "8\n<<>><<>>",
"output": "4"
},
{
"input": "7\n<<><>>>",
"output": "5"
},
{
"input": "5\n><<>>",
"output": "2"
},
{
"input": "10\n<<<<<>>>>>",
"output": "10"
},
{
"input": "6\n><<<<<",
"output": "0"
},
{
"input": "8\n<<><><>>",
"output": "4"
},
{
"input": "10\n<<<<><<<><",
"output": "4"
},
{
"input": "12\n<<<>>>><<>>>",
"output": "6"
},
{
"input": "4\n><>>",
"output": "2"
},
{
"input": "11\n<<><<>><<>>",
"output": "4"
}
] | 1,492,256,473
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 5,222,400
|
'''input
4
>><<
'''
n = int(input())
s = input()
t = "<"
for x in range(n):
if t != s[x]:
break
else:
x += 1
s = s[::-1]
t = ">"
for y in range(n):
if t != s[x]:
break
else:
x += 1
print(x + y)
|
Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
|
```python
'''input
4
>><<
'''
n = int(input())
s = input()
t = "<"
for x in range(n):
if t != s[x]:
break
else:
x += 1
s = s[::-1]
t = ">"
for y in range(n):
if t != s[x]:
break
else:
x += 1
print(x + y)
```
| 0
|
|
471
|
B
|
MUH and Important Things
|
PROGRAMMING
| 1,300
|
[
"implementation",
"sortings"
] | null | null |
It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are *n* tasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.
Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the *n* tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of tasks. The second line contains *n* integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=2000), where *h**i* is the difficulty of the *i*-th task. The larger number *h**i* is, the more difficult the *i*-th task is.
|
In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line *n* distinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.
If there are multiple possible answers, you can print any of them.
|
[
"4\n1 3 3 1\n",
"5\n2 4 1 4 8\n"
] |
[
"YES\n1 4 2 3 \n4 1 2 3 \n4 1 3 2 \n",
"NO"
] |
In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.
In the second sample there are only two sequences of tasks that meet the conditions — [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.
| 1,000
|
[
{
"input": "4\n1 3 3 1",
"output": "YES\n1 4 2 3 \n4 1 2 3 \n4 1 3 2 "
},
{
"input": "5\n2 4 1 4 8",
"output": "NO"
},
{
"input": "8\n1 5 4 12 7 2 10 11",
"output": "NO"
},
{
"input": "6\n5 1 2 5 2 4",
"output": "YES\n2 3 5 6 1 4 \n2 5 3 6 1 4 \n2 5 3 6 4 1 "
},
{
"input": "1\n1083",
"output": "NO"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5",
"output": "YES\n1 2 3 4 5 6 7 8 9 10 \n2 1 3 4 5 6 7 8 9 10 \n2 3 1 4 5 6 7 8 9 10 "
},
{
"input": "6\n3 8 3 9 3 10",
"output": "YES\n1 3 5 2 4 6 \n3 1 5 2 4 6 \n3 5 1 2 4 6 "
},
{
"input": "19\n895 1302 724 952 340 952 939 1302 724 952 939 340 340 1844 770 976 435 1302 1302",
"output": "YES\n5 12 13 17 3 9 15 1 7 11 4 6 10 16 2 8 18 19 14 \n12 5 13 17 3 9 15 1 7 11 4 6 10 16 2 8 18 19 14 \n12 13 5 17 3 9 15 1 7 11 4 6 10 16 2 8 18 19 14 "
},
{
"input": "7\n766 766 1477 766 107 1774 990",
"output": "YES\n5 1 2 4 7 3 6 \n5 2 1 4 7 3 6 \n5 2 4 1 7 3 6 "
},
{
"input": "11\n1552 1010 1552 1248 1550 388 1541 1010 613 1821 388",
"output": "YES\n6 11 9 2 8 4 7 5 1 3 10 \n11 6 9 2 8 4 7 5 1 3 10 \n11 6 9 8 2 4 7 5 1 3 10 "
},
{
"input": "15\n688 848 1462 688 12 1336 1336 1113 1462 1074 659 1384 12 12 1074",
"output": "YES\n5 13 14 11 1 4 2 10 15 8 6 7 12 3 9 \n13 5 14 11 1 4 2 10 15 8 6 7 12 3 9 \n13 14 5 11 1 4 2 10 15 8 6 7 12 3 9 "
},
{
"input": "19\n65 117 159 402 117 402 65 1016 1850 1265 854 159 347 1501 117 805 854 117 1265",
"output": "YES\n1 7 2 5 15 18 3 12 13 4 6 16 11 17 8 10 19 14 9 \n7 1 2 5 15 18 3 12 13 4 6 16 11 17 8 10 19 14 9 \n7 1 5 2 15 18 3 12 13 4 6 16 11 17 8 10 19 14 9 "
},
{
"input": "3\n1282 101 420",
"output": "NO"
},
{
"input": "3\n1 1 1",
"output": "YES\n1 2 3 \n2 1 3 \n2 3 1 "
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1",
"output": "NO"
}
] | 1,587,196,900
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
n=int(input())
l=list(map(int,input().split()))
s=[]
for i in range(n):
x=[l[i],i]
s.append(x)
s.sort()
flag=0
sw=[]
ans=[]
for i in range(n):
x=s[i][1]
ans.append(x)
for i in range(n-1):
if s[i][0]==s[i+1][0]:
sw.append((i,i+1))
if len(sw)==2:
print('YES')
flag=1
break
if flag==1:
for i in range(3):
for j in range(n):
print(ans[j]+1,end=' ')
print()
for x,y in sw:
ans[x],ans[y]=ans[y],ans[x]
else:
print('NO')
|
Title: MUH and Important Things
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are *n* tasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.
Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the *n* tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of tasks. The second line contains *n* integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=2000), where *h**i* is the difficulty of the *i*-th task. The larger number *h**i* is, the more difficult the *i*-th task is.
Output Specification:
In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line *n* distinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.
If there are multiple possible answers, you can print any of them.
Demo Input:
['4\n1 3 3 1\n', '5\n2 4 1 4 8\n']
Demo Output:
['YES\n1 4 2 3 \n4 1 2 3 \n4 1 3 2 \n', 'NO']
Note:
In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.
In the second sample there are only two sequences of tasks that meet the conditions — [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.
|
```python
n=int(input())
l=list(map(int,input().split()))
s=[]
for i in range(n):
x=[l[i],i]
s.append(x)
s.sort()
flag=0
sw=[]
ans=[]
for i in range(n):
x=s[i][1]
ans.append(x)
for i in range(n-1):
if s[i][0]==s[i+1][0]:
sw.append((i,i+1))
if len(sw)==2:
print('YES')
flag=1
break
if flag==1:
for i in range(3):
for j in range(n):
print(ans[j]+1,end=' ')
print()
for x,y in sw:
ans[x],ans[y]=ans[y],ans[x]
else:
print('NO')
```
| 0
|
|
96
|
A
|
Football
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Football
|
2
|
256
|
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
|
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
|
Print "YES" if the situation is dangerous. Otherwise, print "NO".
|
[
"001001\n",
"1000000001\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,691,503,696
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 44
| 122
| 0
|
t = input()
count1 = 0
count0 = 0
y = "NO"
for x in t:
if x == "1":
count1 += 1
count0 = 0
elif x == "0":
count0 += 1
count1 = 0
if count0 >= 7 or count1 >= 7:
y = "YES"
break
print(y)
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
t = input()
count1 = 0
count0 = 0
y = "NO"
for x in t:
if x == "1":
count1 += 1
count0 = 0
elif x == "0":
count0 += 1
count1 = 0
if count0 >= 7 or count1 >= 7:
y = "YES"
break
print(y)
```
| 3.9695
|
435
|
A
|
Queue on Bus Stop
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
|
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
|
[
"4 3\n2 3 2 1\n",
"3 4\n1 2 1\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "4 3\n2 3 2 1",
"output": "3"
},
{
"input": "3 4\n1 2 1",
"output": "1"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "6 4\n1 3 2 3 4 1",
"output": "5"
},
{
"input": "6 8\n6 1 1 1 4 5",
"output": "3"
},
{
"input": "10 10\n1 10 1 10 1 1 7 8 6 7",
"output": "8"
},
{
"input": "100 100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "63"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "10 2\n2 2 1 1 1 1 1 2 1 2",
"output": "8"
},
{
"input": "10 3\n1 3 1 1 3 2 2 2 3 3",
"output": "9"
},
{
"input": "10 4\n2 1 1 1 3 4 4 4 1 2",
"output": "6"
},
{
"input": "10 5\n2 2 3 4 4 1 5 3 1 2",
"output": "7"
},
{
"input": "100 3\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "83"
},
{
"input": "100 7\n4 7 4 7 7 4 7 3 5 6 3 5 4 3 7 2 7 2 4 1 6 3 3 7 4 4 5 4 3 6 4 3 2 2 1 4 4 1 7 3 7 7 1 3 1 5 4 1 5 3 5 2 2 1 5 5 1 5 2 7 5 5 1 5 5 4 6 5 1 3 5 6 7 4 1 3 3 4 3 2 7 6 5 7 2 7 1 1 2 2 3 1 3 7 1 3 2 1 1 7",
"output": "71"
},
{
"input": "100 10\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "64"
},
{
"input": "100 15\n3 12 8 3 11 14 12 14 1 11 13 3 5 13 4 14 2 11 7 8 12 9 15 7 15 1 4 11 6 12 1 3 8 13 1 8 14 4 3 14 1 3 1 6 10 15 13 11 12 1 14 13 11 14 11 3 12 7 3 15 14 4 5 6 5 14 7 14 6 2 6 12 6 13 13 1 9 13 15 11 6 3 15 11 9 4 15 8 15 12 1 15 10 10 4 1 15 1 4 1",
"output": "71"
},
{
"input": "100 30\n7 14 22 16 11 13 7 29 20 19 22 6 12 16 1 8 27 21 22 3 15 27 20 12 4 19 1 26 26 22 25 17 29 25 16 29 29 28 16 26 25 14 16 20 5 21 5 15 19 13 17 21 17 19 23 13 1 25 6 30 16 19 12 10 28 8 15 13 14 24 19 30 12 19 22 1 3 14 16 3 20 26 15 19 9 10 19 27 2 16 10 22 15 13 19 3 24 9 8 13",
"output": "71"
},
{
"input": "100 40\n39 19 13 36 11 21 32 12 1 2 39 26 32 39 24 1 4 19 10 4 16 39 32 34 13 24 30 35 3 10 8 18 13 12 39 27 31 40 37 20 17 17 37 5 10 12 22 17 7 1 31 13 11 10 2 6 22 16 2 4 9 27 6 35 22 16 22 30 33 2 26 20 35 19 40 37 19 17 21 28 37 28 40 4 5 4 35 19 26 36 19 12 21 20 21 30 9 16 9 32",
"output": "65"
},
{
"input": "100 50\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "60"
},
{
"input": "100 60\n34 21 39 17 48 46 23 56 46 52 50 39 55 48 54 38 32 38 24 26 44 12 28 9 25 26 10 52 42 60 41 3 16 60 44 29 27 55 19 19 19 57 45 59 29 35 5 14 50 47 57 48 16 7 12 36 58 31 37 58 30 50 19 11 10 41 59 57 49 41 33 9 12 11 53 50 60 51 21 9 44 23 1 16 4 15 17 57 15 17 46 50 18 52 43 24 47 50 19 18",
"output": "74"
},
{
"input": "100 90\n74 65 49 41 3 79 61 83 50 40 13 57 90 14 62 77 36 10 3 5 5 40 50 75 32 26 3 71 79 54 88 50 46 20 42 59 30 36 83 86 60 62 82 68 62 80 18 65 28 28 81 74 62 33 61 35 33 83 90 72 6 6 51 4 22 20 29 10 8 3 84 69 12 17 24 16 12 64 80 74 68 59 1 59 15 59 37 58 79 83 51 56 81 14 37 45 19 31 61 90",
"output": "67"
},
{
"input": "100 99\n69 46 76 47 71 9 66 46 78 17 96 83 56 96 29 3 43 48 79 23 93 61 19 9 29 72 15 84 93 46 71 87 11 43 96 44 54 75 3 66 2 95 46 32 69 52 79 38 57 53 37 60 71 82 28 31 84 58 89 40 62 74 22 50 45 38 99 67 24 28 28 12 69 88 33 10 31 71 46 7 42 81 54 81 96 44 8 1 20 24 28 19 54 35 69 32 71 13 66 15",
"output": "68"
},
{
"input": "90 100\n25 52 88 89 36 17 57 64 66 11 89 61 54 92 48 51 18 42 44 92 6 14 67 100 16 21 17 88 85 73 33 11 94 84 56 72 4 80 90 78 96 5 62 70 54 70 94 80 10 91 100 89 98 87 69 74 88 63 53 79 38 94 89 52 21 82 67 79 100 81 2 40 30 69 34 15 12 33 87 52 95 18 51 30 15 39 30 99 46 84",
"output": "67"
},
{
"input": "5 100\n14 67 15 28 21",
"output": "2"
},
{
"input": "10 100\n2 17 53 94 95 57 36 47 68 48",
"output": "7"
},
{
"input": "1 100\n18",
"output": "1"
},
{
"input": "100 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "30 100\n56 7 99 83 2 65 35 53 99 36 42 57 13 37 68 52 87 11 50 23 86 24 32 39 97 6 64 1 18 86",
"output": "18"
},
{
"input": "60 100\n18 75 43 88 45 43 20 59 59 79 62 39 53 21 28 46 54 53 97 81 18 15 2 95 84 9 36 70 30 76 17 19 83 40 45 32 31 70 23 14 44 35 79 84 97 96 99 60 3 73 64 83 6 12 67 86 70 89 18 61",
"output": "40"
},
{
"input": "1 73\n24",
"output": "1"
},
{
"input": "4 4\n1 4 4 4",
"output": "4"
},
{
"input": "2 6\n1 4",
"output": "1"
},
{
"input": "2 5\n5 5",
"output": "2"
}
] | 1,570,719,268
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
l=list(map(int,input().split(" ")))
n=l[0]
m=l[1]
l=list(map(int,input().split(" ")))
i=0
s=0
cnt=0
while(i<n):
print(s,cnt)
s=s+l[i]
if(s<m):
i+=1
elif(s==m):
s=0
cnt+=1
i+=1
else:
s=0
cnt+=1
print(cnt)
|
Title: Queue on Bus Stop
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
Output Specification:
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
Demo Input:
['4 3\n2 3 2 1\n', '3 4\n1 2 1\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
l=list(map(int,input().split(" ")))
n=l[0]
m=l[1]
l=list(map(int,input().split(" ")))
i=0
s=0
cnt=0
while(i<n):
print(s,cnt)
s=s+l[i]
if(s<m):
i+=1
elif(s==m):
s=0
cnt+=1
i+=1
else:
s=0
cnt+=1
print(cnt)
```
| 0
|
|
912
|
A
|
Tricky Alchemy
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
|
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
|
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
|
[
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] |
[
"2\n",
"1\n",
"2147483648\n"
] |
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
| 500
|
[
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output": "12"
},
{
"input": "3555165 6693472\n1499112 556941 3075290",
"output": "3089339"
},
{
"input": "0 0\n1000000000 1000000000 1000000000",
"output": "7000000000"
},
{
"input": "1 1\n0 1 0",
"output": "0"
},
{
"input": "117708228 562858833\n118004008 360437130 154015822",
"output": "738362681"
},
{
"input": "999998118 700178721\n822106746 82987112 547955384",
"output": "1753877029"
},
{
"input": "566568710 765371101\n60614022 80126928 809950465",
"output": "1744607222"
},
{
"input": "448858599 829062060\n764716760 97644201 203890025",
"output": "1178219122"
},
{
"input": "626115781 966381948\n395190569 820194184 229233367",
"output": "1525971878"
},
{
"input": "803372962 103701834\n394260597 837711458 623172928",
"output": "3426388098"
},
{
"input": "980630143 241021722\n24734406 928857659 312079781",
"output": "1624075280"
},
{
"input": "862920032 378341609\n360240924 241342224 337423122",
"output": "974174021"
},
{
"input": "40177212 515661496\n64343660 963892207 731362684",
"output": "3694721078"
},
{
"input": "217434393 579352456\n694817470 981409480 756706026",
"output": "4825785129"
},
{
"input": "394691574 716672343\n398920207 72555681 150645586",
"output": "475704521"
},
{
"input": "276981463 853992230\n29394015 90072954 839552440",
"output": "1754738044"
},
{
"input": "843552056 919184611\n341530221 423649259 101547519",
"output": "263157645"
},
{
"input": "20809236 56504497\n972004030 441166533 495487081",
"output": "4235488636"
},
{
"input": "198066417 825228166\n602477839 532312735 520830423",
"output": "2808777834"
},
{
"input": "80356306 962548053\n601547868 549830008 914769984",
"output": "4004161345"
},
{
"input": "257613487 394835231\n642087093 567347282 308709545",
"output": "2692548667"
},
{
"input": "139903376 532155119\n641157122 289897263 629020178",
"output": "3077110809"
},
{
"input": "612127849 669475006\n271630930 676010757 22959739",
"output": "682559736"
},
{
"input": "0 0\n0 0 0",
"output": "0"
},
{
"input": "1000000000 1000000000\n499999998 4 333333332",
"output": "0"
},
{
"input": "1000000000 1000000000\n1000000000 1000000000 1000000000",
"output": "5000000000"
},
{
"input": "4 3\n1 0 1",
"output": "0"
},
{
"input": "4 12\n1 2 3",
"output": "0"
},
{
"input": "4 20\n1 2 1",
"output": "0"
},
{
"input": "100 10\n2 3 4",
"output": "5"
},
{
"input": "6 0\n1 1 1",
"output": "4"
},
{
"input": "25 5\n3 3 3",
"output": "7"
},
{
"input": "48 27\n22 39 20",
"output": "107"
},
{
"input": "4 0\n1 1 1",
"output": "4"
}
] | 1,536,160,743
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 124
| 307,200
|
YC, BC = map(int, input().split())
YB, GB, BB = map(int, input().split())
YCR = 2 * YB + GB
BCR = GB + 3 * BB
print(max(0, YCR-YC) + max(0, BCR-BC))
|
Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
|
```python
YC, BC = map(int, input().split())
YB, GB, BB = map(int, input().split())
YCR = 2 * YB + GB
BCR = GB + 3 * BB
print(max(0, YCR-YC) + max(0, BCR-BC))
```
| 3
|
|
777
|
E
|
Hanoi Factory
|
PROGRAMMING
| 2,000
|
[
"brute force",
"data structures",
"dp",
"greedy",
"sortings"
] | null | null |
Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the rings for this wonderful game? Once upon a time, the ruler of the ancient Egypt ordered the workers of Hanoi Factory to create as high tower as possible. They were not ready to serve such a strange order so they had to create this new tower using already produced rings.
There are *n* rings in factory's stock. The *i*-th ring has inner radius *a**i*, outer radius *b**i* and height *h**i*. The goal is to select some subset of rings and arrange them such that the following conditions are satisfied:
- Outer radiuses form a non-increasing sequence, i.e. one can put the *j*-th ring on the *i*-th ring only if *b**j*<=≤<=*b**i*. - Rings should not fall one into the the other. That means one can place ring *j* on the ring *i* only if *b**j*<=><=*a**i*. - The total height of all rings used should be maximum possible.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of rings in factory's stock.
The *i*-th of the next *n* lines contains three integers *a**i*, *b**i* and *h**i* (1<=≤<=*a**i*,<=*b**i*,<=*h**i*<=≤<=109, *b**i*<=><=*a**i*) — inner radius, outer radius and the height of the *i*-th ring respectively.
|
Print one integer — the maximum height of the tower that can be obtained.
|
[
"3\n1 5 1\n2 6 2\n3 7 3\n",
"4\n1 2 1\n1 3 3\n4 6 2\n5 7 1\n"
] |
[
"6\n",
"4\n"
] |
In the first sample, the optimal solution is to take all the rings and put them on each other in order 3, 2, 1.
In the second sample, one can put the ring 3 on the ring 4 and get the tower of height 3, or put the ring 1 on the ring 2 and get the tower of height 4.
| 2,500
|
[
{
"input": "3\n1 5 1\n2 6 2\n3 7 3",
"output": "6"
},
{
"input": "4\n1 2 1\n1 3 3\n4 6 2\n5 7 1",
"output": "4"
},
{
"input": "1\n1 2 1",
"output": "1"
},
{
"input": "5\n6 10 4\n9 20 19\n8 11 18\n18 20 1\n19 20 8",
"output": "50"
},
{
"input": "5\n21 25 26\n14 30 22\n6 29 30\n13 23 21\n10 11 5",
"output": "99"
},
{
"input": "5\n45 73 26\n59 73 29\n30 74 9\n59 63 72\n73 84 79",
"output": "215"
},
{
"input": "9\n1 5 1000000000\n23 27 834000000\n6 10 1000000000\n26 30 234500000\n5 9 1000000000\n2 6 1000000000\n3 7 1000000000\n24 29 132400000\n25 28 23400000",
"output": "5000000000"
},
{
"input": "10\n312 838 197\n713 996 320\n427 767 520\n490 800 465\n575 681 995\n781 843 908\n545 615 400\n36 667 129\n85 641 27\n396 539 350",
"output": "3961"
},
{
"input": "15\n3 14 531\n28 29 17\n7 10 512\n20 21 264\n8 9 52\n16 19 759\n1 30 391\n2 27 861\n17 18 357\n15 22 428\n24 25 588\n23 26 221\n6 13 413\n11 12 667\n4 5 513",
"output": "2863"
},
{
"input": "15\n7 8 940377876\n5 26 602265950\n18 19 193066910\n23 24 754826087\n2 3 555140266\n6 25 830980693\n27 28 636793242\n20 21 809913904\n1 30 868457738\n10 17 698644057\n9 22 443624155\n13 16 941205435\n14 15 42481040\n4 29 305295425\n11 12 765921496",
"output": "4732954493"
},
{
"input": "15\n10 11 552706512\n24 25 708893662\n18 19 731182554\n12 13 110677554\n28 29 938846280\n1 30 898802132\n16 17 89748120\n2 3 85944653\n6 7 141693677\n14 15 432017874\n4 5 498032180\n22 23 906122257\n8 9 561584264\n20 21 98373932\n26 27 195705303",
"output": "1837648412"
},
{
"input": "15\n1 30 902215054\n12 13 463027806\n9 14 846985104\n20 21 770345346\n17 22 514758030\n2 15 438137019\n18 19 278467923\n16 29 613043570\n10 11 765879041\n6 7 960155075\n26 27 323764310\n24 25 724736351\n4 5 107208241\n3 8 992951987\n23 28 70277204",
"output": "3293459135"
},
{
"input": "3\n1 3 1\n4 5 1\n2 5 1",
"output": "3"
},
{
"input": "5\n1 3 1000000000\n4 5 1000000000\n2 5 1000000000\n4 6 1000000000\n5 6 1000000000",
"output": "5000000000"
}
] | 1,488,317,842
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 4,608,000
|
n=int(input())
j=[]
for i in range(n):
ch=input()
l=ch.split(' ')
j.append(l)
i=n-1
k=[]
while i>=0:
test=True
s=int(j[i][2])
i=i-1
while (test==True)and(i>=0):
if int(j[i][1])>int(j[i+1][0]):
s=s+int(j[i][2])
else:
test=False
i=i-1
k.append(s)
if test==False:
i=i+1
print(max(k))
|
Title: Hanoi Factory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the rings for this wonderful game? Once upon a time, the ruler of the ancient Egypt ordered the workers of Hanoi Factory to create as high tower as possible. They were not ready to serve such a strange order so they had to create this new tower using already produced rings.
There are *n* rings in factory's stock. The *i*-th ring has inner radius *a**i*, outer radius *b**i* and height *h**i*. The goal is to select some subset of rings and arrange them such that the following conditions are satisfied:
- Outer radiuses form a non-increasing sequence, i.e. one can put the *j*-th ring on the *i*-th ring only if *b**j*<=≤<=*b**i*. - Rings should not fall one into the the other. That means one can place ring *j* on the ring *i* only if *b**j*<=><=*a**i*. - The total height of all rings used should be maximum possible.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of rings in factory's stock.
The *i*-th of the next *n* lines contains three integers *a**i*, *b**i* and *h**i* (1<=≤<=*a**i*,<=*b**i*,<=*h**i*<=≤<=109, *b**i*<=><=*a**i*) — inner radius, outer radius and the height of the *i*-th ring respectively.
Output Specification:
Print one integer — the maximum height of the tower that can be obtained.
Demo Input:
['3\n1 5 1\n2 6 2\n3 7 3\n', '4\n1 2 1\n1 3 3\n4 6 2\n5 7 1\n']
Demo Output:
['6\n', '4\n']
Note:
In the first sample, the optimal solution is to take all the rings and put them on each other in order 3, 2, 1.
In the second sample, one can put the ring 3 on the ring 4 and get the tower of height 3, or put the ring 1 on the ring 2 and get the tower of height 4.
|
```python
n=int(input())
j=[]
for i in range(n):
ch=input()
l=ch.split(' ')
j.append(l)
i=n-1
k=[]
while i>=0:
test=True
s=int(j[i][2])
i=i-1
while (test==True)and(i>=0):
if int(j[i][1])>int(j[i+1][0]):
s=s+int(j[i][2])
else:
test=False
i=i-1
k.append(s)
if test==False:
i=i+1
print(max(k))
```
| 0
|
|
932
|
A
|
Palindromic Supersequence
|
PROGRAMMING
| 800
|
[
"constructive algorithms"
] | null | null |
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
|
First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
|
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
|
[
"aba\n",
"ab\n"
] |
[
"aba",
"aabaa"
] |
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
| 500
|
[
{
"input": "aba",
"output": "abaaba"
},
{
"input": "ab",
"output": "abba"
},
{
"input": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpa",
"output": "krnyoixirslfszfqivgkaflgkctvbvksipwomqxlyqxhlbceuhbjbfnhofcgpgwdseffycthmlpcqejgskwjkbkbbmifnurnwyhevsoqzmtvzgfiqajfrgyuzxnrtxectcnlyoisbglpdbjbslxlpoymrcxmdtqhcnlvtqdwftuzgbdxsyscwbrguostbelnvtaqdmkmihmoxqtqlxvlsssisvqvvzotoyqryuyqwoknnqcqggysrqpkrccvyhxsjmhoqoyocwcriplarjoyiqrmmpmueqbsbljddwrumauczfziodpudheexalbwpiypmdjlmwtgdrzhpxneofhqzjdmurgvmrwdotuwyknlrbvuvtnhiouvqitgyfgfieonbaapyhwpcrmehxcpkijzfiayfvoxkpaapkxovfyaifzjikpcxhemrcpwhypaabnoeifgfygtiqvuoihntvuvbrlnkywutodwrmvgrumdjzqhfoenxphzrdgtwmljdm..."
},
{
"input": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadco",
"output": "mgrfmzxqpejcixxppqgvuawutgrmezjkteofjbnrvzzkvjtacfxjjokisavsgrslryxfqgrmdsqwptajbqzvethuljbdatxghfzqrwvfgakwmoawlzqjypmhllbbuuhbpriqsnibywlgjlxowyzagrfnqafvcqwktkcjwejevzbnxhsfmwojshcdypnvbuhhuzqmgovmvgwiizatoxgblyudipahfbkewmuneoqhjmbpdtwnznblwvtjrniwlbyblhppndspojrouffazpoxtqdfpjuhitvijrohavpqatofxwmksvjcvhdecxwwmosqiczjpkfafqlboxosnjgzgdraehzdltthemeusxhiiimrdrugabnxwsygsktkcslhjebfexucsyvlwrptebkjhefsvfrmcqqdlanbetrgzwylizmrystvpgrkhlicfadcoocdafcilhkrgpvtsyrmzilywzgrtebnaldqqcmrfvsfehjkbetprwlvyscuxef..."
},
{
"input": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxar",
"output": "hdmasfcjuigrwjchmjslmpynewnzpphmudzcbxzdexjuhktdtcoibzvevsmwaxakrtdfoivkvoooypyemiidadquqepxwqkesdnakxkbzrcjkgvwwxtqxvfpxcwitljyehldgsjytmekimkkndjvnzqtjykiymkmdzpwakxdtkzcqcatlevppgfhyykgmipuodjrnfjzhcmjdbzvhywprbwdcfxiffpzbjbmbyijkqnosslqbfvvicxvoeuzruraetglthgourzhfpnubzvblfzmmbgepjjyshchthulxarraxluhthchsyjjpegbmmzflbvzbunpfhzruoghtlgtearurzueovxcivvfbqlssonqkjiybmbjbzpffixfcdwbrpwyhvzbdjmchzjfnrjdoupimgkyyhfgppveltacqczktdxkawpzdmkmyikyjtqznvjdnkkmikemtyjsgdlheyjltiwcxpfvxqtxwwvgkjcrzbkxkandsekqwxpequ..."
},
{
"input": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjg",
"output": "fggbyzobbmxtwdajawqdywnppflkkmtxzjvxopqvliwdwhzepcuiwelhbuotlkvesexnwkytonfrpqcxzzqzdvsmbsjcxxeugavekozfjlolrtqgwzqxsfgrnvrgfrqpixhsskbpzghndesvwptpvvkasfalzsetopervpwzmkgpcexqnvtnoulprwnowmsorscecvvvrjfwumcjqyrounqsgdruxttvtmrkivtxauhosokdiahsyrftzsgvgyveqwkzhqstbgywrvmsgfcfyuxpphvmyydzpohgdicoxbtjnsbyhoidnkrialowvlvmjpxcfeygqzphmbcjkupojsmmuqlydixbaluwezvnfasjfxilbyllwyipsmovdzosuwotcxerzcfuvxprtziseshjfcosalyqglpotxvxaanpocypsiyazsejjoximnbvqucftuvdksaxutvjeunodbipsumlaymjnzljurefjggjferujlznjmyalmuspib..."
},
{
"input": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgm",
"output": "qyyxqkbxsvfnjzttdqmpzinbdgayllxpfrpopwciejjjzadguurnnhvixgueukugkkjyghxknedojvmdrskswiotgatsajowionuiumuhyggjuoympuxyfahwftwufvocdguxmxabbxnfviscxtilzzauizsgugwcqtbqgoosefhkumhodwpgolfdkbuiwlzjydonwbgyzzrjwxnceltqgqelrrljmzdbftmaogiuosaqhngmdzxzlmyrwefzhqawmkdckfnyyjgdjgadtfjvrkdwysqofcgyqrnyzutycvspzbjmmesobvhshtqlrytztyieknnkporrbcmlopgtknlmsstzkigreqwgsvagmvbrvwypoxttmzzsgmmgszzmttxopywvrbvmgavsgwqergikztssmlnktgpolmcbrropknnkeiytztyrlqthshvbosemmjbzpsvcytuzynrqygcfoqsywdkrvjftdagjdgjyynfkcdkmwaqhzfewry..."
},
{
"input": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyj",
"output": "scvlhflaqvniyiyofonowwcuqajuwscdrzhbvasymvqfnthzvtjcfuaftrbjghhvslcohwpxkggrbtatjtgehuqtorwinwvrtdldyoeeozxwippuahgkuehvsmyqtodqvlufqqmqautaqirvwzvtodzxtgxiinubhrbeoiybidutrqamsdnasctxatzkvkjkrmavdravnsxyngjlugwftmhmcvvxdbfndurrbmcpuoigjpssqcortmqoqttrabhoqvopjkxvpbqdqsilvlplhgqazauyvnodsxtwnomlinjpozwhrgrkqwmlwcwdkxjxjftexiavwrejvdjcfptterblxysjcheesyqsbgdrzjxbfjqgjgmvccqcyjjycqccvmgjgqjfbxjzrdgbsqyseehcjsyxlbrettpfcjdvjerwvaixetfjxjxkdwcwlmwqkrgrhwzopjnilmonwtxsdonvyuazaqghlplvlisqdqbpvxkjpovqohbarttqoqm..."
},
{
"input": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgd",
"output": "oohkqxxtvxzmvfjjxyjwlbqmeqwwlienzkdbhswgfbkhfygltsucdijozwaiewpixapyazfztksjeoqjugjfhdbqzuezbuajfvvffkwprroyivfoocvslejffgxuiofisenroxoeixmdbzonmreikpflciwsbafrdqfvdfojgoziiibqhwwsvhnzmptgirqqulkgmyzrfekzqqujmdumxkudsgexisupedisgmdgebvlvrpyfrbrqjknrxyzfpwmsxjxismgddgmsixjxsmwpfzyxrnkjqrbrfyprvlvbegdmgsidepusixegsdukxmudmjuqqzkefrzymgkluqqrigtpmznhvswwhqbiiizogjofdvfqdrfabswiclfpkiermnozbdmxieoxornesifoiuxgffjelsvcoofviyorrpwkffvvfjaubzeuzqbdhfjgujqoejsktzfzaypaxipweiawzojidcustlgyfhkbfgwshbdkzneilwwqemqblw..."
},
{
"input": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohlig",
"output": "gilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlwzjyeyfktduzdjnivcldjjarfzmmdbyytvipbbnjqolfnlqjpidotxxfobgtgpvjmpddcyddwdcjsxxumuoyznhpvpqccgqnuouzojntanfwctthcgynrukcvshsuuqrxfdvqqggaatwytikkitywtaaggqqvdfxrquushsvckurnygchttcwfnatnjozuounqgccqpvphnzyoumuxxsjcdwddycddpmjvpgtgbofxxtodipjqlnfloqjnbbpivtyybdmmzfrajjdlcvinjdzudtkfyeyjzwlzcnrgkdxrnjwyfwtnzawmeljknllaqhrgdagttixpljuupcivnlejkemrqnafdigjzxiohliggilhoixzjgidfanqrmekjelnvicpuujlpxittgadgrhqallnkjlemwazntwfywjnrxdkgrnczlw..."
},
{
"input": "abcab",
"output": "abcabbacba"
},
{
"input": "baaaaaaa",
"output": "baaaaaaaaaaaaaab"
},
{
"input": "baaaaaa",
"output": "baaaaaaaaaaaab"
},
{
"input": "baaaaaaaaa",
"output": "baaaaaaaaaaaaaaaaaab"
},
{
"input": "baaaaaaaa",
"output": "baaaaaaaaaaaaaaaab"
}
] | 1,603,513,435
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 108
| 0
|
s=input()
x=s[::-1]
if(s==x):
print(s)
else:
print(s+x)
|
Title: Palindromic Supersequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *A*. Find a string *B*, where *B* is a palindrome and *A* is a subsequence of *B*.
A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".
A palindrome is a string that reads the same forward or backward.
The length of string *B* should be at most 104. It is guaranteed that there always exists such string.
You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104.
Input Specification:
First line contains a string *A* (1<=≤<=|*A*|<=≤<=103) consisting of lowercase Latin letters, where |*A*| is a length of *A*.
Output Specification:
Output single line containing *B* consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string *B* should not exceed 104. If there are many possible *B*, print any of them.
Demo Input:
['aba\n', 'ab\n']
Demo Output:
['aba', 'aabaa']
Note:
In the first example, "aba" is a subsequence of "aba" which is a palindrome.
In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.
|
```python
s=input()
x=s[::-1]
if(s==x):
print(s)
else:
print(s+x)
```
| 3
|
|
825
|
A
|
Binary Protocol
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
|
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=89) — length of the string *s*.
The second line contains string *s* — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
|
Print the decoded number.
|
[
"3\n111\n",
"9\n110011101\n"
] |
[
"3\n",
"2031\n"
] |
none
| 0
|
[
{
"input": "3\n111",
"output": "3"
},
{
"input": "9\n110011101",
"output": "2031"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n100",
"output": "100"
},
{
"input": "5\n10001",
"output": "1001"
},
{
"input": "14\n11001100011000",
"output": "202002000"
},
{
"input": "31\n1000011111111100011110111111111",
"output": "100090049"
},
{
"input": "53\n10110111011110111110111111011111110111111110111111111",
"output": "123456789"
},
{
"input": "89\n11111111101111111110111111111011111111101111111110111111111011111111101111111110111111111",
"output": "999999999"
},
{
"input": "10\n1000000000",
"output": "1000000000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "4\n1110",
"output": "30"
},
{
"input": "8\n10101010",
"output": "11110"
}
] | 1,504,644,067
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 13
| 62
| 0
|
input()
count = 0
out = ''
for l in input():
if l == '1':
count +=1
else:
out=out+str(count)
count = 0
out=out+str(count)
print(out)
|
Title: Binary Protocol
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=89) — length of the string *s*.
The second line contains string *s* — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
Output Specification:
Print the decoded number.
Demo Input:
['3\n111\n', '9\n110011101\n']
Demo Output:
['3\n', '2031\n']
Note:
none
|
```python
input()
count = 0
out = ''
for l in input():
if l == '1':
count +=1
else:
out=out+str(count)
count = 0
out=out+str(count)
print(out)
```
| 3
|
|
675
|
B
|
Restoring Painting
|
PROGRAMMING
| 1,400
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
- The painting is a square 3<=×<=3, each cell contains a single integer from 1 to *n*, and different cells may contain either different or equal integers. - The sum of integers in each of four squares 2<=×<=2 is equal to the sum of integers in the top left square 2<=×<=2. - Four elements *a*, *b*, *c* and *d* are known and are located as shown on the picture below.
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
|
The first line of the input contains five integers *n*, *a*, *b*, *c* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=*n*) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
|
Print one integer — the number of distinct valid squares.
|
[
"2 1 1 1 2\n",
"3 3 1 2 3\n"
] |
[
"2\n",
"6\n"
] |
Below are all the possible paintings for the first sample. <img class="tex-graphics" src="https://espresso.codeforces.com/c4c53d4e7b6814d8aad7b72604b6089d61dadb48.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/46a6ad6a5d3db202f3779b045b9dc77fc2348cf1.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the second sample, only paintings displayed below satisfy all the rules. <img class="tex-graphics" src="https://espresso.codeforces.com/776f231305f8ce7c33e79e887722ce46aa8b6e61.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/2fce9e9a31e70f1e46ea26f11d7305b3414e9b6b.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/be084a4d1f7e475be1183f7dff10e9c89eb175ef.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/96afdb4a35ac14f595d29bea2282f621098902f4.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/79ca8d720334a74910514f017ecf1d0166009a03.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/ad3c37e950bf5702d54f05756db35c831da59ad9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 1 1 1 2",
"output": "2"
},
{
"input": "3 3 1 2 3",
"output": "6"
},
{
"input": "1 1 1 1 1",
"output": "1"
},
{
"input": "1000 522 575 426 445",
"output": "774000"
},
{
"input": "99000 52853 14347 64237 88869",
"output": "1296306000"
},
{
"input": "100000 2 2 2 2",
"output": "10000000000"
},
{
"input": "2 1 1 2 2",
"output": "0"
},
{
"input": "10 9 10 8 10",
"output": "70"
},
{
"input": "100 19 16 35 83",
"output": "1700"
},
{
"input": "1000 102 583 606 929",
"output": "150000"
},
{
"input": "10000 1816 3333 6908 7766",
"output": "4750000"
},
{
"input": "100000 80015 84290 50777 30497",
"output": "1696900000"
},
{
"input": "100000 64022 49026 55956 88430",
"output": "6866200000"
},
{
"input": "100000 10263 46628 10268 22948",
"output": "5095500000"
},
{
"input": "100000 81311 81584 51625 57276",
"output": "4600600000"
},
{
"input": "100000 77594 3226 21255 8541",
"output": "1291800000"
},
{
"input": "100000 65131 35523 58220 87645",
"output": "5478900000"
},
{
"input": "100000 83958 32567 91083 95317",
"output": "3012500000"
},
{
"input": "100000 36851 54432 21164 85520",
"output": "1806300000"
},
{
"input": "100000 55732 17473 23832 75148",
"output": "7422500000"
},
{
"input": "100000 60789 25296 49585 25237",
"output": "4015900000"
},
{
"input": "100000 92060 77234 58709 36956",
"output": "2637100000"
},
{
"input": "100000 87223 66046 27153 40823",
"output": "1470700000"
},
{
"input": "100000 3809 35468 34556 51158",
"output": "5173900000"
},
{
"input": "100000 35038 37363 95275 88903",
"output": "0"
},
{
"input": "100000 45274 9250 36558 49486",
"output": "6848000000"
},
{
"input": "100000 1 1 1 1",
"output": "10000000000"
},
{
"input": "100000 1 1 1 100000",
"output": "100000"
},
{
"input": "100000 1 1 100000 1",
"output": "100000"
},
{
"input": "100000 1 1 100000 100000",
"output": "0"
},
{
"input": "100000 1 100000 1 1",
"output": "100000"
},
{
"input": "100000 1 100000 1 100000",
"output": "0"
},
{
"input": "100000 1 100000 100000 1",
"output": "10000000000"
},
{
"input": "100000 1 100000 100000 100000",
"output": "100000"
},
{
"input": "100000 100000 1 1 1",
"output": "100000"
},
{
"input": "100000 100000 1 1 100000",
"output": "10000000000"
},
{
"input": "100000 100000 1 100000 1",
"output": "0"
},
{
"input": "100000 100000 1 100000 100000",
"output": "100000"
},
{
"input": "100000 100000 100000 1 1",
"output": "0"
},
{
"input": "100000 100000 100000 1 100000",
"output": "100000"
},
{
"input": "100000 100000 100000 100000 1",
"output": "100000"
},
{
"input": "100000 100000 100000 100000 100000",
"output": "10000000000"
},
{
"input": "3 3 3 1 1",
"output": "0"
},
{
"input": "10 1 2 5 10",
"output": "0"
},
{
"input": "5 1 1 5 5",
"output": "0"
},
{
"input": "4 4 4 1 1",
"output": "0"
},
{
"input": "10 10 10 1 1",
"output": "0"
},
{
"input": "5 5 5 1 1",
"output": "0"
},
{
"input": "100 100 100 1 1",
"output": "0"
},
{
"input": "3 1 1 3 3",
"output": "0"
},
{
"input": "10 2 10 1 10",
"output": "0"
},
{
"input": "7 7 7 1 1",
"output": "0"
},
{
"input": "5 5 3 4 1",
"output": "0"
},
{
"input": "7 1 1 7 7",
"output": "0"
},
{
"input": "100 1 1 100 100",
"output": "0"
},
{
"input": "123 1 2 3 100",
"output": "2829"
},
{
"input": "10 1 1 10 10",
"output": "0"
},
{
"input": "803 525 6 623 8",
"output": "0"
}
] | 1,469,919,335
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 218
| 0
|
n,a,b,c,d=map(int,input().split())
M=0
for E in range(1,n+1):
F=E+b-c
H=E+a-d
K=E+a+b-c-d
if 0<F<=n and 0<H<=n and 0<K<=n:M+=n
print(M)
|
Title: Restoring Painting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
- The painting is a square 3<=×<=3, each cell contains a single integer from 1 to *n*, and different cells may contain either different or equal integers. - The sum of integers in each of four squares 2<=×<=2 is equal to the sum of integers in the top left square 2<=×<=2. - Four elements *a*, *b*, *c* and *d* are known and are located as shown on the picture below.
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input Specification:
The first line of the input contains five integers *n*, *a*, *b*, *c* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=*n*) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output Specification:
Print one integer — the number of distinct valid squares.
Demo Input:
['2 1 1 1 2\n', '3 3 1 2 3\n']
Demo Output:
['2\n', '6\n']
Note:
Below are all the possible paintings for the first sample. <img class="tex-graphics" src="https://espresso.codeforces.com/c4c53d4e7b6814d8aad7b72604b6089d61dadb48.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/46a6ad6a5d3db202f3779b045b9dc77fc2348cf1.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the second sample, only paintings displayed below satisfy all the rules. <img class="tex-graphics" src="https://espresso.codeforces.com/776f231305f8ce7c33e79e887722ce46aa8b6e61.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/2fce9e9a31e70f1e46ea26f11d7305b3414e9b6b.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/be084a4d1f7e475be1183f7dff10e9c89eb175ef.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/96afdb4a35ac14f595d29bea2282f621098902f4.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/79ca8d720334a74910514f017ecf1d0166009a03.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/ad3c37e950bf5702d54f05756db35c831da59ad9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n,a,b,c,d=map(int,input().split())
M=0
for E in range(1,n+1):
F=E+b-c
H=E+a-d
K=E+a+b-c-d
if 0<F<=n and 0<H<=n and 0<K<=n:M+=n
print(M)
```
| 3
|
|
950
|
A
|
Left-handers, Right-handers and Ambidexters
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
|
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
|
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
|
[
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] |
[
"6\n",
"14\n",
"0\n"
] |
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
| 500
|
[
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "100 100 100",
"output": "300"
},
{
"input": "1 1 1",
"output": "2"
},
{
"input": "30 70 35",
"output": "130"
},
{
"input": "89 44 76",
"output": "208"
},
{
"input": "0 100 100",
"output": "200"
},
{
"input": "100 0 100",
"output": "200"
},
{
"input": "100 1 100",
"output": "200"
},
{
"input": "1 100 100",
"output": "200"
},
{
"input": "100 100 0",
"output": "200"
},
{
"input": "100 100 1",
"output": "200"
},
{
"input": "1 2 1",
"output": "4"
},
{
"input": "0 0 100",
"output": "100"
},
{
"input": "0 100 0",
"output": "0"
},
{
"input": "100 0 0",
"output": "0"
},
{
"input": "10 8 7",
"output": "24"
},
{
"input": "45 47 16",
"output": "108"
},
{
"input": "59 43 100",
"output": "202"
},
{
"input": "34 1 30",
"output": "62"
},
{
"input": "14 81 1",
"output": "30"
},
{
"input": "53 96 94",
"output": "242"
},
{
"input": "62 81 75",
"output": "218"
},
{
"input": "21 71 97",
"output": "188"
},
{
"input": "49 82 73",
"output": "204"
},
{
"input": "88 19 29",
"output": "96"
},
{
"input": "89 4 62",
"output": "132"
},
{
"input": "58 3 65",
"output": "126"
},
{
"input": "27 86 11",
"output": "76"
},
{
"input": "35 19 80",
"output": "134"
},
{
"input": "4 86 74",
"output": "156"
},
{
"input": "32 61 89",
"output": "182"
},
{
"input": "68 60 98",
"output": "226"
},
{
"input": "37 89 34",
"output": "142"
},
{
"input": "92 9 28",
"output": "74"
},
{
"input": "79 58 98",
"output": "234"
},
{
"input": "35 44 88",
"output": "166"
},
{
"input": "16 24 19",
"output": "58"
},
{
"input": "74 71 75",
"output": "220"
},
{
"input": "83 86 99",
"output": "268"
},
{
"input": "97 73 15",
"output": "176"
},
{
"input": "77 76 73",
"output": "226"
},
{
"input": "48 85 55",
"output": "188"
},
{
"input": "1 2 2",
"output": "4"
},
{
"input": "2 2 2",
"output": "6"
},
{
"input": "2 1 2",
"output": "4"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "3 2 1",
"output": "6"
},
{
"input": "1 2 3",
"output": "6"
},
{
"input": "1 3 2",
"output": "6"
},
{
"input": "2 1 3",
"output": "6"
},
{
"input": "2 3 1",
"output": "6"
},
{
"input": "3 1 2",
"output": "6"
},
{
"input": "99 99 99",
"output": "296"
},
{
"input": "99 99 100",
"output": "298"
},
{
"input": "99 100 99",
"output": "298"
},
{
"input": "99 100 100",
"output": "298"
},
{
"input": "100 99 99",
"output": "298"
},
{
"input": "100 99 100",
"output": "298"
},
{
"input": "100 100 99",
"output": "298"
},
{
"input": "89 32 23",
"output": "110"
},
{
"input": "4 5 0",
"output": "8"
},
{
"input": "3 0 3",
"output": "6"
},
{
"input": "0 0 2",
"output": "2"
},
{
"input": "97 97 0",
"output": "194"
},
{
"input": "1 4 0",
"output": "2"
},
{
"input": "5 2 0",
"output": "4"
},
{
"input": "0 5 10",
"output": "14"
},
{
"input": "0 1 2",
"output": "2"
},
{
"input": "5 2 3",
"output": "10"
},
{
"input": "5 5 0",
"output": "10"
},
{
"input": "0 0 10",
"output": "10"
},
{
"input": "0 1 1",
"output": "2"
},
{
"input": "0 0 1",
"output": "0"
}
] | 1,520,960,157
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 5,632,000
|
from math import*
A = []
A = input().split()
B=[]
B.insert(0,int(A[0]))
B.insert(1,int(A[1]))
B.insert(2,int(A[2]))
count = min(B[0], B[1])
razn=abs(B[0]-B[1])
if (B[2]>=razn):
count += razn
B[2] = B[2]/2
count += int(B[2])
else:
count += B[2]
print(count)
|
Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
|
```python
from math import*
A = []
A = input().split()
B=[]
B.insert(0,int(A[0]))
B.insert(1,int(A[1]))
B.insert(2,int(A[2]))
count = min(B[0], B[1])
razn=abs(B[0]-B[1])
if (B[2]>=razn):
count += razn
B[2] = B[2]/2
count += int(B[2])
else:
count += B[2]
print(count)
```
| 0
|
|
527
|
A
|
Playing with Paper
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
|
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
|
Print a single integer — the number of ships that Vasya will make.
|
[
"2 1\n",
"10 7\n",
"1000000000000 1\n"
] |
[
"2\n",
"6\n",
"1000000000000\n"
] |
Pictures to the first and second sample test.
| 500
|
[
{
"input": "2 1",
"output": "2"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "1000000000000 1",
"output": "1000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "1000 700",
"output": "6"
},
{
"input": "959986566087 524054155168",
"output": "90"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "7 6",
"output": "7"
},
{
"input": "1000 999",
"output": "1000"
},
{
"input": "1000 998",
"output": "500"
},
{
"input": "1000 997",
"output": "336"
},
{
"input": "42 1",
"output": "42"
},
{
"input": "1000 1",
"output": "1000"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "13 8",
"output": "6"
},
{
"input": "987 610",
"output": "15"
},
{
"input": "442 42",
"output": "22"
},
{
"input": "754 466",
"output": "13"
},
{
"input": "1000000000000 999999999999",
"output": "1000000000000"
},
{
"input": "1000000000000 999999999998",
"output": "500000000000"
},
{
"input": "941 14",
"output": "74"
},
{
"input": "998 2",
"output": "499"
},
{
"input": "1000 42",
"output": "32"
},
{
"input": "1000 17",
"output": "66"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "5 3",
"output": "4"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "293 210",
"output": "17"
},
{
"input": "787878787878 424242424242",
"output": "8"
},
{
"input": "956722026041 591286729879",
"output": "58"
},
{
"input": "956722026041 365435296162",
"output": "58"
},
{
"input": "628625247282 464807889701",
"output": "102"
},
{
"input": "695928431619 424778620208",
"output": "167"
},
{
"input": "1000000000000 42",
"output": "23809523821"
},
{
"input": "987654345678 23",
"output": "42941493300"
},
{
"input": "10000000001 2",
"output": "5000000002"
},
{
"input": "1000000000000 2",
"output": "500000000000"
},
{
"input": "1000000000000 3",
"output": "333333333336"
},
{
"input": "100000000000 3",
"output": "33333333336"
},
{
"input": "100000000000 23",
"output": "4347826109"
},
{
"input": "999999999997 7",
"output": "142857142861"
},
{
"input": "8589934592 4294967296",
"output": "2"
}
] | 1,468,050,868
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 2,000
| 0
|
d = [int(x) for x in input().split()]
a,b = d[0], d[1]
sq = 0
while a != b:
if a > b:
sq+=1
a-=b
else:
sq+=1
b-=a
if a == b:
print(sq+1)
else:
print(sq)
|
Title: Playing with Paper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
Input Specification:
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
Output Specification:
Print a single integer — the number of ships that Vasya will make.
Demo Input:
['2 1\n', '10 7\n', '1000000000000 1\n']
Demo Output:
['2\n', '6\n', '1000000000000\n']
Note:
Pictures to the first and second sample test.
|
```python
d = [int(x) for x in input().split()]
a,b = d[0], d[1]
sq = 0
while a != b:
if a > b:
sq+=1
a-=b
else:
sq+=1
b-=a
if a == b:
print(sq+1)
else:
print(sq)
```
| 0
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,567,289,139
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 9
| 280
| 0
|
n = int(input())
name = []
cnt = []
maxs = 0
maxn = ''
for _ in range(n):
s = input().split()
if s[0] not in name:
name.append(s[0])
cnt.append(int(s[1]))
else:
cnt[name.index(s[0])] += int(s[1])
if cnt[name.index(s[0])] > maxs:
maxs = cnt[name.index(s[0])]
maxn = s[0]
print(maxn)
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
n = int(input())
name = []
cnt = []
maxs = 0
maxn = ''
for _ in range(n):
s = input().split()
if s[0] not in name:
name.append(s[0])
cnt.append(int(s[1]))
else:
cnt[name.index(s[0])] += int(s[1])
if cnt[name.index(s[0])] > maxs:
maxs = cnt[name.index(s[0])]
maxn = s[0]
print(maxn)
```
| 0
|
584
|
D
|
Dima and Lisa
|
PROGRAMMING
| 1,800
|
[
"brute force",
"math",
"number theory"
] | null | null |
Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.
More formally, you are given an odd numer *n*. Find a set of numbers *p**i* (1<=≤<=*i*<=≤<=*k*), such that
1. 1<=≤<=*k*<=≤<=31. *p**i* is a prime1.
The numbers *p**i* do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.
|
The single line contains an odd number *n* (3<=≤<=*n*<=<<=109).
|
In the first line print *k* (1<=≤<=*k*<=≤<=3), showing how many numbers are in the representation you found.
In the second line print numbers *p**i* in any order. If there are multiple possible solutions, you can print any of them.
|
[
"27\n"
] |
[
"3\n5 11 11\n"
] |
A prime is an integer strictly larger than one that is divisible only by one and by itself.
| 2,000
|
[
{
"input": "27",
"output": "3\n2 2 23"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "25",
"output": "2\n2 23"
},
{
"input": "9",
"output": "2\n2 7"
},
{
"input": "91",
"output": "2\n2 89"
},
{
"input": "57",
"output": "3\n2 2 53"
},
{
"input": "31",
"output": "1\n31"
},
{
"input": "555",
"output": "3\n3 5 547"
},
{
"input": "700000001",
"output": "1\n700000001"
},
{
"input": "5",
"output": "1\n5"
},
{
"input": "7",
"output": "1\n7"
},
{
"input": "11",
"output": "1\n11"
},
{
"input": "13",
"output": "1\n13"
},
{
"input": "15",
"output": "2\n2 13"
},
{
"input": "17",
"output": "1\n17"
},
{
"input": "19",
"output": "1\n19"
},
{
"input": "21",
"output": "2\n2 19"
},
{
"input": "23",
"output": "1\n23"
},
{
"input": "29",
"output": "1\n29"
},
{
"input": "79",
"output": "1\n79"
},
{
"input": "99",
"output": "2\n2 97"
},
{
"input": "27",
"output": "3\n2 2 23"
},
{
"input": "55",
"output": "2\n2 53"
},
{
"input": "79",
"output": "1\n79"
},
{
"input": "585",
"output": "3\n3 5 577"
},
{
"input": "245",
"output": "3\n2 2 241"
},
{
"input": "793",
"output": "3\n3 3 787"
},
{
"input": "133",
"output": "2\n2 131"
},
{
"input": "681",
"output": "3\n2 2 677"
},
{
"input": "981399",
"output": "2\n2 981397"
},
{
"input": "867773",
"output": "1\n867773"
},
{
"input": "654141",
"output": "3\n3 11 654127"
},
{
"input": "202985",
"output": "3\n2 2 202981"
},
{
"input": "784533",
"output": "3\n3 17 784513"
},
{
"input": "370359",
"output": "3\n19 79 370261"
},
{
"input": "396831",
"output": "3\n19 79 396733"
},
{
"input": "492211",
"output": "3\n19 79 492113"
},
{
"input": "838347",
"output": "3\n19 79 838249"
},
{
"input": "1098945",
"output": "3\n19 79 1098847"
},
{
"input": "1313565",
"output": "3\n19 79 1313467"
},
{
"input": "1349631",
"output": "3\n19 79 1349533"
},
{
"input": "1357299",
"output": "3\n19 79 1357201"
},
{
"input": "1357323",
"output": "3\n13 109 1357201"
},
{
"input": "1357329",
"output": "3\n19 109 1357201"
},
{
"input": "1388581",
"output": "3\n19 79 1388483"
},
{
"input": "5275",
"output": "2\n2 5273"
},
{
"input": "9515",
"output": "3\n2 2 9511"
},
{
"input": "7847",
"output": "3\n3 3 7841"
},
{
"input": "7077",
"output": "3\n3 5 7069"
},
{
"input": "9531",
"output": "3\n3 7 9521"
},
{
"input": "7865",
"output": "3\n5 7 7853"
},
{
"input": "9675",
"output": "3\n3 11 9661"
},
{
"input": "8909",
"output": "3\n3 13 8893"
},
{
"input": "7147",
"output": "3\n5 13 7129"
},
{
"input": "8487",
"output": "3\n3 17 8467"
},
{
"input": "436273289",
"output": "3\n3 277 436273009"
},
{
"input": "649580445",
"output": "3\n3 271 649580171"
},
{
"input": "944193065",
"output": "3\n7 251 944192807"
},
{
"input": "630045387",
"output": "3\n11 239 630045137"
},
{
"input": "931103229",
"output": "3\n3 223 931103003"
},
{
"input": "950664039",
"output": "3\n3 197 950663839"
},
{
"input": "996104777",
"output": "3\n7 173 996104597"
},
{
"input": "997255617",
"output": "3\n7 157 997255453"
},
{
"input": "999962901",
"output": "3\n19 109 999962773"
},
{
"input": "999995529",
"output": "3\n19 79 999995431"
},
{
"input": "999995339",
"output": "3\n5 43 999995291"
},
{
"input": "999998367",
"output": "3\n5 23 999998339"
},
{
"input": "999999891",
"output": "3\n3 5 999999883"
},
{
"input": "999999935",
"output": "3\n3 3 999999929"
},
{
"input": "999999755",
"output": "3\n2 2 999999751"
},
{
"input": "999999759",
"output": "2\n2 999999757"
},
{
"input": "999999191",
"output": "1\n999999191"
},
{
"input": "999999999",
"output": "3\n3 59 999999937"
},
{
"input": "409449117",
"output": "3\n2 2 409449113"
},
{
"input": "882499837",
"output": "3\n3 3 882499831"
},
{
"input": "765615965",
"output": "3\n5 23 765615937"
},
{
"input": "648732093",
"output": "3\n3 11 648732079"
},
{
"input": "826815517",
"output": "3\n3 11 826815503"
},
{
"input": "4898941",
"output": "2\n2 4898939"
},
{
"input": "182982365",
"output": "3\n5 13 182982347"
},
{
"input": "66098493",
"output": "3\n3 41 66098449"
},
{
"input": "539149213",
"output": "1\n539149213"
},
{
"input": "655957385",
"output": "3\n3 13 655957369"
},
{
"input": "199999581",
"output": "3\n19 79 199999483"
},
{
"input": "199998345",
"output": "3\n19 79 199998247"
},
{
"input": "199991935",
"output": "3\n19 79 199991837"
},
{
"input": "199986207",
"output": "3\n19 79 199986109"
},
{
"input": "499991589",
"output": "3\n19 79 499991491"
},
{
"input": "499984689",
"output": "3\n19 79 499984591"
},
{
"input": "499984159",
"output": "3\n19 79 499984061"
},
{
"input": "499966179",
"output": "3\n19 79 499966081"
},
{
"input": "999995529",
"output": "3\n19 79 999995431"
},
{
"input": "999995085",
"output": "3\n19 79 999994987"
},
{
"input": "999991817",
"output": "3\n11 137 999991669"
},
{
"input": "999991797",
"output": "3\n19 109 999991669"
},
{
"input": "999991791",
"output": "3\n13 109 999991669"
},
{
"input": "748859699",
"output": "3\n3 3 748859693"
},
{
"input": "323845235",
"output": "3\n3 3 323845229"
},
{
"input": "462409937",
"output": "3\n2 2 462409933"
},
{
"input": "618047403",
"output": "3\n3 13 618047387"
},
{
"input": "501148647",
"output": "3\n2 2 501148643"
},
{
"input": "998017623",
"output": "2\n2 998017621"
},
{
"input": "436273289",
"output": "3\n3 277 436273009"
},
{
"input": "999999965",
"output": "3\n5 23 999999937"
},
{
"input": "5",
"output": "1\n5"
},
{
"input": "1000037",
"output": "1\n1000037"
},
{
"input": "989898987",
"output": "3\n2 2 989898983"
},
{
"input": "999999999",
"output": "3\n3 59 999999937"
},
{
"input": "100000003",
"output": "3\n3 11 99999989"
}
] | 1,675,704,523
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 109
| 77
| 1,638,400
|
import sys
import math
input = sys.stdin.readline
def p(n):
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
else:
return True
n = int(input())
if p(n):
print(1)
print(n)
quit()
if p(n - 2):
print(2)
print(n - 2, 2)
quit()
z = [0] * 1000
for i in range(2, 1001):
if p(i):
z[i - 1] = 1
for i in range(2, 1000):
if z[i - 1]:
for j in range(2, 1000):
if z[j - 1] and n - i - j > 0 and p(n - i - j):
print(3)
print(i, j, n - j - i)
quit()
|
Title: Dima and Lisa
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.
More formally, you are given an odd numer *n*. Find a set of numbers *p**i* (1<=≤<=*i*<=≤<=*k*), such that
1. 1<=≤<=*k*<=≤<=31. *p**i* is a prime1.
The numbers *p**i* do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.
Input Specification:
The single line contains an odd number *n* (3<=≤<=*n*<=<<=109).
Output Specification:
In the first line print *k* (1<=≤<=*k*<=≤<=3), showing how many numbers are in the representation you found.
In the second line print numbers *p**i* in any order. If there are multiple possible solutions, you can print any of them.
Demo Input:
['27\n']
Demo Output:
['3\n5 11 11\n']
Note:
A prime is an integer strictly larger than one that is divisible only by one and by itself.
|
```python
import sys
import math
input = sys.stdin.readline
def p(n):
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
else:
return True
n = int(input())
if p(n):
print(1)
print(n)
quit()
if p(n - 2):
print(2)
print(n - 2, 2)
quit()
z = [0] * 1000
for i in range(2, 1001):
if p(i):
z[i - 1] = 1
for i in range(2, 1000):
if z[i - 1]:
for j in range(2, 1000):
if z[j - 1] and n - i - j > 0 and p(n - i - j):
print(3)
print(i, j, n - j - i)
quit()
```
| 3
|
|
588
|
B
|
Duff in Love
|
PROGRAMMING
| 1,300
|
[
"math"
] | null | null |
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
|
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
|
Print the answer in one line.
|
[
"10\n",
"12\n"
] |
[
"10\n",
"6\n"
] |
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely.
| 1,000
|
[
{
"input": "10",
"output": "10"
},
{
"input": "12",
"output": "6"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "3",
"output": "3"
},
{
"input": "31",
"output": "31"
},
{
"input": "97",
"output": "97"
},
{
"input": "1000000000000",
"output": "10"
},
{
"input": "15",
"output": "15"
},
{
"input": "894",
"output": "894"
},
{
"input": "271",
"output": "271"
},
{
"input": "2457",
"output": "273"
},
{
"input": "2829",
"output": "2829"
},
{
"input": "5000",
"output": "10"
},
{
"input": "20",
"output": "10"
},
{
"input": "68",
"output": "34"
},
{
"input": "3096",
"output": "258"
},
{
"input": "1024",
"output": "2"
},
{
"input": "1048576",
"output": "2"
},
{
"input": "413933789280",
"output": "25870861830"
},
{
"input": "817634153013",
"output": "817634153013"
},
{
"input": "56517269141",
"output": "56517269141"
},
{
"input": "30707328551",
"output": "30707328551"
},
{
"input": "279564127218",
"output": "10354226934"
},
{
"input": "491159577042",
"output": "18191095446"
},
{
"input": "734337660466",
"output": "734337660466"
},
{
"input": "808453785117",
"output": "808453785117"
},
{
"input": "55926835837",
"output": "55926835837"
},
{
"input": "294809951965",
"output": "294809951965"
},
{
"input": "537988035389",
"output": "76855433627"
},
{
"input": "822722434952",
"output": "205680608738"
},
{
"input": "699511759613",
"output": "699511759613"
},
{
"input": "942689843037",
"output": "104743315893"
},
{
"input": "663634158717",
"output": "663634158717"
},
{
"input": "213612977250",
"output": "11730"
},
{
"input": "999999999989",
"output": "999999999989"
},
{
"input": "999999999988",
"output": "499999999994"
},
{
"input": "87178291200",
"output": "30030"
},
{
"input": "927668721948",
"output": "463834360974"
},
{
"input": "562436815639",
"output": "37927"
},
{
"input": "302981118597",
"output": "35853"
},
{
"input": "5",
"output": "5"
},
{
"input": "9",
"output": "3"
},
{
"input": "36",
"output": "6"
},
{
"input": "2231",
"output": "2231"
},
{
"input": "27648",
"output": "6"
},
{
"input": "40320",
"output": "210"
},
{
"input": "648000",
"output": "30"
},
{
"input": "999966000289",
"output": "999983"
},
{
"input": "999985999949",
"output": "999985999949"
},
{
"input": "991921850317",
"output": "9973"
}
] | 1,636,452,640
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 4,505,600
|
import math
def sq(n):
return int(math.sqrt(n))+1
def judge(k):
for i in range(2,sq(n)):
if(k%(i*i) == 0):
return False
return True
while True:
n=eval(input())
ans=0
f = 0
for i in range(1,sq(n)):
k = n / i
if(n%i==0 and judge(k)):
f=1
ans=k
break
if(f !=1):
for i in range(1,sq(n)):
if(n %i==0 and judge(i)):
f=1
ans=i
break
print("{:.0f}".format(ans))
|
Title: Duff in Love
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
Input Specification:
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
Output Specification:
Print the answer in one line.
Demo Input:
['10\n', '12\n']
Demo Output:
['10\n', '6\n']
Note:
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely.
|
```python
import math
def sq(n):
return int(math.sqrt(n))+1
def judge(k):
for i in range(2,sq(n)):
if(k%(i*i) == 0):
return False
return True
while True:
n=eval(input())
ans=0
f = 0
for i in range(1,sq(n)):
k = n / i
if(n%i==0 and judge(k)):
f=1
ans=k
break
if(f !=1):
for i in range(1,sq(n)):
if(n %i==0 and judge(i)):
f=1
ans=i
break
print("{:.0f}".format(ans))
```
| -1
|
|
150
|
C
|
Smart Cheater
|
PROGRAMMING
| 2,200
|
[
"data structures",
"math",
"probabilities"
] | null | null |
I guess there's not much point in reminding you that Nvodsk winters aren't exactly hot. That increased the popularity of the public transport dramatically. The route of bus 62 has exactly *n* stops (stop 1 goes first on its way and stop *n* goes last). The stops are positioned on a straight line and their coordinates are 0<==<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*.
Each day exactly *m* people use bus 62. For each person we know the number of the stop where he gets on the bus and the number of the stop where he gets off the bus. A ticket from stop *a* to stop *b* (*a*<=<<=*b*) costs *x**b*<=-<=*x**a* rubles. However, the conductor can choose no more than one segment NOT TO SELL a ticket for. We mean that conductor should choose C and D (С <= D) and sell a ticket for the segments [*A*, *C*] and [*D*, *B*], or not sell the ticket at all. The conductor and the passenger divide the saved money between themselves equally. The conductor's "untaxed income" is sometimes interrupted by inspections that take place as the bus drives on some segment of the route located between two consecutive stops. The inspector fines the conductor by *c* rubles for each passenger who doesn't have the ticket for this route's segment.
You know the coordinated of all stops *x**i*; the numbers of stops where the *i*-th passenger gets on and off, *a**i* and *b**i* (*a**i*<=<<=*b**i*); the fine *c*; and also *p**i* — the probability of inspection on segment between the *i*-th and the *i*<=+<=1-th stop. The conductor asked you to help him make a plan of selling tickets that maximizes the mathematical expectation of his profit.
|
The first line contains three integers *n*, *m* and *c* (2<=≤<=*n*<=≤<=150<=000, 1<=≤<=*m*<=≤<=300<=000, 1<=≤<=*c*<=≤<=10<=000).
The next line contains *n* integers *x**i* (0<=≤<=*x**i*<=≤<=109, *x*1<==<=0, *x**i*<=<<=*x**i*<=+<=1) — the coordinates of the stops on the bus's route.
The third line contains *n*<=-<=1 integer *p**i* (0<=≤<=*p**i*<=≤<=100) — the probability of inspection in percents on the segment between stop *i* and stop *i*<=+<=1.
Then follow *m* lines that describe the bus's passengers. Each line contains exactly two integers *a**i* and *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) — the numbers of stops where the *i*-th passenger gets on and off.
|
Print the single real number — the maximum expectation of the conductor's profit. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"3 3 10\n0 10 100\n100 0\n1 2\n2 3\n1 3\n",
"10 8 187\n0 10 30 70 150 310 630 1270 2550 51100\n13 87 65 0 100 44 67 3 4\n1 10\n2 9\n3 8\n1 5\n6 10\n2 7\n4 10\n4 5\n"
] |
[
"90.000000000\n",
"76859.990000000\n"
] |
A comment to the first sample:
The first and third passengers get tickets from stop 1 to stop 2. The second passenger doesn't get a ticket. There always is inspection on the segment 1-2 but both passengers have the ticket for it. There never is an inspection on the segment 2-3, that's why the second passenger gets away with the cheating. Our total profit is (0 + 90 / 2 + 90 / 2) = 90.
| 1,000
|
[] | 1,668,610,127
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
import sys
from math import gcd
from collections import defaultdict
input = sys.stdin.readline
get_int = lambda: int(input().rstrip())
get_arr = lambda: [int(w) for w in input().split()]
get_str = lambda: input().rstrip()
n, m, c = get_arr()
xs = get_arr()
ps = get_arr()
ost = [0] * (n + 1)
for _ in range(m):
a, b = get_arr()
ost[a - 1] += 1
ost[b - 1] -= 1
ct = 0
mn = 0
cv = 0
ans = 0
for i in range(n - 1):
ct += ost[i]
ev = ct * (xs[i + 1] - xs[i]) / 2 - ct * c * ps[i] / 100
# print(i, ct, ev)
cv += ev
ans = max(ans, cv - mn)
mn = min(cv, mn)
print(ans)
|
Title: Smart Cheater
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
I guess there's not much point in reminding you that Nvodsk winters aren't exactly hot. That increased the popularity of the public transport dramatically. The route of bus 62 has exactly *n* stops (stop 1 goes first on its way and stop *n* goes last). The stops are positioned on a straight line and their coordinates are 0<==<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*.
Each day exactly *m* people use bus 62. For each person we know the number of the stop where he gets on the bus and the number of the stop where he gets off the bus. A ticket from stop *a* to stop *b* (*a*<=<<=*b*) costs *x**b*<=-<=*x**a* rubles. However, the conductor can choose no more than one segment NOT TO SELL a ticket for. We mean that conductor should choose C and D (С <= D) and sell a ticket for the segments [*A*, *C*] and [*D*, *B*], or not sell the ticket at all. The conductor and the passenger divide the saved money between themselves equally. The conductor's "untaxed income" is sometimes interrupted by inspections that take place as the bus drives on some segment of the route located between two consecutive stops. The inspector fines the conductor by *c* rubles for each passenger who doesn't have the ticket for this route's segment.
You know the coordinated of all stops *x**i*; the numbers of stops where the *i*-th passenger gets on and off, *a**i* and *b**i* (*a**i*<=<<=*b**i*); the fine *c*; and also *p**i* — the probability of inspection on segment between the *i*-th and the *i*<=+<=1-th stop. The conductor asked you to help him make a plan of selling tickets that maximizes the mathematical expectation of his profit.
Input Specification:
The first line contains three integers *n*, *m* and *c* (2<=≤<=*n*<=≤<=150<=000, 1<=≤<=*m*<=≤<=300<=000, 1<=≤<=*c*<=≤<=10<=000).
The next line contains *n* integers *x**i* (0<=≤<=*x**i*<=≤<=109, *x*1<==<=0, *x**i*<=<<=*x**i*<=+<=1) — the coordinates of the stops on the bus's route.
The third line contains *n*<=-<=1 integer *p**i* (0<=≤<=*p**i*<=≤<=100) — the probability of inspection in percents on the segment between stop *i* and stop *i*<=+<=1.
Then follow *m* lines that describe the bus's passengers. Each line contains exactly two integers *a**i* and *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) — the numbers of stops where the *i*-th passenger gets on and off.
Output Specification:
Print the single real number — the maximum expectation of the conductor's profit. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['3 3 10\n0 10 100\n100 0\n1 2\n2 3\n1 3\n', '10 8 187\n0 10 30 70 150 310 630 1270 2550 51100\n13 87 65 0 100 44 67 3 4\n1 10\n2 9\n3 8\n1 5\n6 10\n2 7\n4 10\n4 5\n']
Demo Output:
['90.000000000\n', '76859.990000000\n']
Note:
A comment to the first sample:
The first and third passengers get tickets from stop 1 to stop 2. The second passenger doesn't get a ticket. There always is inspection on the segment 1-2 but both passengers have the ticket for it. There never is an inspection on the segment 2-3, that's why the second passenger gets away with the cheating. Our total profit is (0 + 90 / 2 + 90 / 2) = 90.
|
```python
import sys
from math import gcd
from collections import defaultdict
input = sys.stdin.readline
get_int = lambda: int(input().rstrip())
get_arr = lambda: [int(w) for w in input().split()]
get_str = lambda: input().rstrip()
n, m, c = get_arr()
xs = get_arr()
ps = get_arr()
ost = [0] * (n + 1)
for _ in range(m):
a, b = get_arr()
ost[a - 1] += 1
ost[b - 1] -= 1
ct = 0
mn = 0
cv = 0
ans = 0
for i in range(n - 1):
ct += ost[i]
ev = ct * (xs[i + 1] - xs[i]) / 2 - ct * c * ps[i] / 100
# print(i, ct, ev)
cv += ev
ans = max(ans, cv - mn)
mn = min(cv, mn)
print(ans)
```
| 0
|
|
729
|
B
|
Spotlights
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation"
] | null | null |
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
|
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
|
Print one integer — the number of good positions for placing the spotlight.
|
[
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] |
[
"9\n",
"20\n"
] |
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
| 1,000
|
[
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 1\n1\n0\n0",
"output": "2"
},
{
"input": "5 7\n0 0 0 0 0 0 1\n0 0 0 0 0 0 1\n0 0 0 1 0 0 0\n0 0 0 0 0 0 0\n0 0 0 0 0 0 0",
"output": "25"
},
{
"input": "10 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "5 7\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1\n0 1 1 1 1 1 1\n0 0 0 0 0 0 0\n0 1 1 1 1 0 1",
"output": "26"
},
{
"input": "10 20\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "121"
},
{
"input": "1 2\n0 1",
"output": "1"
},
{
"input": "1 2\n1 0",
"output": "1"
},
{
"input": "1 2\n1 1",
"output": "0"
},
{
"input": "2 1\n1\n0",
"output": "1"
},
{
"input": "2 1\n0\n1",
"output": "1"
},
{
"input": "2 1\n1\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "4 4\n1 1 1 1\n1 0 0 1\n1 0 0 1\n1 1 1 1",
"output": "16"
}
] | 1,551,188,786
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 11,059,200
|
n,m=map(int,input().split())
M=[[0 for i in range(m)] for i in range(n)]
L=[list(map(int,input().split())) for i in range(n)]
for i in range(n) :
p=0
for j in range(m) :
if L[i][j]==1 :
p=1
else :
if p==1 :
M[i][j]+=1
p=0
for j in range(m-1,-1,-1) :
if L[i][j]==1 :
p=1
else :
if p==1 :
M[i][j]+=1
for i in range(m) :
p=0
for j in range(n) :
if L[j][i]==1 :
p=1
else :
if p==1 :
M[j][i]+=1
p=0
for j in range(n-1,-1,-1) :
if L[j][i]==1 :
p=1
else :
if p==1 :
M[j][i]+=1
k=0
for i in range(n) :
for j in range(m):
k+=M[i][j]
print(k)
|
Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
|
```python
n,m=map(int,input().split())
M=[[0 for i in range(m)] for i in range(n)]
L=[list(map(int,input().split())) for i in range(n)]
for i in range(n) :
p=0
for j in range(m) :
if L[i][j]==1 :
p=1
else :
if p==1 :
M[i][j]+=1
p=0
for j in range(m-1,-1,-1) :
if L[i][j]==1 :
p=1
else :
if p==1 :
M[i][j]+=1
for i in range(m) :
p=0
for j in range(n) :
if L[j][i]==1 :
p=1
else :
if p==1 :
M[j][i]+=1
p=0
for j in range(n-1,-1,-1) :
if L[j][i]==1 :
p=1
else :
if p==1 :
M[j][i]+=1
k=0
for i in range(n) :
for j in range(m):
k+=M[i][j]
print(k)
```
| 0
|
|
617
|
A
|
Elephant
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
|
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
|
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
|
[
"5\n",
"12\n"
] |
[
"1\n",
"3\n"
] |
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
| 500
|
[
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "534204",
"output": "106841"
},
{
"input": "469569",
"output": "93914"
},
{
"input": "502877",
"output": "100576"
},
{
"input": "942212",
"output": "188443"
},
{
"input": "97",
"output": "20"
},
{
"input": "53",
"output": "11"
},
{
"input": "89",
"output": "18"
},
{
"input": "574",
"output": "115"
},
{
"input": "716",
"output": "144"
},
{
"input": "729",
"output": "146"
},
{
"input": "8901",
"output": "1781"
},
{
"input": "3645",
"output": "729"
},
{
"input": "4426",
"output": "886"
},
{
"input": "46573",
"output": "9315"
},
{
"input": "86380",
"output": "17276"
},
{
"input": "94190",
"output": "18838"
},
{
"input": "999990",
"output": "199998"
},
{
"input": "999991",
"output": "199999"
},
{
"input": "999992",
"output": "199999"
},
{
"input": "999993",
"output": "199999"
},
{
"input": "999994",
"output": "199999"
},
{
"input": "999995",
"output": "199999"
},
{
"input": "999996",
"output": "200000"
},
{
"input": "999997",
"output": "200000"
},
{
"input": "999998",
"output": "200000"
}
] | 1,695,937,600
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
x= int (input())
print ((((x//5)//4)//3)//2)
|
Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
|
```python
x= int (input())
print ((((x//5)//4)//3)//2)
```
| 0
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,645,849,458
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 92
| 0
|
s=input()
l=set(s)
m=max(l)
print(s.count(m)*m)
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
s=input()
l=set(s)
m=max(l)
print(s.count(m)*m)
```
| 3
|
|
580
|
C
|
Kefa and Park
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go.
|
The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree.
|
A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats.
|
[
"4 1\n1 1 0 0\n1 2\n1 3\n1 4\n",
"7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n"
] |
[
"2\n",
"2\n"
] |
Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7.
| 1,500
|
[
{
"input": "4 1\n1 1 0 0\n1 2\n1 3\n1 4",
"output": "2"
},
{
"input": "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7",
"output": "2"
},
{
"input": "3 2\n1 1 1\n1 2\n2 3",
"output": "0"
},
{
"input": "5 2\n1 1 0 1 1\n1 2\n2 3\n3 4\n4 5",
"output": "1"
},
{
"input": "6 1\n1 0 1 1 0 0\n1 2\n1 3\n1 4\n1 5\n1 6",
"output": "3"
},
{
"input": "7 3\n1 1 1 1 1 0 1\n1 2\n1 3\n2 4\n3 5\n5 6\n6 7",
"output": "2"
},
{
"input": "15 2\n1 0 1 0 1 0 0 0 0 0 0 0 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n4 8\n4 9\n5 10\n5 11\n6 12\n6 13\n7 14\n7 15",
"output": "8"
},
{
"input": "2 1\n1 1\n2 1",
"output": "0"
},
{
"input": "12 3\n1 0 1 0 1 1 1 1 0 0 0 0\n6 7\n12 1\n9 7\n1 4\n10 7\n7 1\n11 8\n5 1\n3 7\n5 8\n4 2",
"output": "7"
}
] | 1,693,151,934
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n, m = map(int, input().split())
cats = list(map(int, input().split()))
adjList = dict()
parent = dict()
for i in range (n) :
adjList[i+1] = []
for _ in range (n-1) :
a, b = map(int, input().split())
adjList[a].append(b)
adjList[b].append(a)
if a>b : parent[a] = b
else : parent[b] = a
# print(adjList)
stk = [1]
visited = [0 for _ in range (n)]
currConsecCats = 0
maxConsecCats = 0
ans = 0
catss = [(0,0) for _ in range (n)]
while len(stk)>0 :
current = stk.pop()
currConsecCats = 0
maxConsecCats = 0
if current > 1 :
currConsecCats = catss[parent[current]-1][0]
maxConsecCats = catss[parent[current]-1][1]
if currConsecCats!=0 and cats[current-1]==0 :
maxConsecCats = max(maxConsecCats, currConsecCats)
currConsecCats = 0
else :
currConsecCats += cats[current-1]
maxConsecCats = max(maxConsecCats, currConsecCats)
catss[current-1] = (currConsecCats, maxConsecCats)
if len(adjList[current])==1 and maxConsecCats <= m : ans += 1
visited[current-1] = 1
# print(current, currConsecCats, maxConsecCats)
for neighbor in adjList[current] :
if visited[neighbor-1]==0 : stk.append(neighbor)
print(ans)
|
Title: Kefa and Park
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go.
Input Specification:
The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree.
Output Specification:
A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats.
Demo Input:
['4 1\n1 1 0 0\n1 2\n1 3\n1 4\n', '7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n']
Demo Output:
['2\n', '2\n']
Note:
Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7.
|
```python
n, m = map(int, input().split())
cats = list(map(int, input().split()))
adjList = dict()
parent = dict()
for i in range (n) :
adjList[i+1] = []
for _ in range (n-1) :
a, b = map(int, input().split())
adjList[a].append(b)
adjList[b].append(a)
if a>b : parent[a] = b
else : parent[b] = a
# print(adjList)
stk = [1]
visited = [0 for _ in range (n)]
currConsecCats = 0
maxConsecCats = 0
ans = 0
catss = [(0,0) for _ in range (n)]
while len(stk)>0 :
current = stk.pop()
currConsecCats = 0
maxConsecCats = 0
if current > 1 :
currConsecCats = catss[parent[current]-1][0]
maxConsecCats = catss[parent[current]-1][1]
if currConsecCats!=0 and cats[current-1]==0 :
maxConsecCats = max(maxConsecCats, currConsecCats)
currConsecCats = 0
else :
currConsecCats += cats[current-1]
maxConsecCats = max(maxConsecCats, currConsecCats)
catss[current-1] = (currConsecCats, maxConsecCats)
if len(adjList[current])==1 and maxConsecCats <= m : ans += 1
visited[current-1] = 1
# print(current, currConsecCats, maxConsecCats)
for neighbor in adjList[current] :
if visited[neighbor-1]==0 : stk.append(neighbor)
print(ans)
```
| 0
|
|
999
|
A
|
Mishka and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
|
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
|
Print one integer — the maximum number of problems Mishka can solve.
|
[
"8 4\n4 2 3 1 5 1 6 4\n",
"5 2\n3 1 2 1 3\n",
"5 100\n12 34 55 43 21\n"
] |
[
"5\n",
"0\n",
"5\n"
] |
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
| 0
|
[
{
"input": "8 4\n4 2 3 1 5 1 6 4",
"output": "5"
},
{
"input": "5 2\n3 1 2 1 3",
"output": "0"
},
{
"input": "5 100\n12 34 55 43 21",
"output": "5"
},
{
"input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62",
"output": "100"
},
{
"input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98",
"output": "98"
},
{
"input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2",
"output": "39"
},
{
"input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16",
"output": "99"
},
{
"input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1",
"output": "1"
},
{
"input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "3 2\n1 4 1",
"output": "2"
},
{
"input": "1 2\n100",
"output": "0"
},
{
"input": "7 4\n4 2 3 4 4 2 3",
"output": "7"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "1 2\n15",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "5 3\n3 4 3 2 1",
"output": "4"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "1 5\n1",
"output": "1"
},
{
"input": "6 6\n7 1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n6 5 5 5 5",
"output": "4"
},
{
"input": "1 4\n2",
"output": "1"
},
{
"input": "9 4\n1 2 1 2 4 2 1 2 1",
"output": "9"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n5",
"output": "1"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4",
"output": "71"
},
{
"input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90",
"output": "63"
},
{
"input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8",
"output": "61"
},
{
"input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22",
"output": "72"
},
{
"input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2",
"output": "52"
},
{
"input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90",
"output": "60"
},
{
"input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10",
"output": "56"
},
{
"input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90",
"output": "61"
},
{
"input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24",
"output": "99"
},
{
"input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50",
"output": "99"
},
{
"input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48",
"output": "99"
},
{
"input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57",
"output": "99"
},
{
"input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93",
"output": "99"
},
{
"input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33",
"output": "3"
},
{
"input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51",
"output": "3"
},
{
"input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49",
"output": "6"
},
{
"input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8",
"output": "66"
},
{
"input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95",
"output": "0"
},
{
"input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66",
"output": "0"
},
{
"input": "2 8\n8 8",
"output": "2"
},
{
"input": "1 6\n3",
"output": "1"
},
{
"input": "1 5\n5",
"output": "1"
}
] | 1,616,618,937
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 307,200
|
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
i = 0
j = n-1
z = 0
l = True
r = True
while l or r:
if a[i] <= k:
i += 1
else:
l = False
if a[j] <= k:
j -= 1
z += 1
else:
r = False
if j <= i:
break
if j <= i:
print(n)
else:
print(i+z)
|
Title: Mishka and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Output Specification:
Print one integer — the maximum number of problems Mishka can solve.
Demo Input:
['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n']
Demo Output:
['5\n', '0\n', '5\n']
Note:
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
|
```python
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
i = 0
j = n-1
z = 0
l = True
r = True
while l or r:
if a[i] <= k:
i += 1
else:
l = False
if a[j] <= k:
j -= 1
z += 1
else:
r = False
if j <= i:
break
if j <= i:
print(n)
else:
print(i+z)
```
| 0
|
|
29
|
A
|
Spit Problem
|
PROGRAMMING
| 1,000
|
[
"brute force"
] |
A. Spit Problem
|
2
|
256
|
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
|
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
|
[
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n7594 7675\n-5412 -6872",
"output": "NO"
},
{
"input": "11\n-1536 3809\n-2406 -8438\n-1866 395\n5636 -490\n-6867 -7030\n7525 3575\n-6796 2908\n3884 4629\n-2862 -6122\n-8984 6122\n7137 -326",
"output": "YES"
},
{
"input": "12\n-9765 1132\n-1382 -215\n-9405 7284\n-2040 3947\n-9360 3150\n6425 9386\n806 -2278\n-2121 -7284\n5663 -1608\n-8377 9297\n6245 708\n8470 6024",
"output": "YES"
},
{
"input": "15\n8122 -9991\n-4068 -3386\n8971 3731\n3458 5161\n-8700 7562\n2691 8735\n-1510 -3892\n5183 -3753\n-7018 6637\n-7454 3386\n-818 -6377\n6771 -8647\n-7357 -1246\n-6186 1922\n9889 -3627",
"output": "YES"
},
{
"input": "20\n-5264 6424\n-3664 -7459\n-2780 -9859\n-3317 6842\n5681 -8092\n1555 1904\n-6684 1414\n6593 -1253\n-5708 -1202\n335 1733\n-926 7579\n3459 -1904\n-4486 4006\n6201 3616\n2847 -5255\n8438 7057\n8171 6042\n-9102 3545\n7731 -233\n6264 6563",
"output": "YES"
},
{
"input": "30\n-398 -1774\n313 -6974\n2346 -4657\n8552 -9647\n-5265 1538\n8195 4864\n-5641 -5219\n-1394 8563\n-1190 1992\n-4669 -1156\n7574 256\n9206 -2414\n4140 -549\n-294 2169\n7029 -2871\n3808 -9799\n3141 5690\n4648 -2680\n-5990 9800\n-2299 1697\n6077 -7177\n-400 -9724\n-4644 -2392\n-2198 -9531\n-2105 9386\n-8165 -4201\n-1589 -7916\n2518 -7840\n4173 -6949\n-3368 -9943",
"output": "NO"
}
] | 1,633,041,376
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 6,656,000
|
n=int(input())
lx=set()
for i in range(n):
x,d=map(int,input().split())
if (x+d,x) in lx:
print("YES")
exit()
lx.add((x+d,x))
print("NO")
|
Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
n=int(input())
lx=set()
for i in range(n):
x,d=map(int,input().split())
if (x+d,x) in lx:
print("YES")
exit()
lx.add((x+d,x))
print("NO")
```
| 0
|
181
|
B
|
Number of Triplets
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force"
] | null | null |
You are given *n* points on a plane. All points are different.
Find the number of different groups of three points (*A*,<=*B*,<=*C*) such that point *B* is the middle of segment *AC*.
The groups of three points are considered unordered, that is, if point *B* is the middle of segment *AC*, then groups (*A*,<=*B*,<=*C*) and (*C*,<=*B*,<=*A*) are considered the same.
|
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=3000) — the number of points.
Next *n* lines contain the points. The *i*-th line contains coordinates of the *i*-th point: two space-separated integers *x**i*,<=*y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000).
It is guaranteed that all given points are different.
|
Print the single number — the answer to the problem.
|
[
"3\n1 1\n2 2\n3 3\n",
"3\n0 0\n-1 0\n0 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 1,000
|
[
{
"input": "3\n1 1\n2 2\n3 3",
"output": "1"
},
{
"input": "3\n0 0\n-1 0\n0 1",
"output": "0"
},
{
"input": "4\n0 0\n1 0\n2 0\n3 0",
"output": "2"
},
{
"input": "5\n0 -1\n0 -2\n0 -3\n0 -4\n0 -5",
"output": "4"
},
{
"input": "7\n1 1\n-1 -1\n1 0\n0 1\n-1 0\n0 -1\n0 0",
"output": "3"
},
{
"input": "9\n1 1\n1 0\n0 1\n0 0\n-1 0\n-1 1\n-1 -1\n1 -1\n0 -1",
"output": "8"
},
{
"input": "10\n2 1\n-1 0\n-2 -1\n-1 1\n0 2\n2 -2\n0 0\n-2 -2\n0 -2\n-2 1",
"output": "4"
},
{
"input": "10\n-2 1\n2 -2\n-1 -2\n0 0\n2 -1\n0 -2\n2 2\n0 2\n-1 -1\n1 -2",
"output": "4"
},
{
"input": "10\n0 1\n-1 -1\n1 1\n-1 0\n1 -1\n-2 -1\n-2 2\n-2 0\n0 -2\n0 -1",
"output": "5"
},
{
"input": "10\n2 1\n-1 1\n0 0\n-3 1\n-2 -3\n-1 -2\n-1 -1\n1 2\n3 -2\n0 -2",
"output": "1"
},
{
"input": "20\n-3 -3\n0 4\n-3 1\n1 1\n-1 2\n-4 4\n3 -1\n-3 0\n0 2\n4 0\n2 3\n2 4\n4 -3\n-4 3\n-1 1\n1 3\n-2 4\n1 -2\n1 -1\n3 0",
"output": "10"
},
{
"input": "20\n-3 -3\n0 4\n-3 1\n1 1\n-1 2\n-4 4\n3 -1\n-3 0\n0 2\n4 0\n2 3\n2 4\n4 -3\n-4 3\n-1 1\n1 3\n-2 4\n1 -2\n1 -1\n3 0",
"output": "10"
},
{
"input": "20\n-1 18\n-2 5\n-5 4\n2 -33\n9 -18\n0 0\n11 -22\n2 0\n-1 2\n-4 41\n1 6\n1 -2\n6 -12\n0 1\n-3 6\n3 -6\n3 -8\n-1 4\n2 -5\n1 0",
"output": "21"
},
{
"input": "40\n-8 24\n2 -1\n1 -18\n72 -70\n5 -4\n-308 436\n-19 40\n36 -35\n-178 265\n-1 2\n-7 30\n-1 0\n3 -2\n200 -285\n17 -16\n-35 74\n0 -4\n-86 106\n-1 4\n-7 6\n0 1\n-5 4\n-2 3\n6 -5\n-4 5\n181 -262\n76 -118\n0 0\n-7 18\n-58 104\n-5 6\n-6 12\n-3 4\n1 0\n11 -10\n-86 130\n-3 6\n153 -236\n-183 270\n-33 64",
"output": "57"
},
{
"input": "3\n3 3\n1 2\n1 1",
"output": "0"
},
{
"input": "3\n0 0\n0 -1\n0 1",
"output": "1"
}
] | 1,641,265,774
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 716
| 77,619,200
|
N = 1000
n = int(input())
x = []
y = []
f = [[0 for x in range(3*N)] for y in range(3*N)]
s = 0
for i in range(n):
u, v = map(int, input().split())
x.append(u)
y.append(v)
f[u+N][v+N] = 1
for i in range(n):
for j in range(i+1, n):
u = x[i]+x[j]
v = y[i]+y[j]
s += not(u % 2 or v % 2) and f[u//2+N][v//2+N]
print(s)
|
Title: Number of Triplets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on a plane. All points are different.
Find the number of different groups of three points (*A*,<=*B*,<=*C*) such that point *B* is the middle of segment *AC*.
The groups of three points are considered unordered, that is, if point *B* is the middle of segment *AC*, then groups (*A*,<=*B*,<=*C*) and (*C*,<=*B*,<=*A*) are considered the same.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=3000) — the number of points.
Next *n* lines contain the points. The *i*-th line contains coordinates of the *i*-th point: two space-separated integers *x**i*,<=*y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000).
It is guaranteed that all given points are different.
Output Specification:
Print the single number — the answer to the problem.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n0 0\n-1 0\n0 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
N = 1000
n = int(input())
x = []
y = []
f = [[0 for x in range(3*N)] for y in range(3*N)]
s = 0
for i in range(n):
u, v = map(int, input().split())
x.append(u)
y.append(v)
f[u+N][v+N] = 1
for i in range(n):
for j in range(i+1, n):
u = x[i]+x[j]
v = y[i]+y[j]
s += not(u % 2 or v % 2) and f[u//2+N][v//2+N]
print(s)
```
| 3
|
|
291
|
A
|
Spyke Talks
|
PROGRAMMING
| 800
|
[
"*special",
"implementation",
"sortings"
] | null | null |
Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way.
|
Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place.
|
[
"6\n0 1 7 1 7 10\n",
"3\n1 1 1\n",
"1\n0\n"
] |
[
"2\n",
"-1\n",
"0\n"
] |
In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed.
| 500
|
[
{
"input": "6\n0 1 7 1 7 10",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n2 2 1 1 3",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "10\n4 21 3 21 21 1 1 2 2 3",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "6\n6 6 0 8 0 0",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 0 1 0 1",
"output": "-1"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 0 3 0 0 3 0 0 0 0 0 0 3 0 0 3 0 0 0 0 0 0 0 3 0 0 0 0 0",
"output": "-1"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 0",
"output": "0"
},
{
"input": "2\n1000000000 1000000000",
"output": "1"
},
{
"input": "5\n1 0 0 0 1",
"output": "1"
},
{
"input": "15\n380515742 842209759 945171461 664384656 945171461 474872104 0 0 131648973 131648973 474872104 842209759 664384656 0 380515742",
"output": "6"
},
{
"input": "123\n0 6361 8903 10428 0 258 0 10422 0 0 2642 1958 0 0 0 0 0 8249 1958 0 0 2642 0 0 0 11566 4709 1847 3998 0 1331 0 0 10289 2739 6135 3450 0 0 10994 6069 4337 5854 1331 5854 0 630 630 11244 5928 2706 0 683 214 0 9080 0 0 0 10422 683 11566 10994 0 0 3450 11244 11542 3998 1847 2708 9871 2739 2001 0 12216 6069 0 5928 0 10289 1307 0 1307 8903 0 6361 6135 6632 10428 0 0 632 258 9080 12216 4709 4967 2706 0 11542 2001 6632 0 8249 214 0 10301 4967 10301 7296 7296 10914 2708 4337 0 0 632 0 10914 0 9871 0",
"output": "40"
},
{
"input": "10\n0 3 2 3 2 0 1 3 3 0",
"output": "-1"
},
{
"input": "20\n0 1 2 0 0 0 0 5 3 4 0 0 1 1 3 0 4 0 1 0",
"output": "-1"
},
{
"input": "47\n1 6 0 6 1 1 6 4 3 6 5 3 6 3 2 2 5 1 4 7 3 5 6 1 6 7 4 5 6 3 3 3 7 4 1 6 1 1 7 1 3 1 5 5 1 3 6",
"output": "-1"
},
{
"input": "74\n0 0 0 0 0 37 0 0 0 0 0 0 0 8 0 0 9 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 9 0 7 0 0 0 0 0 0 19 19 0 0 0 0 0 0 0 0 0 0 0 0 17 0 30 0 0 0 0 0 0 30 0 0 0 0 0 0 0 37 0",
"output": "5"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n10000 10000 1 1 10000",
"output": "-1"
}
] | 1,659,244,589
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 92
| 0
|
n = int(input())
vals = [int(x) for x in input().strip().split()]
count = 0
for i in vals:
counter = 0
for j in vals:
if i == j:
counter+=1
count+=1
if counter % 2 != 0 and counter != 1:
print(-1)
exit()
print((count-n)//2)
|
Title: Spyke Talks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way.
Output Specification:
Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place.
Demo Input:
['6\n0 1 7 1 7 10\n', '3\n1 1 1\n', '1\n0\n']
Demo Output:
['2\n', '-1\n', '0\n']
Note:
In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed.
|
```python
n = int(input())
vals = [int(x) for x in input().strip().split()]
count = 0
for i in vals:
counter = 0
for j in vals:
if i == j:
counter+=1
count+=1
if counter % 2 != 0 and counter != 1:
print(-1)
exit()
print((count-n)//2)
```
| 0
|
|
691
|
E
|
Xor-sequences
|
PROGRAMMING
| 1,900
|
[
"matrices"
] | null | null |
You are given *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n*.
A sequence of integers *x*1,<=<=*x*2,<=<=...,<=<=*x**k* is called a "xor-sequence" if for every 1<=<=≤<=<=*i*<=<=≤<=<=*k*<=-<=1 the number of ones in the binary representation of the number *x**i* *x**i*<=<=+<=<=1's is a multiple of 3 and for all 1<=≤<=*i*<=≤<=*k*. The symbol is used for the binary exclusive or operation.
How many "xor-sequences" of length *k* exist? Output the answer modulo 109<=+<=7.
Note if *a*<==<=[1,<=1] and *k*<==<=1 then the answer is 2, because you should consider the ones from *a* as different.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1018) — the number of given integers and the length of the "xor-sequences".
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1018).
|
Print the only integer *c* — the number of "xor-sequences" of length *k* modulo 109<=+<=7.
|
[
"5 2\n15 1 2 4 8\n",
"5 1\n15 1 2 4 8\n"
] |
[
"13\n",
"5\n"
] |
none
| 0
|
[
{
"input": "5 2\n15 1 2 4 8",
"output": "13"
},
{
"input": "5 1\n15 1 2 4 8",
"output": "5"
},
{
"input": "10 1\n44 65 23 44 100 19 19 23 19 40",
"output": "10"
},
{
"input": "10 2\n93 93 85 48 44 98 93 100 98 98",
"output": "52"
},
{
"input": "10 100\n22 0 41 63 22 41 17 22 15 42",
"output": "205668186"
},
{
"input": "10 1000000000\n454240622 216977025 454240622 509843007 509843007 26552516 488949284 708817573 453191950 447767457",
"output": "108319885"
},
{
"input": "100 576460752303423487\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "959277620"
},
{
"input": "100 576460752303423488\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "927761335"
}
] | 1,633,449,818
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 3,000
| 29,491,200
|
import sys
input = sys.stdin.readline
def matmul(a, b, n):
c = []
for i in range(n):
for j in range(n):
x = 0
for k in range(n):
x += a[n * i + k] * b[n * k + j] % mod
x %= mod
c.append(x)
return c
n, k = map(int, input().split())
mod = pow(10, 9) + 7
a = list(map(int, input().split()))
b = [set() for _ in range(n)]
for i in range(n):
a0 = bin(a[i])
for j in range(1, len(a0)):
if a0[-j] == "1":
b[i].add(j)
x = [0] * (n * n)
for i in range(n):
x[n * i + i] = 1
y = []
for bi in b:
for bj in b:
c = 0
for l in bj:
if not l in bi:
c += 1
for l in bi:
if not l in bj:
c += 1
y.append(0 if c % 3 else 1)
p2 = 1
k -= 1
while k:
if k & p2:
k ^= p2
x = matmul(x, y, n)
y = matmul(y, y, n)
p2 *= 2
ans = sum(x) % mod
print(ans)
|
Title: Xor-sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n*.
A sequence of integers *x*1,<=<=*x*2,<=<=...,<=<=*x**k* is called a "xor-sequence" if for every 1<=<=≤<=<=*i*<=<=≤<=<=*k*<=-<=1 the number of ones in the binary representation of the number *x**i* *x**i*<=<=+<=<=1's is a multiple of 3 and for all 1<=≤<=*i*<=≤<=*k*. The symbol is used for the binary exclusive or operation.
How many "xor-sequences" of length *k* exist? Output the answer modulo 109<=+<=7.
Note if *a*<==<=[1,<=1] and *k*<==<=1 then the answer is 2, because you should consider the ones from *a* as different.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1018) — the number of given integers and the length of the "xor-sequences".
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1018).
Output Specification:
Print the only integer *c* — the number of "xor-sequences" of length *k* modulo 109<=+<=7.
Demo Input:
['5 2\n15 1 2 4 8\n', '5 1\n15 1 2 4 8\n']
Demo Output:
['13\n', '5\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
def matmul(a, b, n):
c = []
for i in range(n):
for j in range(n):
x = 0
for k in range(n):
x += a[n * i + k] * b[n * k + j] % mod
x %= mod
c.append(x)
return c
n, k = map(int, input().split())
mod = pow(10, 9) + 7
a = list(map(int, input().split()))
b = [set() for _ in range(n)]
for i in range(n):
a0 = bin(a[i])
for j in range(1, len(a0)):
if a0[-j] == "1":
b[i].add(j)
x = [0] * (n * n)
for i in range(n):
x[n * i + i] = 1
y = []
for bi in b:
for bj in b:
c = 0
for l in bj:
if not l in bi:
c += 1
for l in bi:
if not l in bj:
c += 1
y.append(0 if c % 3 else 1)
p2 = 1
k -= 1
while k:
if k & p2:
k ^= p2
x = matmul(x, y, n)
y = matmul(y, y, n)
p2 *= 2
ans = sum(x) % mod
print(ans)
```
| 0
|
|
900
|
B
|
Position in Fraction
|
PROGRAMMING
| 1,300
|
[
"math",
"number theory"
] | null | null |
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
|
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
|
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
|
[
"1 2 0\n",
"2 3 7\n"
] |
[
"2",
"-1"
] |
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
| 1,000
|
[
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
},
{
"input": "7 31 3",
"output": "15"
},
{
"input": "8880 81608 9",
"output": "161"
},
{
"input": "4942 62768 5",
"output": "122"
},
{
"input": "69168 84860 4",
"output": "107"
},
{
"input": "971 1883 3",
"output": "130"
},
{
"input": "1636 3269 6",
"output": "150"
},
{
"input": "6873 7769 3",
"output": "163"
},
{
"input": "13805 15538 3",
"output": "164"
},
{
"input": "10958 21926 3",
"output": "117"
},
{
"input": "8 51 0",
"output": "10"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "1 9 0",
"output": "-1"
},
{
"input": "53 101 6",
"output": "-1"
},
{
"input": "1 10001 9",
"output": "5"
},
{
"input": "25102 31579 2",
"output": "174"
},
{
"input": "38790 39359 0",
"output": "212"
},
{
"input": "47117 78718 0",
"output": "213"
},
{
"input": "1 57 0",
"output": "1"
},
{
"input": "1 3 0",
"output": "-1"
},
{
"input": "1 100 0",
"output": "1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "99971 99989 0",
"output": "9"
},
{
"input": "567 1580 0",
"output": "6"
},
{
"input": "45 97 0",
"output": "39"
},
{
"input": "35 111 4",
"output": "-1"
},
{
"input": "1 2 5",
"output": "1"
},
{
"input": "1 7 0",
"output": "-1"
}
] | 1,565,868,623
| 23
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n = input()
c = [-9] + [0] * n
p, q = 0, 0
for i in map(int, raw_input().split()):
if p < i:
q = p
p = i
c[p] -= 1
elif q < i:
q = i
c[p] += 1
print c.index(max(c))
|
Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction.
|
```python
n = input()
c = [-9] + [0] * n
p, q = 0, 0
for i in map(int, raw_input().split()):
if p < i:
q = p
p = i
c[p] -= 1
elif q < i:
q = i
c[p] += 1
print c.index(max(c))
```
| -1
|
|
426
|
A
|
Sereja and Mugs
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
|
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
|
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
|
[
"3 4\n1 1 1\n",
"3 4\n3 1 3\n",
"3 4\n4 4 4\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "3 4\n1 1 1",
"output": "YES"
},
{
"input": "3 4\n3 1 3",
"output": "YES"
},
{
"input": "3 4\n4 4 4",
"output": "NO"
},
{
"input": "2 1\n1 10",
"output": "YES"
},
{
"input": "3 12\n5 6 6",
"output": "YES"
},
{
"input": "4 10\n6 3 8 7",
"output": "NO"
},
{
"input": "5 16\n3 3 2 7 9",
"output": "YES"
},
{
"input": "6 38\n9 10 3 8 10 6",
"output": "YES"
},
{
"input": "7 12\n4 4 5 2 2 4 9",
"output": "NO"
},
{
"input": "8 15\n8 10 4 2 10 9 7 6",
"output": "NO"
},
{
"input": "9 22\n1 3 5 9 7 6 1 10 1",
"output": "NO"
},
{
"input": "10 30\n9 10 4 5 5 7 1 7 7 2",
"output": "NO"
},
{
"input": "38 83\n9 9 3 10 2 4 6 10 9 5 1 8 7 4 7 2 6 5 3 1 10 8 4 8 3 7 1 2 7 6 8 6 5 2 3 1 1 2",
"output": "NO"
},
{
"input": "84 212\n6 2 3 1 2 7 5 1 7 2 9 10 9 5 2 5 4 10 9 9 1 9 8 8 9 4 9 4 8 2 1 8 4 5 10 7 6 2 1 10 10 7 9 4 5 9 5 10 10 3 6 6 4 4 4 8 5 4 9 1 9 9 1 7 9 2 10 9 10 8 3 3 9 3 9 10 1 8 9 2 6 9 7 2",
"output": "NO"
},
{
"input": "8 50\n8 8 8 4 4 6 10 10",
"output": "YES"
},
{
"input": "7 24\n1 4 9 1 2 3 6",
"output": "YES"
},
{
"input": "47 262\n3 7 6 4 10 3 5 7 2 9 3 2 2 10 8 7 3 10 6 3 1 1 4 10 2 9 2 10 6 4 3 6 3 6 9 7 8 8 3 3 10 5 2 10 7 10 9",
"output": "YES"
},
{
"input": "42 227\n3 6 1 9 4 10 4 10 7 8 10 10 8 7 10 4 6 8 7 7 6 9 3 6 5 5 2 7 2 7 4 4 6 6 4 3 9 3 6 4 7 2",
"output": "NO"
},
{
"input": "97 65\n3 10 2 6 1 4 7 5 10 3 10 4 5 5 1 6 10 7 4 5 3 9 9 8 6 9 2 3 6 8 5 5 5 5 5 3 10 4 1 8 8 9 8 4 1 4 9 3 6 3 1 4 8 3 10 8 6 4 5 4 3 2 2 4 3 6 4 6 2 3 3 3 7 5 1 8 1 4 5 1 1 6 4 2 1 7 8 6 1 1 5 6 5 10 6 7 5",
"output": "NO"
},
{
"input": "94 279\n2 5 9 5 10 3 1 8 1 7 1 8 1 6 7 8 4 9 5 10 3 7 6 8 8 5 6 8 10 9 4 1 3 3 4 7 8 2 6 6 5 1 3 7 1 7 2 2 2 8 4 1 1 5 9 4 1 2 3 10 1 4 9 9 6 8 8 1 9 10 4 1 8 5 8 9 4 8 2 1 1 9 4 5 6 1 2 5 6 7 3 1 4 6",
"output": "NO"
},
{
"input": "58 70\n8 2 10 2 7 3 8 3 8 7 6 2 4 10 10 6 10 3 7 6 4 3 5 5 5 3 8 10 3 4 8 4 2 6 8 9 6 9 4 3 5 2 2 6 10 6 2 1 7 5 6 4 1 9 10 2 4 5",
"output": "NO"
},
{
"input": "6 14\n3 9 2 1 4 2",
"output": "YES"
},
{
"input": "78 400\n5 9 3 4 7 4 1 4 6 3 9 1 8 3 3 6 10 2 1 9 6 1 8 10 1 6 4 5 2 1 5 9 6 10 3 6 5 2 4 10 6 9 3 8 10 7 2 8 8 2 10 1 4 5 2 8 6 4 4 3 5 2 3 10 1 9 8 5 6 7 9 1 8 8 5 4 2 4",
"output": "YES"
},
{
"input": "41 181\n5 3 10 4 2 5 9 3 1 6 6 10 4 3 9 8 5 9 2 5 4 6 6 3 7 9 10 3 10 6 10 5 6 1 6 9 9 1 2 4 3",
"output": "NO"
},
{
"input": "2 4\n4 4",
"output": "YES"
},
{
"input": "29 71\n4 8 9 4 8 10 4 10 2 9 3 9 1 2 9 5 9 7 1 10 4 1 1 9 8 7 4 6 7",
"output": "NO"
},
{
"input": "49 272\n4 10 8 7 5 6 9 7 2 6 6 2 10 7 5 6 5 3 6 4 3 7 9 3 7 7 4 10 5 6 7 3 6 4 6 7 7 2 5 5 7 3 7 9 3 6 6 2 1",
"output": "YES"
},
{
"input": "91 486\n1 3 5 4 4 7 3 9 3 4 5 4 5 4 7 9 5 8 4 10 9 1 1 9 9 1 6 2 5 4 7 4 10 3 2 10 9 3 4 5 1 3 4 2 10 9 10 9 10 2 4 6 2 5 3 6 4 9 10 3 9 8 1 2 5 9 2 10 4 6 10 8 10 9 1 2 5 8 6 6 6 1 10 3 9 3 5 6 1 5 5",
"output": "YES"
},
{
"input": "80 78\n1 9 4 9 8 3 7 10 4 9 2 1 4 4 9 5 9 1 2 6 5 2 4 8 4 6 9 6 7 10 1 9 10 4 7 1 7 10 8 9 10 5 2 6 7 7 7 7 7 8 2 5 1 7 2 3 2 5 10 6 3 4 5 2 6 3 4 2 7 9 9 3 8 8 2 3 7 1 5 10",
"output": "NO"
},
{
"input": "53 245\n5 6 9 9 2 3 2 5 10 9 3 5 6 3 10 10 9 4 9 7 10 9 7 7 3 4 9 3 7 3 8 6 8 9 3 8 9 1 3 1 9 10 3 9 3 1 6 6 3 8 7 8 9",
"output": "NO"
},
{
"input": "50 271\n6 9 10 1 1 1 8 3 6 6 3 2 5 9 7 5 7 9 10 9 4 6 6 2 6 6 9 5 1 6 5 8 3 2 5 10 10 1 4 1 4 6 1 8 7 8 9 4 7 5",
"output": "YES"
},
{
"input": "38 214\n5 8 4 5 1 9 9 2 6 3 4 3 5 7 7 7 3 10 1 5 10 4 2 2 10 10 6 6 6 7 1 6 10 5 7 4 5 10",
"output": "YES"
},
{
"input": "100 1\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 100\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 989\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 990\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "100 1000\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "2 1\n1 1",
"output": "YES"
},
{
"input": "2 1\n2 2",
"output": "NO"
},
{
"input": "2 1000\n1 1",
"output": "YES"
},
{
"input": "10 89\n10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "10 44\n1 10 2 3 4 5 6 7 8 9",
"output": "NO"
}
] | 1,585,130,826
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 42
| 155
| 0
|
n, s = map(int, input().split())
a = [int(s) for s in input().split()]
a = sorted(a)
if sum(a) - a[len(a)-1] <= s:
print('YES')
else:
print('NO')
|
Title: Sereja and Mugs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
Input Specification:
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
Output Specification:
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
Demo Input:
['3 4\n1 1 1\n', '3 4\n3 1 3\n', '3 4\n4 4 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n, s = map(int, input().split())
a = [int(s) for s in input().split()]
a = sorted(a)
if sum(a) - a[len(a)-1] <= s:
print('YES')
else:
print('NO')
```
| 3
|
|
748
|
A
|
Santa Claus and a Place in a Class
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
|
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
|
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
|
[
"4 3 9\n",
"4 3 24\n",
"2 4 4\n"
] |
[
"2 2 L\n",
"4 3 R\n",
"1 2 R\n"
] |
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
| 500
|
[
{
"input": "4 3 9",
"output": "2 2 L"
},
{
"input": "4 3 24",
"output": "4 3 R"
},
{
"input": "2 4 4",
"output": "1 2 R"
},
{
"input": "3 10 24",
"output": "2 2 R"
},
{
"input": "10 3 59",
"output": "10 3 L"
},
{
"input": "10000 10000 160845880",
"output": "8043 2940 R"
},
{
"input": "1 1 1",
"output": "1 1 L"
},
{
"input": "1 1 2",
"output": "1 1 R"
},
{
"input": "1 10000 1",
"output": "1 1 L"
},
{
"input": "1 10000 20000",
"output": "1 10000 R"
},
{
"input": "10000 1 1",
"output": "1 1 L"
},
{
"input": "10000 1 10000",
"output": "5000 1 R"
},
{
"input": "10000 1 20000",
"output": "10000 1 R"
},
{
"input": "3 2 1",
"output": "1 1 L"
},
{
"input": "3 2 2",
"output": "1 1 R"
},
{
"input": "3 2 3",
"output": "1 2 L"
},
{
"input": "3 2 4",
"output": "1 2 R"
},
{
"input": "3 2 5",
"output": "2 1 L"
},
{
"input": "3 2 6",
"output": "2 1 R"
},
{
"input": "3 2 7",
"output": "2 2 L"
},
{
"input": "3 2 8",
"output": "2 2 R"
},
{
"input": "3 2 9",
"output": "3 1 L"
},
{
"input": "3 2 10",
"output": "3 1 R"
},
{
"input": "3 2 11",
"output": "3 2 L"
},
{
"input": "3 2 12",
"output": "3 2 R"
},
{
"input": "300 2000 1068628",
"output": "268 314 R"
},
{
"input": "300 2000 584756",
"output": "147 378 R"
},
{
"input": "300 2000 268181",
"output": "68 91 L"
},
{
"input": "10000 9999 186450844",
"output": "9324 4745 R"
},
{
"input": "10000 9999 197114268",
"output": "9857 6990 R"
},
{
"input": "10000 9999 112390396",
"output": "5621 818 R"
},
{
"input": "10000 10000 1",
"output": "1 1 L"
},
{
"input": "10000 10000 2",
"output": "1 1 R"
},
{
"input": "10000 10000 100000001",
"output": "5001 1 L"
},
{
"input": "10000 10000 199999999",
"output": "10000 10000 L"
},
{
"input": "10000 10000 200000000",
"output": "10000 10000 R"
},
{
"input": "1 2 1",
"output": "1 1 L"
},
{
"input": "1 2 2",
"output": "1 1 R"
},
{
"input": "1 2 3",
"output": "1 2 L"
},
{
"input": "1 2 4",
"output": "1 2 R"
},
{
"input": "2 1 1",
"output": "1 1 L"
},
{
"input": "2 1 2",
"output": "1 1 R"
},
{
"input": "2 1 3",
"output": "2 1 L"
},
{
"input": "2 1 4",
"output": "2 1 R"
},
{
"input": "4 3 7",
"output": "2 1 L"
},
{
"input": "1 1 1",
"output": "1 1 L"
}
] | 1,538,411,558
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 124
| 0
|
s1 = input()
li=s1.split()
m = int(s1[0])
n = int(s1[2])
san = int(li[2])
# print(m,n,san)
sumdes = 0
for i in range(m):
sumdes = sumdes + 2*n
if san <= sumdes:
lane = i + 1
break
# print(lane,sumdes)
desk = n
for i in range(n):
if san > sumdes-2:
break
sumdes = sumdes - 2
desk = desk - 1
if san%2 == 1:
taraf = 'L'
else:
taraf= 'R'
print(lane,desk,taraf)
|
Title: Santa Claus and a Place in a Class
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output Specification:
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Demo Input:
['4 3 9\n', '4 3 24\n', '2 4 4\n']
Demo Output:
['2 2 L\n', '4 3 R\n', '1 2 R\n']
Note:
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
|
```python
s1 = input()
li=s1.split()
m = int(s1[0])
n = int(s1[2])
san = int(li[2])
# print(m,n,san)
sumdes = 0
for i in range(m):
sumdes = sumdes + 2*n
if san <= sumdes:
lane = i + 1
break
# print(lane,sumdes)
desk = n
for i in range(n):
if san > sumdes-2:
break
sumdes = sumdes - 2
desk = desk - 1
if san%2 == 1:
taraf = 'L'
else:
taraf= 'R'
print(lane,desk,taraf)
```
| -1
|
|
991
|
A
|
If at first you don't succeed...
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
|
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
|
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
|
[
"10 10 5 20\n",
"2 2 0 4\n",
"2 2 2 1\n"
] |
[
"5",
"-1",
"-1"
] |
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.
| 500
|
[
{
"input": "10 10 5 20",
"output": "5"
},
{
"input": "2 2 0 4",
"output": "-1"
},
{
"input": "2 2 2 1",
"output": "-1"
},
{
"input": "98 98 97 100",
"output": "1"
},
{
"input": "1 5 2 10",
"output": "-1"
},
{
"input": "5 1 2 10",
"output": "-1"
},
{
"input": "6 7 5 8",
"output": "-1"
},
{
"input": "6 7 5 9",
"output": "1"
},
{
"input": "6 7 5 7",
"output": "-1"
},
{
"input": "50 50 1 100",
"output": "1"
},
{
"input": "8 3 2 12",
"output": "3"
},
{
"input": "10 19 6 25",
"output": "2"
},
{
"input": "1 0 0 99",
"output": "98"
},
{
"input": "0 1 0 98",
"output": "97"
},
{
"input": "1 1 0 97",
"output": "95"
},
{
"input": "1 1 1 96",
"output": "95"
},
{
"input": "0 0 0 0",
"output": "-1"
},
{
"input": "100 0 0 0",
"output": "-1"
},
{
"input": "0 100 0 0",
"output": "-1"
},
{
"input": "100 100 0 0",
"output": "-1"
},
{
"input": "0 0 100 0",
"output": "-1"
},
{
"input": "100 0 100 0",
"output": "-1"
},
{
"input": "0 100 100 0",
"output": "-1"
},
{
"input": "100 100 100 0",
"output": "-1"
},
{
"input": "0 0 0 100",
"output": "100"
},
{
"input": "100 0 0 100",
"output": "-1"
},
{
"input": "0 100 0 100",
"output": "-1"
},
{
"input": "100 100 0 100",
"output": "-1"
},
{
"input": "0 0 100 100",
"output": "-1"
},
{
"input": "100 0 100 100",
"output": "-1"
},
{
"input": "0 100 100 100",
"output": "-1"
},
{
"input": "100 100 100 100",
"output": "-1"
},
{
"input": "10 45 7 52",
"output": "4"
},
{
"input": "38 1 1 68",
"output": "30"
},
{
"input": "8 45 2 67",
"output": "16"
},
{
"input": "36 36 18 65",
"output": "11"
},
{
"input": "10 30 8 59",
"output": "27"
},
{
"input": "38 20 12 49",
"output": "3"
},
{
"input": "8 19 4 38",
"output": "15"
},
{
"input": "36 21 17 72",
"output": "32"
},
{
"input": "14 12 12 89",
"output": "75"
},
{
"input": "38 6 1 44",
"output": "1"
},
{
"input": "13 4 6 82",
"output": "-1"
},
{
"input": "5 3 17 56",
"output": "-1"
},
{
"input": "38 5 29 90",
"output": "-1"
},
{
"input": "22 36 18 55",
"output": "15"
},
{
"input": "13 0 19 75",
"output": "-1"
},
{
"input": "62 65 10 89",
"output": "-1"
},
{
"input": "2 29 31 72",
"output": "-1"
},
{
"input": "1 31 19 55",
"output": "-1"
},
{
"input": "1 25 28 88",
"output": "-1"
},
{
"input": "34 32 28 33",
"output": "-1"
},
{
"input": "43 36 1 100",
"output": "22"
},
{
"input": "16 39 55 70",
"output": "-1"
},
{
"input": "2 3 0 91",
"output": "86"
},
{
"input": "55 29 12 48",
"output": "-1"
},
{
"input": "7 33 20 88",
"output": "-1"
},
{
"input": "40 38 27 99",
"output": "48"
},
{
"input": "18 28 14 84",
"output": "52"
},
{
"input": "34 25 25 92",
"output": "58"
},
{
"input": "4 24 5 76",
"output": "-1"
},
{
"input": "5 22 16 96",
"output": "-1"
},
{
"input": "1 1 0 4",
"output": "2"
},
{
"input": "5 5 3 1",
"output": "-1"
},
{
"input": "0 0 0 1",
"output": "1"
},
{
"input": "2 3 0 8",
"output": "3"
},
{
"input": "5 5 2 5",
"output": "-1"
},
{
"input": "1 2 1 3",
"output": "1"
},
{
"input": "3 0 0 4",
"output": "1"
},
{
"input": "0 0 0 5",
"output": "5"
},
{
"input": "5 5 0 3",
"output": "-1"
},
{
"input": "5 6 1 7",
"output": "-1"
},
{
"input": "10 10 10 11",
"output": "1"
},
{
"input": "0 0 0 10",
"output": "10"
},
{
"input": "5 15 5 30",
"output": "15"
},
{
"input": "3 2 0 7",
"output": "2"
}
] | 1,685,201,943
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
a,b,c,n = map(int,input().split())
x = a+b-c
if n-x>0:
print(n-x)
else:
print(-1)
|
Title: If at first you don't succeed...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
Input Specification:
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
Output Specification:
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
Demo Input:
['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n']
Demo Output:
['5', '-1', '-1']
Note:
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.
|
```python
a,b,c,n = map(int,input().split())
x = a+b-c
if n-x>0:
print(n-x)
else:
print(-1)
```
| 0
|
|
989
|
B
|
A Tide of Riverscape
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"strings"
] | null | null |
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
|
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
|
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
|
[
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] |
[
"1000100010\n",
"1001101000\n",
"No\n"
] |
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
| 1,000
|
[
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"input": "17 10\n..1.100..1..0.100",
"output": "00101000010000100"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "3 1\n.0.",
"output": "001"
},
{
"input": "3 1\n00.",
"output": "001"
},
{
"input": "3 2\n0..",
"output": "001"
},
{
"input": "3 2\n0.0",
"output": "No"
},
{
"input": "3 2\n1..",
"output": "100"
},
{
"input": "3 2\n.1.",
"output": "011"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "3 3\n1..",
"output": "No"
},
{
"input": "3 3\n.00",
"output": "No"
},
{
"input": "5 3\n0.000",
"output": "01000"
},
{
"input": "10 6\n10010.1001",
"output": "No"
},
{
"input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.",
"output": "000001000011001000010001000111000110000000010000110010000100010001110001101"
},
{
"input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001",
"output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001"
},
{
"input": "5 4\n.101.",
"output": "01011"
},
{
"input": "4 2\n101.",
"output": "1011"
},
{
"input": "5 4\n.1011",
"output": "01011"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "5 3\n00.11",
"output": "00011"
},
{
"input": "10 8\n1111.00000",
"output": "1111000000"
},
{
"input": "10 3\n11111111.1",
"output": "1111111101"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "6 4\n11..10",
"output": "110010"
},
{
"input": "4 2\n.111",
"output": "0111"
},
{
"input": "3 2\n01.",
"output": "011"
},
{
"input": "5 4\n10.00",
"output": "10000"
},
{
"input": "10 9\n1........0",
"output": "1000000000"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "8 4\n111111..",
"output": "11111100"
},
{
"input": "3 2\n0.1",
"output": "001"
},
{
"input": "4 1\n111.",
"output": "1110"
},
{
"input": "3 1\n01.",
"output": "010"
},
{
"input": "10 7\n000....111",
"output": "0000000111"
}
] | 1,609,345,910
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 59
| 155
| 1,331,200
|
n,p=map(int, input().split())
s=input()
t=s.replace('.', '0')
if(all(t[i] == t[i + p] for i in range(n - p))):
k=-1
for r in range(p - 1, -1, -1):
x=s[r::p].rfind('.')
if(x!=-1 and s[r::p]!='.'):
k=max(k,x*p+r)
if(k==-1):
print('No')
else:
print(t[:k]+'1'+t[(k+1):])
else:
print(t)
|
Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
|
```python
n,p=map(int, input().split())
s=input()
t=s.replace('.', '0')
if(all(t[i] == t[i + p] for i in range(n - p))):
k=-1
for r in range(p - 1, -1, -1):
x=s[r::p].rfind('.')
if(x!=-1 and s[r::p]!='.'):
k=max(k,x*p+r)
if(k==-1):
print('No')
else:
print(t[:k]+'1'+t[(k+1):])
else:
print(t)
```
| 3
|
|
576
|
A
|
Vasya and Petya's Game
|
PROGRAMMING
| 1,500
|
[
"math",
"number theory"
] | null | null |
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
|
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
|
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
|
[
"4\n",
"6\n"
] |
[
"3\n2 4 3 \n",
"4\n2 4 3 5 \n"
] |
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.
| 500
|
[
{
"input": "4",
"output": "3\n2 4 3 "
},
{
"input": "6",
"output": "4\n2 4 3 5 "
},
{
"input": "1",
"output": "0"
},
{
"input": "15",
"output": "9\n2 4 8 3 9 5 7 11 13 "
},
{
"input": "19",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "20",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "37",
"output": "19\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 "
},
{
"input": "211",
"output": "61\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 169 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 "
},
{
"input": "557",
"output": "123\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 "
},
{
"input": "907",
"output": "179\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..."
},
{
"input": "953",
"output": "186\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..."
},
{
"input": "289",
"output": "78\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 "
},
{
"input": "400",
"output": "97\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 "
},
{
"input": "900",
"output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..."
},
{
"input": "625",
"output": "136\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 "
},
{
"input": "729",
"output": "152\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..."
},
{
"input": "784",
"output": "160\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 ..."
},
{
"input": "31",
"output": "17\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 31 "
},
{
"input": "44",
"output": "21\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 41 43 "
},
{
"input": "160",
"output": "50\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 "
},
{
"input": "322",
"output": "83\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 "
},
{
"input": "894",
"output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 ..."
},
{
"input": "998",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..."
},
{
"input": "1000",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..."
},
{
"input": "1000",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 ..."
},
{
"input": "30",
"output": "16\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 "
}
] | 1,572,446,527
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 93
| 0
|
from math import sqrt
n = int(input())
ans = []
my_dict = {}
def is_prime(n):
if n == 2:
return True
elif n < 2:
return False
for i in range(2, int(sqrt(n))+1):
if n%i == 0:
return False
return True
def divisors(n):
divisores = []
for i in range(2, n):
if n%i == 0:
divisores.append(i)
return divisores
for i in range(2, n+1):
divisores = divisors(i)
if len(divisores) > 1:
for j in divisores:
if j not in my_dict and is_prime(j):
my_dict[j] = 1
ans.append(str(j))
#my_dict = add_multiples_to_my_dict(my_dict, j, ans)
else:
continue
else:
my_dict[i] = 1
ans.append(str(i))
print(len(ans))
print(" ".join(ans))
|
Title: Vasya and Petya's Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
Input Specification:
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
Output Specification:
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Demo Input:
['4\n', '6\n']
Demo Output:
['3\n2 4 3 \n', '4\n2 4 3 5 \n']
Note:
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.
|
```python
from math import sqrt
n = int(input())
ans = []
my_dict = {}
def is_prime(n):
if n == 2:
return True
elif n < 2:
return False
for i in range(2, int(sqrt(n))+1):
if n%i == 0:
return False
return True
def divisors(n):
divisores = []
for i in range(2, n):
if n%i == 0:
divisores.append(i)
return divisores
for i in range(2, n+1):
divisores = divisors(i)
if len(divisores) > 1:
for j in divisores:
if j not in my_dict and is_prime(j):
my_dict[j] = 1
ans.append(str(j))
#my_dict = add_multiples_to_my_dict(my_dict, j, ans)
else:
continue
else:
my_dict[i] = 1
ans.append(str(i))
print(len(ans))
print(" ".join(ans))
```
| 0
|
|
393
|
A
|
Nineteen
|
PROGRAMMING
| 0
|
[] | null | null |
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
|
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
|
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
|
[
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] |
[
"2",
"2",
"2"
] |
none
| 500
|
[
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihnrhimihrjinjiehmtjimnrss",
"output": "1"
},
{
"input": "rrrteiehtesisntnjirtitijnjjjthrsmhtneirjimniemmnrhirssjnhetmnmjejjnjjritjttnnrhnjs",
"output": "2"
},
{
"input": "mmrehtretseihsrjmtsenemniehssnisijmsnntesismmtmthnsieijjjnsnhisi",
"output": "2"
},
{
"input": "hshretttnntmmiertrrnjihnrmshnthirnnirrheinnnrjiirshthsrsijtrrtrmnjrrjnresnintnmtrhsnjrinsseimn",
"output": "1"
},
{
"input": "snmmensntritetnmmmerhhrmhnehehtesmhthseemjhmnrti",
"output": "2"
},
{
"input": "rmeetriiitijmrenmeiijt",
"output": "0"
},
{
"input": "ihimeitimrmhriemsjhrtjtijtesmhemnmmrsetmjttthtjhnnmirtimne",
"output": "1"
},
{
"input": "rhtsnmnesieernhstjnmmirthhieejsjttsiierhihhrrijhrrnejsjer",
"output": "2"
},
{
"input": "emmtjsjhretehmiiiestmtmnmissjrstnsnjmhimjmststsitemtttjrnhsrmsenjtjim",
"output": "2"
},
{
"input": "nmehhjrhirniitshjtrrtitsjsntjhrstjehhhrrerhemehjeermhmhjejjesnhsiirheijjrnrjmminneeehtm",
"output": "3"
},
{
"input": "hsntijjetmehejtsitnthietssmeenjrhhetsnjrsethisjrtrhrierjtmimeenjnhnijeesjttrmn",
"output": "3"
},
{
"input": "jnirirhmirmhisemittnnsmsttesjhmjnsjsmntisheneiinsrjsjirnrmnjmjhmistntersimrjni",
"output": "1"
},
{
"input": "neithjhhhtmejjnmieishethmtetthrienrhjmjenrmtejerernmthmsnrthhtrimmtmshm",
"output": "2"
},
{
"input": "sithnrsnemhijsnjitmijjhejjrinejhjinhtisttteermrjjrtsirmessejireihjnnhhemiirmhhjeet",
"output": "3"
},
{
"input": "jrjshtjstteh",
"output": "0"
},
{
"input": "jsihrimrjnnmhttmrtrenetimemjnshnimeiitmnmjishjjneisesrjemeshjsijithtn",
"output": "2"
},
{
"input": "hhtjnnmsemermhhtsstejehsssmnesereehnnsnnremjmmieethmirjjhn",
"output": "2"
},
{
"input": "tmnersmrtsehhntsietttrehrhneiireijnijjejmjhei",
"output": "1"
},
{
"input": "mtstiresrtmesritnjriirehtermtrtseirtjrhsejhhmnsineinsjsin",
"output": "2"
},
{
"input": "ssitrhtmmhtnmtreijteinimjemsiiirhrttinsnneshintjnin",
"output": "1"
},
{
"input": "rnsrsmretjiitrjthhritniijhjmm",
"output": "0"
},
{
"input": "hntrteieimrimteemenserntrejhhmijmtjjhnsrsrmrnsjseihnjmehtthnnithirnhj",
"output": "3"
},
{
"input": "nmmtsmjrntrhhtmimeresnrinstjnhiinjtnjjjnthsintmtrhijnrnmtjihtinmni",
"output": "0"
},
{
"input": "eihstiirnmteejeehimttrijittjsntjejmessstsemmtristjrhenithrrsssihnthheehhrnmimssjmejjreimjiemrmiis",
"output": "2"
},
{
"input": "srthnimimnemtnmhsjmmmjmmrsrisehjseinemienntetmitjtnnneseimhnrmiinsismhinjjnreehseh",
"output": "3"
},
{
"input": "etrsmrjehntjjimjnmsresjnrthjhehhtreiijjminnheeiinseenmmethiemmistsei",
"output": "3"
},
{
"input": "msjeshtthsieshejsjhsnhejsihisijsertenrshhrthjhiirijjneinjrtrmrs",
"output": "1"
},
{
"input": "mehsmstmeejrhhsjihntjmrjrihssmtnensttmirtieehimj",
"output": "1"
},
{
"input": "mmmsermimjmrhrhejhrrejermsneheihhjemnehrhihesnjsehthjsmmjeiejmmnhinsemjrntrhrhsmjtttsrhjjmejj",
"output": "2"
},
{
"input": "rhsmrmesijmmsnsmmhertnrhsetmisshriirhetmjihsmiinimtrnitrseii",
"output": "1"
},
{
"input": "iihienhirmnihh",
"output": "0"
},
{
"input": "ismtthhshjmhisssnmnhe",
"output": "0"
},
{
"input": "rhsmnrmhejshinnjrtmtsssijimimethnm",
"output": "0"
},
{
"input": "eehnshtiriejhiirntminrirnjihmrnittnmmnjejjhjtennremrnssnejtntrtsiejjijisermj",
"output": "3"
},
{
"input": "rnhmeesnhttrjintnhnrhristjrthhrmehrhjmjhjehmstrijemjmmistes",
"output": "2"
},
{
"input": "ssrmjmjeeetrnimemrhimes",
"output": "0"
},
{
"input": "n",
"output": "0"
},
{
"input": "ni",
"output": "0"
},
{
"input": "nine",
"output": "0"
},
{
"input": "nineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteen",
"output": "13"
},
{
"input": "ninetee",
"output": "0"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwynd",
"output": "0"
},
{
"input": "zenudggmyopddhszhrbmftgzmjorabhgojdtfnzxjkayjlkgczsyshczutkdch",
"output": "0"
},
{
"input": "rtzxovxqfapkdmelxiyjroohufhbakpmmvaxq",
"output": "0"
},
{
"input": "zninetneeineteeniwnteeennieteenineteenineteenineteenineteenineteenineteenineteenineteeninetzeenz",
"output": "13"
},
{
"input": "nnnnnnniiiiiiiiiiiitttttttttteeeeeeeeeeeeeeeeee",
"output": "3"
},
{
"input": "ttttiiiieeeeeeeeeeeennnnnnnnn",
"output": "4"
},
{
"input": "ttttttttteeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiinnnnnnn",
"output": "3"
},
{
"input": "nnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiiiiitttttttttttttttttttt",
"output": "4"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiittttttttttttttttnnnnnnn",
"output": "3"
},
{
"input": "nineeen",
"output": "0"
},
{
"input": "nineteeeeeeeeeeeeeeeeettttttttttttttttttiiiiiiiiiiiiiiiiii",
"output": "0"
},
{
"input": "nineteenieteenieteenieteenieteenieteenieteen",
"output": "4"
},
{
"input": "nineteenineteenineteenineteenineteen",
"output": "5"
}
] | 1,602,611,917
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 124
| 0
|
import math
a= input()
b= a.count('n')
c= a.count('i')
d= a.count('e')
e= a.count('t')
print( min(math.ceil((b+2)//3),c,d//3,e))
|
Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none
|
```python
import math
a= input()
b= a.count('n')
c= a.count('i')
d= a.count('e')
e= a.count('t')
print( min(math.ceil((b+2)//3),c,d//3,e))
```
| 0
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,696,346,428
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 186
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Tue Oct 3 23:19:43 2023
@author: 袁兆瑞
"""
a = input()
print(a[0].capitalize() + a[1:])
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Tue Oct 3 23:19:43 2023
@author: 袁兆瑞
"""
a = input()
print(a[0].capitalize() + a[1:])
```
| 3
|
|
10
|
A
|
Power Consumption Calculation
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Power Consumption Calculation
|
1
|
256
|
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
|
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
|
Output the answer to the problem.
|
[
"1 3 2 1 5 10\n0 10\n",
"2 8 4 2 5 10\n20 30\n50 100\n"
] |
[
"30",
"570"
] |
none
| 0
|
[
{
"input": "1 3 2 1 5 10\n0 10",
"output": "30"
},
{
"input": "2 8 4 2 5 10\n20 30\n50 100",
"output": "570"
},
{
"input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436",
"output": "8445"
},
{
"input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945",
"output": "52870"
},
{
"input": "5 41 20 33 43 4\n46 465\n598 875\n967 980\n1135 1151\n1194 1245",
"output": "46995"
},
{
"input": "6 88 28 100 53 36\n440 445\n525 614\n644 844\n1238 1261\n1305 1307\n1425 1434",
"output": "85540"
},
{
"input": "7 46 61 55 28 59\n24 26\n31 61\n66 133\n161 612\n741 746\n771 849\n1345 1357",
"output": "67147"
},
{
"input": "8 83 18 30 28 5\n196 249\n313 544\n585 630\n718 843\n1040 1194\n1207 1246\n1268 1370\n1414 1422",
"output": "85876"
},
{
"input": "9 31 65 27 53 54\n164 176\n194 210\n485 538\n617 690\n875 886\n888 902\n955 957\n1020 1200\n1205 1282",
"output": "38570"
},
{
"input": "30 3 1 58 44 7\n11 13\n14 32\n37 50\n70 74\n101 106\n113 129\n184 195\n197 205\n213 228\n370 394\n443 446\n457 460\n461 492\n499 585\n602 627\n709 776\n812 818\n859 864\n910 913\n918 964\n1000 1010\n1051 1056\n1063 1075\n1106 1145\n1152 1189\n1211 1212\n1251 1259\n1272 1375\n1412 1417\n1430 1431",
"output": "11134"
},
{
"input": "30 42 3 76 28 26\n38 44\n55 66\n80 81\n84 283\n298 314\n331 345\n491 531\n569 579\n597 606\n612 617\n623 701\n723 740\n747 752\n766 791\n801 827\n842 846\n853 891\n915 934\n945 949\n955 964\n991 1026\n1051 1059\n1067 1179\n1181 1191\n1214 1226\n1228 1233\n1294 1306\n1321 1340\n1371 1374\n1375 1424",
"output": "59043"
},
{
"input": "30 46 5 93 20 46\n12 34\n40 41\n54 58\n100 121\n162 182\n220 349\n358 383\n390 398\n401 403\n408 409\n431 444\n466 470\n471 535\n556 568\n641 671\n699 709\n767 777\n786 859\n862 885\n912 978\n985 997\n1013 1017\n1032 1038\n1047 1048\n1062 1080\n1094 1097\n1102 1113\n1122 1181\n1239 1280\n1320 1369",
"output": "53608"
},
{
"input": "30 50 74 77 4 57\n17 23\n24 61\n67 68\n79 87\n93 101\n104 123\n150 192\n375 377\n398 414\n461 566\n600 633\n642 646\n657 701\n771 808\n812 819\n823 826\n827 833\n862 875\n880 891\n919 920\n928 959\n970 1038\n1057 1072\n1074 1130\n1165 1169\n1171 1230\n1265 1276\n1279 1302\n1313 1353\n1354 1438",
"output": "84067"
},
{
"input": "30 54 76 95 48 16\n9 11\n23 97\n112 116\n126 185\n214 223\n224 271\n278 282\n283 348\n359 368\n373 376\n452 463\n488 512\n532 552\n646 665\n681 685\n699 718\n735 736\n750 777\n791 810\n828 838\n841 858\n874 1079\n1136 1171\n1197 1203\n1210 1219\n1230 1248\n1280 1292\n1324 1374\n1397 1435\n1438 1439",
"output": "79844"
},
{
"input": "30 58 78 12 41 28\n20 26\n27 31\n35 36\n38 99\n103 104\n106 112\n133 143\n181 246\n248 251\n265 323\n350 357\n378 426\n430 443\n466 476\n510 515\n517 540\n542 554\n562 603\n664 810\n819 823\n826 845\n869 895\n921 973\n1002 1023\n1102 1136\n1143 1148\n1155 1288\n1316 1388\n1394 1403\n1434 1437",
"output": "82686"
},
{
"input": "30 62 80 97 25 47\n19 20\n43 75\n185 188\n199 242\n252 258\n277 310\n316 322\n336 357\n398 399\n404 436\n443 520\n549 617\n637 649\n679 694\n705 715\n725 730\n731 756\n768 793\n806 833\n834 967\n1003 1079\n1088 1097\n1100 1104\n1108 1121\n1127 1164\n1240 1263\n1274 1307\n1367 1407\n1419 1425\n1433 1437",
"output": "92356"
},
{
"input": "30 100 48 14 9 7\n26 55\n75 107\n145 146\n158 167\n197 199\n201 213\n222 238\n257 265\n271 338\n357 439\n454 462\n463 562\n633 660\n670 691\n696 829\n844 864\n880 888\n891 894\n895 900\n950 973\n1004 1007\n1018 1021\n1045 1049\n1062 1073\n1138 1184\n1240 1272\n1278 1315\n1403 1410\n1412 1418\n1426 1434",
"output": "107490"
},
{
"input": "30 3 50 32 1 18\n46 60\n66 94\n99 141\n162 239\n309 334\n400 550\n594 630\n659 663\n664 701\n735 736\n741 755\n756 801\n823 854\n865 915\n937 964\n984 986\n990 1007\n1008 1031\n1052 1064\n1073 1080\n1190 1191\n1192 1254\n1265 1266\n1272 1273\n1274 1277\n1284 1302\n1308 1312\n1358 1383\n1388 1389\n1390 1401",
"output": "25258"
},
{
"input": "30 7 52 49 46 37\n2 73\n129 130\n131 140\n143 150\n187 190\n217 238\n248 271\n375 426\n465 470\n475 503\n506 541\n569 571\n578 585\n680 683\n690 738\n742 761\n868 877\n890 892\n904 915\n935 961\n1076 1100\n1103 1114\n1232 1244\n1246 1247\n1257 1261\n1271 1276\n1280 1298\n1346 1360\n1373 1382\n1407 1419",
"output": "23992"
}
] | 1,694,109,847
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
n,p1,p2,p3,t1,t2=map(int,input().split())
p=0
t=0
for i in range(n):
a,b=map(int,input().split())
d=a-t
if i==0:
p=p+p1*(b-a)
if i!=0 and i<n:
if d<=t1:
p=p+((b-a)+d)*p1
if d>t1 and d<=t1+t2:
p=p+((b-a)+t1)*p1+p2*(d-t1)
if d>t1+t2:
p=p+p1*((b-a)+t1)+(d-t1-t2)*p3+p2*t2
t=b
print(p)
|
Title: Power Consumption Calculation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
Input Specification:
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
Output Specification:
Output the answer to the problem.
Demo Input:
['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n']
Demo Output:
['30', '570']
Note:
none
|
```python
n,p1,p2,p3,t1,t2=map(int,input().split())
p=0
t=0
for i in range(n):
a,b=map(int,input().split())
d=a-t
if i==0:
p=p+p1*(b-a)
if i!=0 and i<n:
if d<=t1:
p=p+((b-a)+d)*p1
if d>t1 and d<=t1+t2:
p=p+((b-a)+t1)*p1+p2*(d-t1)
if d>t1+t2:
p=p+p1*((b-a)+t1)+(d-t1-t2)*p3+p2*t2
t=b
print(p)
```
| 3.954
|
5
|
C
|
Longest Regular Bracket Sequence
|
PROGRAMMING
| 1,900
|
[
"constructive algorithms",
"data structures",
"dp",
"greedy",
"sortings",
"strings"
] |
C. Longest Regular Bracket Sequence
|
2
|
256
|
This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
|
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
|
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
|
[
")((())))(()())\n",
"))(\n"
] |
[
"6 2\n",
"0 1\n"
] |
none
| 0
|
[
{
"input": ")((())))(()())",
"output": "6 2"
},
{
"input": "))(",
"output": "0 1"
},
{
"input": "()(())()",
"output": "8 1"
},
{
"input": "((((()(((",
"output": "2 1"
},
{
"input": "))))()())))",
"output": "4 1"
},
{
"input": "(()())()(())()()())())()((()(()(())()()())((()(())()(()()()()))()(())()(((()())()(()((())()(())(()))",
"output": "28 1"
},
{
"input": ")(()(((((())))))(((()))))))((()(((((((())())()))))((((((((())))((((((())))))(()((((()))))(()))))))))",
"output": "68 1"
},
{
"input": "())(((((())())((((()))(())))())())(((()(()()()())(())()))((()(())())()()))()(()())))))(()))((())((((",
"output": "80 1"
},
{
"input": "))(((((((()((((())(((((((((()()))))(()())()((())())))((((()))(()()()))))(()()))())))))((())())()))((",
"output": "92 1"
}
] | 1,587,566,265
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
import sys
import math
t = sys.stdin.read().split("\n")
if (len(t) > 1):
#t.pop(-1)
t = t[0]
def check_for_matching(string):
o = []
index = 0
#print("=========================")
#print("String:" + string)
for char in string:
index +=1
if char == "(":
o.append("(")
elif char == ")":
if o!=[]:
o.pop(-1)
if(o == []):
#print("Submatch!")
#print(string[0:index])
#print("Length:" + str(len(string[0:index])))
return [string[0:index], True]
break
else:
#print("invalid sub string!")
#print(string[0:index])
return [string[0:index], False]
return [[],False]
base = t
index = 0
substrings = []
while True:
aux = check_for_matching(t)
if (len(t) == 0):
break
if aux[1] == False:
substrings.append(-1)
index+=1
elif aux[1] == True:
substrings.append(len(aux[0]))
index+=len(aux[0])
t = base[index::]
if -1 in substrings:
answers = []
suma = 0
for value in substrings:
if value == -1:
if suma!=0:
answers.append(suma)
suma = 0
else:
suma+= value
max_length = 0
count = 1
for aux in answers:
if aux == max_length:
count+=1
elif aux > max_length:
max_length = aux
count=1
print("{0} {1}".format(max_length, count))
else:
print("{0} 1".format(sum(substrings)) )
#print(str(max_length) + " " + str(count))
|
Title: Longest Regular Bracket Sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
Input Specification:
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
Demo Input:
[')((())))(()())\n', '))(\n']
Demo Output:
['6 2\n', '0 1\n']
Note:
none
|
```python
import sys
import math
t = sys.stdin.read().split("\n")
if (len(t) > 1):
#t.pop(-1)
t = t[0]
def check_for_matching(string):
o = []
index = 0
#print("=========================")
#print("String:" + string)
for char in string:
index +=1
if char == "(":
o.append("(")
elif char == ")":
if o!=[]:
o.pop(-1)
if(o == []):
#print("Submatch!")
#print(string[0:index])
#print("Length:" + str(len(string[0:index])))
return [string[0:index], True]
break
else:
#print("invalid sub string!")
#print(string[0:index])
return [string[0:index], False]
return [[],False]
base = t
index = 0
substrings = []
while True:
aux = check_for_matching(t)
if (len(t) == 0):
break
if aux[1] == False:
substrings.append(-1)
index+=1
elif aux[1] == True:
substrings.append(len(aux[0]))
index+=len(aux[0])
t = base[index::]
if -1 in substrings:
answers = []
suma = 0
for value in substrings:
if value == -1:
if suma!=0:
answers.append(suma)
suma = 0
else:
suma+= value
max_length = 0
count = 1
for aux in answers:
if aux == max_length:
count+=1
elif aux > max_length:
max_length = aux
count=1
print("{0} {1}".format(max_length, count))
else:
print("{0} 1".format(sum(substrings)) )
#print(str(max_length) + " " + str(count))
```
| -1
|
721
|
B
|
Passwords
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math",
"sortings",
"strings"
] | null | null |
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
|
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
|
[
"5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n",
"4 100\n11\n22\n1\n2\n22\n"
] |
[
"1 15\n",
"3 4\n"
] |
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
| 1,000
|
[
{
"input": "5 2\ncba\nabc\nbb1\nabC\nABC\nabc",
"output": "1 15"
},
{
"input": "4 100\n11\n22\n1\n2\n22",
"output": "3 4"
},
{
"input": "1 1\na1\na1",
"output": "1 1"
},
{
"input": "1 100\na1\na1",
"output": "1 1"
},
{
"input": "2 1\nabc\nAbc\nAbc",
"output": "1 7"
},
{
"input": "2 2\nabc\nAbc\nabc",
"output": "1 2"
},
{
"input": "2 1\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 2\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 1\nab\nabc\nabc",
"output": "7 7"
},
{
"input": "2 2\nab\nabc\nabc",
"output": "2 2"
},
{
"input": "10 3\nOIbV1igi\no\nZS\nQM\n9woLzI\nWreboD\nQ7yl\nA5Rb\nS9Lno72TkP\nfT97o\no",
"output": "1 1"
},
{
"input": "10 3\nHJZNMsT\nLaPcH2C\nlrhqIO\n9cxw\noTC1XwjW\nGHL9Ul6\nUyIs\nPuzwgR4ZKa\nyIByoKR5\nd3QA\nPuzwgR4ZKa",
"output": "25 25"
},
{
"input": "20 5\nvSyC787KlIL8kZ2Uv5sw\nWKWOP\n7i8J3E8EByIq\nNW2VyGweL\nmyR2sRNu\nmXusPP0\nf4jgGxra\n4wHRzRhOCpEt\npPz9kybGb\nOtSpePCRoG5nkjZ2VxRy\nwHYsSttWbJkg\nKBOP9\nQfiOiFyHPPsw3GHo8J8\nxB8\nqCpehZEeEhdq\niOLjICK6\nQ91\nHmCsfMGTFKoFFnv238c\nJKjhg\ngkEUh\nKBOP9",
"output": "3 11"
},
{
"input": "15 2\nw6S9WyU\nMVh\nkgUhQHW\nhGQNOF\nUuym\n7rGQA\nBM8vLPRB\n9E\nDs32U\no\nz1aV2C5T\n8\nzSXjrqQ\n1FO\n3kIt\nBM8vLPRB",
"output": "44 50"
},
{
"input": "20 2\ni\n5Rp6\nE4vsr\nSY\nORXx\nh13C\nk6tzC\ne\nN\nKQf4C\nWZcdL\ndiA3v\n0InQT\nuJkAr\nGCamp\nBuIRd\nY\nM\nxZYx7\n0a5A\nWZcdL",
"output": "36 65"
},
{
"input": "20 2\naWLQ6\nSgQ9r\nHcPdj\n2BNaO\n3TjNb\nnvwFM\nqsKt7\nFnb6N\nLoc0p\njxuLq\nBKAjf\nEKgZB\nBfOSa\nsMIvr\nuIWcR\nIura3\nLAqSf\ntXq3G\n8rQ8I\n8otAO\nsMIvr",
"output": "1 65"
},
{
"input": "20 15\n0ZpQugVlN7\nm0SlKGnohN\nRFXTqhNGcn\n1qm2ZbB\nQXtJWdf78P\nbc2vH\nP21dty2Z1P\nm2c71LFhCk\n23EuP1Dvh3\nanwri5RhQN\n55v6HYv288\n1u5uKOjM5r\n6vg0GC1\nDAPYiA3ns1\nUZaaJ3Gmnk\nwB44x7V4Zi\n4hgB2oyU8P\npYFQpy8gGK\ndbz\nBv\n55v6HYv288",
"output": "6 25"
},
{
"input": "3 1\na\nb\naa\naa",
"output": "13 13"
},
{
"input": "6 3\nab\nac\nad\nabc\nabd\nabe\nabc",
"output": "9 11"
},
{
"input": "4 2\n1\n2\n11\n22\n22",
"output": "8 9"
},
{
"input": "2 1\n1\n12\n12",
"output": "7 7"
},
{
"input": "3 1\nab\nabc\nabd\nabc",
"output": "7 13"
},
{
"input": "2 1\na\nab\nab",
"output": "7 7"
},
{
"input": "5 2\na\nb\nc\nab\naa\naa",
"output": "9 15"
},
{
"input": "6 1\n1\n2\n11\n22\n111\n2222\n22",
"output": "13 19"
},
{
"input": "3 1\n1\n2\n11\n11",
"output": "13 13"
},
{
"input": "10 4\na\nb\nc\nd\ne\nf\nab\ncd\nac\nad\nac",
"output": "12 20"
},
{
"input": "4 2\na\nb\nc\nd\na",
"output": "1 9"
},
{
"input": "4 1\n1\n2\n3\n4\n4",
"output": "1 19"
},
{
"input": "5 1\na\nb\nc\nd\nef\nef",
"output": "25 25"
},
{
"input": "6 4\n1\n2\n22\n33\n44\n555\n555",
"output": "11 11"
},
{
"input": "5 2\na\nb\nc\nd\nab\nab",
"output": "15 15"
},
{
"input": "6 2\n1\n2\n3\n4\n5\n23\n23",
"output": "16 16"
},
{
"input": "4 2\na\nb\naa\nbb\naa",
"output": "8 9"
},
{
"input": "5 4\na\nbb\ncc\ndd\nee\nbb",
"output": "2 10"
},
{
"input": "4 1\na\nb\nc\nab\nab",
"output": "19 19"
},
{
"input": "7 100\na\nb\nc\nd\ne\ng\nab\nab",
"output": "7 7"
},
{
"input": "6 1\na\nb\nc\nd\ne\naa\naa",
"output": "31 31"
},
{
"input": "4 1\na\nas\nasd\nasde\nasde",
"output": "19 19"
},
{
"input": "5 2\n1\n2\n3\n11\n22\n22",
"output": "9 15"
},
{
"input": "10 2\na\nb\nc\nd\nee\nff\ngg\nhh\nii\njj\nii",
"output": "15 30"
},
{
"input": "3 1\na\nab\nbc\nab",
"output": "7 13"
},
{
"input": "6 4\na\nb\nc\nbb\nbc\ncc\ncc",
"output": "4 11"
}
] | 1,576,229,675
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 140
| 0
|
n,k=map(int,input().split())
f=[]
for i in range(n):
s=str(input())
f.append(len(s))
d=len(str(input()))
f.append(d)
f.append(d+1)
f.sort()
best_t=f.index(d)+1+f.index(d)//k*5
wor_t=f.index(d+1)+(f.index(d+1)-1)//k*5-1
print(best_t,wor_t)
|
Title: Passwords
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
Output Specification:
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Demo Input:
['5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n', '4 100\n11\n22\n1\n2\n22\n']
Demo Output:
['1 15\n', '3 4\n']
Note:
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
```python
n,k=map(int,input().split())
f=[]
for i in range(n):
s=str(input())
f.append(len(s))
d=len(str(input()))
f.append(d)
f.append(d+1)
f.sort()
best_t=f.index(d)+1+f.index(d)//k*5
wor_t=f.index(d+1)+(f.index(d+1)-1)//k*5-1
print(best_t,wor_t)
```
| 0
|
|
11
|
A
|
Increasing Sequence
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"implementation",
"math"
] |
A. Increasing Sequence
|
1
|
64
|
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
|
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
|
Output the minimal number of moves needed to make the sequence increasing.
|
[
"4 2\n1 3 3 2\n"
] |
[
"3\n"
] |
none
| 0
|
[
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
{
"input": "2 7\n1 1",
"output": "1"
},
{
"input": "3 3\n18 1 9",
"output": "10"
},
{
"input": "3 3\n15 17 9",
"output": "3"
},
{
"input": "3 3\n10 9 12",
"output": "2"
},
{
"input": "10 3\n2 1 17 10 5 16 8 4 15 17",
"output": "31"
},
{
"input": "10 3\n6 11 4 12 22 15 23 26 24 26",
"output": "13"
},
{
"input": "10 3\n10 24 13 15 18 14 15 26 33 35",
"output": "29"
},
{
"input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580",
"output": "15717"
},
{
"input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327",
"output": "16133"
}
] | 1,574,446,140
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 16
| 1,000
| 1,843,200
|
n,d=map(int,input().split())
a = [int(x) for x in input().split()]
c=0
for i in range(0,n-1):
if(a[i]>=a[i+1]):
while(a[i]>=a[i+1]):
a[i+1]+=d
c+=1
print(c)
|
Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
n,d=map(int,input().split())
a = [int(x) for x in input().split()]
c=0
for i in range(0,n-1):
if(a[i]>=a[i+1]):
while(a[i]>=a[i+1]):
a[i+1]+=d
c+=1
print(c)
```
| 0
|
959
|
A
|
Mahmoud and Ehab and the even-odd game
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
|
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
|
[
"1\n",
"2\n"
] |
[
"Ehab",
"Mahmoud"
] |
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
| 500
|
[
{
"input": "1",
"output": "Ehab"
},
{
"input": "2",
"output": "Mahmoud"
},
{
"input": "10000",
"output": "Mahmoud"
},
{
"input": "33333",
"output": "Ehab"
},
{
"input": "5",
"output": "Ehab"
},
{
"input": "1000000000",
"output": "Mahmoud"
},
{
"input": "999999999",
"output": "Ehab"
},
{
"input": "123123123",
"output": "Ehab"
},
{
"input": "22222221",
"output": "Ehab"
},
{
"input": "22222220",
"output": "Mahmoud"
},
{
"input": "3",
"output": "Ehab"
},
{
"input": "4",
"output": "Mahmoud"
},
{
"input": "6",
"output": "Mahmoud"
},
{
"input": "7",
"output": "Ehab"
},
{
"input": "8",
"output": "Mahmoud"
},
{
"input": "9",
"output": "Ehab"
},
{
"input": "10",
"output": "Mahmoud"
},
{
"input": "536870912",
"output": "Mahmoud"
},
{
"input": "536870913",
"output": "Ehab"
},
{
"input": "536870911",
"output": "Ehab"
}
] | 1,608,695,131
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 155
| 0
|
import sys
n = int(sys.stdin.readline().split()[0])
if n % 2 == 0:
print("Mahmoud")
else:
print("Ehab")
|
Title: Mahmoud and Ehab and the even-odd game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
Output Specification:
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Demo Input:
['1\n', '2\n']
Demo Output:
['Ehab', 'Mahmoud']
Note:
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
|
```python
import sys
n = int(sys.stdin.readline().split()[0])
if n % 2 == 0:
print("Mahmoud")
else:
print("Ehab")
```
| 3
|
|
869
|
B
|
The Eternal Immortality
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
|
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
|
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
|
[
"2 4\n",
"0 10\n",
"107 109\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
| 1,000
|
[
{
"input": "2 4",
"output": "2"
},
{
"input": "0 10",
"output": "0"
},
{
"input": "107 109",
"output": "2"
},
{
"input": "10 13",
"output": "6"
},
{
"input": "998244355 998244359",
"output": "4"
},
{
"input": "999999999000000000 1000000000000000000",
"output": "0"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 15",
"output": "0"
},
{
"input": "24 26",
"output": "0"
},
{
"input": "14 60",
"output": "0"
},
{
"input": "11 79",
"output": "0"
},
{
"input": "1230 1232",
"output": "2"
},
{
"input": "2633 2634",
"output": "4"
},
{
"input": "535 536",
"output": "6"
},
{
"input": "344319135 396746843",
"output": "0"
},
{
"input": "696667767 696667767",
"output": "1"
},
{
"input": "419530302 610096911",
"output": "0"
},
{
"input": "238965115 821731161",
"output": "0"
},
{
"input": "414626436 728903812",
"output": "0"
},
{
"input": "274410639 293308324",
"output": "0"
},
{
"input": "650636673091305697 650636673091305702",
"output": "0"
},
{
"input": "651240548333620923 651240548333620924",
"output": "4"
},
{
"input": "500000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1"
},
{
"input": "0 4",
"output": "4"
},
{
"input": "50000000062000007 50000000062000011",
"output": "0"
},
{
"input": "0 0",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "0 2",
"output": "2"
},
{
"input": "10000000000012 10000000000015",
"output": "0"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "12 23",
"output": "0"
},
{
"input": "0 11",
"output": "0"
},
{
"input": "11111234567890 11111234567898",
"output": "0"
},
{
"input": "0 3",
"output": "6"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999999999999999997 999999999999999999",
"output": "2"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "0 1",
"output": "1"
},
{
"input": "101 1002",
"output": "0"
},
{
"input": "0 100000000000000001",
"output": "0"
},
{
"input": "99999999999999997 99999999999999999",
"output": "2"
},
{
"input": "14 15",
"output": "5"
},
{
"input": "8 19",
"output": "0"
},
{
"input": "12 22",
"output": "0"
},
{
"input": "999999999999996 999999999999999",
"output": "4"
},
{
"input": "1 3",
"output": "6"
},
{
"input": "124 125",
"output": "5"
},
{
"input": "11 32",
"output": "0"
},
{
"input": "0 5",
"output": "0"
},
{
"input": "0 999999",
"output": "0"
},
{
"input": "151151151515 151151151526",
"output": "0"
},
{
"input": "6 107",
"output": "0"
},
{
"input": "5 16",
"output": "0"
},
{
"input": "7 16",
"output": "0"
},
{
"input": "6 19",
"output": "0"
},
{
"input": "11113111111111 13111111111111",
"output": "0"
},
{
"input": "1 1000",
"output": "0"
},
{
"input": "24 25",
"output": "5"
},
{
"input": "0 100000000000",
"output": "0"
},
{
"input": "1 22",
"output": "0"
},
{
"input": "999999999999999996 999999999999999999",
"output": "4"
}
] | 1,508,334,781
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 5,529,600
|
a = int(input(""))
b = int(input(""))
result = 1
diff = b - a
for counter in range (diff+1, b+1):
result= result * counter
result = str(result)
length = len(result)
lastDigit = result[length-1]
print(lastDigit)
|
Title: The Eternal Immortality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
Input Specification:
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output Specification:
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
Demo Input:
['2 4\n', '0 10\n', '107 109\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
|
```python
a = int(input(""))
b = int(input(""))
result = 1
diff = b - a
for counter in range (diff+1, b+1):
result= result * counter
result = str(result)
length = len(result)
lastDigit = result[length-1]
print(lastDigit)
```
| -1
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,682,930,685
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 92
| 0
|
class Solution:
def __init__(self):
pass
def solve(self):
s = input()
t = input()
i = 0; j = 0
while j < len(t):
if t[j] == s[i]:
i += 1
j += 1
print(i + 1)
if __name__ == "__main__":
sol = Solution()
sol.solve()
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
class Solution:
def __init__(self):
pass
def solve(self):
s = input()
t = input()
i = 0; j = 0
while j < len(t):
if t[j] == s[i]:
i += 1
j += 1
print(i + 1)
if __name__ == "__main__":
sol = Solution()
sol.solve()
```
| 3
|
|
337
|
A
|
Puzzles
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
|
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
|
Print a single integer — the least possible difference the teacher can obtain.
|
[
"4 6\n10 12 10 7 5 22\n"
] |
[
"5\n"
] |
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
| 500
|
[
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,670,511,791
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
n,m = map(int,input().strip().split())
l = list(map(int,input().strip().split()))
l.sort()
ans = 100000
for i in range(1,m-n):
ans = min(ans,l[i+n-1] - l[i])
print(ans)
|
Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
|
```python
n,m = map(int,input().strip().split())
l = list(map(int,input().strip().split()))
l.sort()
ans = 100000
for i in range(1,m-n):
ans = min(ans,l[i+n-1] - l[i])
print(ans)
```
| 0
|
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,675,457,394
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 4,505,600
|
T = int(input())
teams = {}
for t in range(T):
team = input()
if team not in teams:
teams[team] = 1
else:
teams[team] += 1
max_score = 0
winner_team = ''
for team in teams:
if teams[team] > max_score:
max_score = teams[team]
winner_team = team
print(winner_team)
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
T = int(input())
teams = {}
for t in range(T):
team = input()
if team not in teams:
teams[team] = 1
else:
teams[team] += 1
max_score = 0
winner_team = ''
for team in teams:
if teams[team] > max_score:
max_score = teams[team]
winner_team = team
print(winner_team)
```
| 3.968608
|
65
|
A
|
Harry Potter and Three Spells
|
PROGRAMMING
| 1,800
|
[
"implementation",
"math"
] |
A. Harry Potter and Three Spells
|
2
|
256
|
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert *a* grams of sand into *b* grams of lead, the second one allows you to convert *c* grams of lead into *d* grams of gold and third one allows you to convert *e* grams of gold into *f* grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
|
The first line contains 6 integers *a*, *b*, *c*, *d*, *e*, *f* (0<=≤<=*a*,<=*b*,<=*c*,<=*d*,<=*e*,<=*f*<=≤<=1000).
|
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
|
[
"100 200 250 150 200 250\n",
"100 50 50 200 200 100\n",
"100 10 200 20 300 30\n",
"0 0 0 0 0 0\n",
"1 1 0 1 1 1\n",
"1 0 1 2 1 2\n",
"100 1 100 1 0 1\n"
] |
[
"Ron\n",
"Hermione\n",
"Hermione\n",
"Hermione\n",
"Ron\n",
"Hermione\n",
"Ron\n"
] |
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
| 500
|
[
{
"input": "100 200 250 150 200 250",
"output": "Ron"
},
{
"input": "100 50 50 200 200 100",
"output": "Hermione"
},
{
"input": "100 10 200 20 300 30",
"output": "Hermione"
},
{
"input": "0 0 0 0 0 0",
"output": "Hermione"
},
{
"input": "1 1 0 1 1 1",
"output": "Ron"
},
{
"input": "1 0 1 2 1 2",
"output": "Hermione"
},
{
"input": "100 1 100 1 0 1",
"output": "Ron"
},
{
"input": "1 1 2 2 1 1",
"output": "Hermione"
},
{
"input": "4 4 1 3 1 4",
"output": "Ron"
},
{
"input": "3 3 2 1 4 4",
"output": "Hermione"
},
{
"input": "5 1 2 9 1 10",
"output": "Ron"
},
{
"input": "63 65 21 41 95 23",
"output": "Hermione"
},
{
"input": "913 0 0 0 0 0",
"output": "Hermione"
},
{
"input": "0 333 0 0 0 0",
"output": "Hermione"
},
{
"input": "842 538 0 0 0 0",
"output": "Hermione"
},
{
"input": "0 0 536 0 0 0",
"output": "Hermione"
},
{
"input": "324 0 495 0 0 0",
"output": "Hermione"
},
{
"input": "0 407 227 0 0 0",
"output": "Hermione"
},
{
"input": "635 63 924 0 0 0",
"output": "Hermione"
},
{
"input": "0 0 0 493 0 0",
"output": "Ron"
},
{
"input": "414 0 0 564 0 0",
"output": "Ron"
},
{
"input": "0 143 0 895 0 0",
"output": "Ron"
},
{
"input": "276 264 0 875 0 0",
"output": "Ron"
},
{
"input": "0 0 532 186 0 0",
"output": "Hermione"
},
{
"input": "510 0 692 825 0 0",
"output": "Hermione"
},
{
"input": "0 777 910 46 0 0",
"output": "Ron"
},
{
"input": "754 329 959 618 0 0",
"output": "Hermione"
},
{
"input": "0 0 0 0 416 0",
"output": "Hermione"
},
{
"input": "320 0 0 0 526 0",
"output": "Hermione"
},
{
"input": "0 149 0 0 6 0",
"output": "Hermione"
},
{
"input": "996 13 0 0 111 0",
"output": "Hermione"
},
{
"input": "0 0 531 0 688 0",
"output": "Hermione"
},
{
"input": "544 0 837 0 498 0",
"output": "Hermione"
},
{
"input": "0 54 680 0 988 0",
"output": "Hermione"
},
{
"input": "684 986 930 0 555 0",
"output": "Hermione"
},
{
"input": "0 0 0 511 534 0",
"output": "Ron"
},
{
"input": "594 0 0 819 304 0",
"output": "Ron"
},
{
"input": "0 55 0 977 230 0",
"output": "Ron"
},
{
"input": "189 291 0 845 97 0",
"output": "Ron"
},
{
"input": "0 0 77 302 95 0",
"output": "Hermione"
},
{
"input": "247 0 272 232 96 0",
"output": "Hermione"
},
{
"input": "0 883 219 748 77 0",
"output": "Ron"
},
{
"input": "865 643 599 98 322 0",
"output": "Hermione"
},
{
"input": "0 0 0 0 0 699",
"output": "Hermione"
},
{
"input": "907 0 0 0 0 99",
"output": "Hermione"
},
{
"input": "0 891 0 0 0 529",
"output": "Hermione"
},
{
"input": "640 125 0 0 0 849",
"output": "Hermione"
},
{
"input": "0 0 698 0 0 702",
"output": "Hermione"
},
{
"input": "58 0 483 0 0 470",
"output": "Hermione"
},
{
"input": "0 945 924 0 0 355",
"output": "Hermione"
},
{
"input": "998 185 209 0 0 554",
"output": "Hermione"
},
{
"input": "0 0 0 914 0 428",
"output": "Ron"
},
{
"input": "412 0 0 287 0 575",
"output": "Ron"
},
{
"input": "0 850 0 509 0 76",
"output": "Ron"
},
{
"input": "877 318 0 478 0 782",
"output": "Ron"
},
{
"input": "0 0 823 740 0 806",
"output": "Hermione"
},
{
"input": "126 0 620 51 0 835",
"output": "Hermione"
},
{
"input": "0 17 946 633 0 792",
"output": "Ron"
},
{
"input": "296 546 493 22 0 893",
"output": "Ron"
},
{
"input": "0 0 0 0 766 813",
"output": "Hermione"
},
{
"input": "319 0 0 0 891 271",
"output": "Hermione"
},
{
"input": "0 252 0 0 261 576",
"output": "Hermione"
},
{
"input": "876 440 0 0 65 362",
"output": "Hermione"
},
{
"input": "0 0 580 0 245 808",
"output": "Hermione"
},
{
"input": "835 0 116 0 9 552",
"output": "Hermione"
},
{
"input": "0 106 748 0 773 840",
"output": "Hermione"
},
{
"input": "935 388 453 0 797 235",
"output": "Hermione"
},
{
"input": "0 0 0 893 293 289",
"output": "Ron"
},
{
"input": "938 0 0 682 55 725",
"output": "Ron"
},
{
"input": "0 710 0 532 389 511",
"output": "Ron"
},
{
"input": "617 861 0 247 920 902",
"output": "Ron"
},
{
"input": "0 0 732 202 68 389",
"output": "Hermione"
},
{
"input": "279 0 254 964 449 143",
"output": "Hermione"
},
{
"input": "0 746 400 968 853 85",
"output": "Ron"
},
{
"input": "565 846 658 828 767 734",
"output": "Ron"
},
{
"input": "6 6 1 6 1 6",
"output": "Ron"
},
{
"input": "3 6 1 6 3 3",
"output": "Ron"
},
{
"input": "6 3 1 3 2 3",
"output": "Ron"
},
{
"input": "3 6 2 2 6 3",
"output": "Hermione"
},
{
"input": "3 2 2 1 3 3",
"output": "Hermione"
},
{
"input": "1 1 1 6 1 1",
"output": "Ron"
},
{
"input": "1 3 1 3 3 2",
"output": "Ron"
},
{
"input": "6 2 6 6 2 3",
"output": "Hermione"
},
{
"input": "2 6 2 1 2 1",
"output": "Hermione"
},
{
"input": "2 3 2 1 6 6",
"output": "Hermione"
},
{
"input": "2 1 2 1 6 2",
"output": "Hermione"
},
{
"input": "6 1 3 1 3 3",
"output": "Hermione"
},
{
"input": "1 2 2 3 2 2",
"output": "Ron"
},
{
"input": "3 3 2 6 3 6",
"output": "Ron"
},
{
"input": "2 1 6 1 2 6",
"output": "Hermione"
},
{
"input": "2 3 1 3 1 6",
"output": "Ron"
},
{
"input": "6 6 2 3 1 3",
"output": "Ron"
},
{
"input": "6 2 6 2 3 1",
"output": "Hermione"
},
{
"input": "1 6 6 2 3 2",
"output": "Ron"
},
{
"input": "6 3 6 2 6 6",
"output": "Hermione"
},
{
"input": "1 3 1 6 6 1",
"output": "Ron"
},
{
"input": "1 1 1 1 6 6",
"output": "Hermione"
},
{
"input": "2 6 2 2 2 3",
"output": "Ron"
},
{
"input": "1 6 1 6 6 3",
"output": "Ron"
},
{
"input": "6 6 3 1 3 3",
"output": "Hermione"
},
{
"input": "2 6 6 1 2 6",
"output": "Ron"
},
{
"input": "3 2 6 6 1 6",
"output": "Ron"
},
{
"input": "1 2 3 2 2 3",
"output": "Ron"
},
{
"input": "2 6 1 1 1 6",
"output": "Ron"
},
{
"input": "1 6 3 6 6 3",
"output": "Ron"
},
{
"input": "3 3 3 2 6 2",
"output": "Hermione"
},
{
"input": "6 2 3 6 2 2",
"output": "Hermione"
},
{
"input": "2 1 2 3 3 2",
"output": "Hermione"
},
{
"input": "6 2 1 1 3 6",
"output": "Hermione"
},
{
"input": "6 6 6 6 3 2",
"output": "Hermione"
},
{
"input": "6 1 1 6 2 3",
"output": "Ron"
},
{
"input": "6 1 3 6 1 2",
"output": "Hermione"
},
{
"input": "2 1 2 6 6 1",
"output": "Hermione"
},
{
"input": "1 2 2 3 2 1",
"output": "Ron"
},
{
"input": "2 2 6 6 6 6",
"output": "Hermione"
},
{
"input": "31 75 12 62 94 137",
"output": "Ron"
},
{
"input": "236 9 184 68 345 332",
"output": "Hermione"
},
{
"input": "362 142 107 565 329 608",
"output": "Ron"
},
{
"input": "731 859 197 474 370 465",
"output": "Ron"
},
{
"input": "452 577 639 31 60 673",
"output": "Hermione"
},
{
"input": "821 294 730 941 101 530",
"output": "Ron"
},
{
"input": "542 11 820 850 791 738",
"output": "Hermione"
},
{
"input": "911 728 910 407 833 594",
"output": "Hermione"
},
{
"input": "632 446 352 317 522 451",
"output": "Hermione"
}
] | 1,685,952,247
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 124
| 92
| 0
|
# LUOGU_RID: 112089745
a, b, c, d, e, f = map(int, input().split());
if a * c * e < b * d * f or a == 0 and b * d > 0 or c == 0 and d > 0:
print("Ron");
else:
print("Hermione");
|
Title: Harry Potter and Three Spells
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A long time ago (probably even in the first book), Nicholas Flamel, a great alchemist and the creator of the Philosopher's Stone, taught Harry Potter three useful spells. The first one allows you to convert *a* grams of sand into *b* grams of lead, the second one allows you to convert *c* grams of lead into *d* grams of gold and third one allows you to convert *e* grams of gold into *f* grams of sand. When Harry told his friends about these spells, Ron Weasley was amazed. After all, if they succeed in turning sand into lead, lead into gold, and then turning part of the gold into sand again and so on, then it will be possible to start with a small amount of sand and get huge amounts of gold! Even an infinite amount of gold! Hermione Granger, by contrast, was skeptical about that idea. She argues that according to the law of conservation of matter getting an infinite amount of matter, even using magic, is impossible. On the contrary, the amount of matter may even decrease during transformation, being converted to magical energy. Though Hermione's theory seems convincing, Ron won't believe her. As far as Ron is concerned, Hermione made up her law of conservation of matter to stop Harry and Ron wasting their time with this nonsense, and to make them go and do homework instead. That's why Ron has already collected a certain amount of sand for the experiments. A quarrel between the friends seems unavoidable...
Help Harry to determine which one of his friends is right, and avoid the quarrel after all. To do this you have to figure out whether it is possible to get the amount of gold greater than any preassigned number from some finite amount of sand.
Input Specification:
The first line contains 6 integers *a*, *b*, *c*, *d*, *e*, *f* (0<=≤<=*a*,<=*b*,<=*c*,<=*d*,<=*e*,<=*f*<=≤<=1000).
Output Specification:
Print "Ron", if it is possible to get an infinitely large amount of gold having a certain finite amount of sand (and not having any gold and lead at all), i.e., Ron is right. Otherwise, print "Hermione".
Demo Input:
['100 200 250 150 200 250\n', '100 50 50 200 200 100\n', '100 10 200 20 300 30\n', '0 0 0 0 0 0\n', '1 1 0 1 1 1\n', '1 0 1 2 1 2\n', '100 1 100 1 0 1\n']
Demo Output:
['Ron\n', 'Hermione\n', 'Hermione\n', 'Hermione\n', 'Ron\n', 'Hermione\n', 'Ron\n']
Note:
Consider the first sample. Let's start with the 500 grams of sand. Apply the first spell 5 times and turn the sand into 1000 grams of lead. Then apply the second spell 4 times to get 600 grams of gold. Let’s take 400 grams from the resulting amount of gold turn them back into sand. We get 500 grams of sand and 200 grams of gold. If we apply the same operations to 500 grams of sand again, we can get extra 200 grams of gold every time. Thus, you can get 200, 400, 600 etc. grams of gold, i.e., starting with a finite amount of sand (500 grams), you can get the amount of gold which is greater than any preassigned number.
In the forth sample it is impossible to get sand, or lead, or gold, applying the spells.
In the fifth sample an infinitely large amount of gold can be obtained by using only the second spell, which allows you to receive 1 gram of gold out of nothing. Note that if such a second spell is available, then the first and the third one do not affect the answer at all.
The seventh sample is more interesting. We can also start with a zero amount of sand there. With the aid of the third spell you can get sand out of nothing. We get 10000 grams of sand in this manner. Let's get 100 grams of lead using the first spell 100 times. Then make 1 gram of gold from them. We managed to receive 1 gram of gold, starting with a zero amount of sand! Clearly, in this manner you can get an infinitely large amount of gold.
|
```python
# LUOGU_RID: 112089745
a, b, c, d, e, f = map(int, input().split());
if a * c * e < b * d * f or a == 0 and b * d > 0 or c == 0 and d > 0:
print("Ron");
else:
print("Hermione");
```
| 3.977
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,697,640,122
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
listt=[]
for m in range(5):
ans=list(map(int,input().split(" ")))
listt.append(ans)
moves=0
row=2
col=2
for i in range(0,5):
for x in range(0,5):
if listt[i][x]==1:
co=x
ro=i
break
while ro != row or co != col:
if ro != row:
if ro<row:
moves+=1
ro+=1
else:
moves+=1
ro-=1
if co != col:
if co<row:
moves+=1
co+=1
else:
moves+=1
co-=1
print(moves)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
listt=[]
for m in range(5):
ans=list(map(int,input().split(" ")))
listt.append(ans)
moves=0
row=2
col=2
for i in range(0,5):
for x in range(0,5):
if listt[i][x]==1:
co=x
ro=i
break
while ro != row or co != col:
if ro != row:
if ro<row:
moves+=1
ro+=1
else:
moves+=1
ro-=1
if co != col:
if co<row:
moves+=1
co+=1
else:
moves+=1
co-=1
print(moves)
```
| 3
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,662,402,456
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 92
| 0
|
a = int(input())
b = int(input())
a_bez_0 = a
b_bez_0 = b
s_bez_0 = a + b # summa bez 0
summa = a + b
i = -1
i2 = -1
i3 = -1
lastA = 0
lastB = 0
lastS = 0
while a > 0:
lastA = a % 10 # posl cifra
if lastA == 0:
a_bez_0 = int(str(a_bez_0)[:i] + str(a_bez_0)[i + 1:]) # virezaem 0 cherez stroku
i = i - 1
a = a // 10 #chislo bez posl cifri
while b > 0:
lastB = b % 10
if lastB == 0:
b_bez_0 = int(str(b_bez_0)[:i2] + str(b_bez_0)[i2 + 1:])
i2 = i2 - 1
b = b // 10
while summa > 0:
lastS = summa % 10
if lastS == 0:
s_bez_0 = int(str(s_bez_0)[:i3] + str(s_bez_0)[i3 + 1:])
i3 = i3 - 1
summa = summa // 10
if a_bez_0 + b_bez_0 == s_bez_0:
print('YES')
else:
print('NO')
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
a = int(input())
b = int(input())
a_bez_0 = a
b_bez_0 = b
s_bez_0 = a + b # summa bez 0
summa = a + b
i = -1
i2 = -1
i3 = -1
lastA = 0
lastB = 0
lastS = 0
while a > 0:
lastA = a % 10 # posl cifra
if lastA == 0:
a_bez_0 = int(str(a_bez_0)[:i] + str(a_bez_0)[i + 1:]) # virezaem 0 cherez stroku
i = i - 1
a = a // 10 #chislo bez posl cifri
while b > 0:
lastB = b % 10
if lastB == 0:
b_bez_0 = int(str(b_bez_0)[:i2] + str(b_bez_0)[i2 + 1:])
i2 = i2 - 1
b = b // 10
while summa > 0:
lastS = summa % 10
if lastS == 0:
s_bez_0 = int(str(s_bez_0)[:i3] + str(s_bez_0)[i3 + 1:])
i3 = i3 - 1
summa = summa // 10
if a_bez_0 + b_bez_0 == s_bez_0:
print('YES')
else:
print('NO')
```
| 3.977
|
803
|
B
|
Distances to Zero
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms"
] | null | null |
You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*.
|
[
"9\n2 1 0 3 0 0 3 2 4\n",
"5\n0 1 2 3 4\n",
"7\n5 6 0 1 -2 3 4\n"
] |
[
"2 1 0 1 0 0 1 2 3 ",
"0 1 2 3 4 ",
"2 1 0 1 2 3 4 "
] |
none
| 0
|
[
{
"input": "9\n2 1 0 3 0 0 3 2 4",
"output": "2 1 0 1 0 0 1 2 3 "
},
{
"input": "5\n0 1 2 3 4",
"output": "0 1 2 3 4 "
},
{
"input": "7\n5 6 0 1 -2 3 4",
"output": "2 1 0 1 2 3 4 "
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "2\n0 1",
"output": "0 1 "
},
{
"input": "2\n1 0",
"output": "1 0 "
},
{
"input": "5\n0 1000000000 1000000000 1000000000 1000000000",
"output": "0 1 2 3 4 "
},
{
"input": "5\n-1000000000 -1000000000 0 1000000000 1000000000",
"output": "2 1 0 1 2 "
},
{
"input": "5\n-1000000000 1000000000 1000000000 1000000000 0",
"output": "4 3 2 1 0 "
},
{
"input": "15\n1000000000 -1000000000 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 0",
"output": "14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 "
},
{
"input": "15\n0 0 0 0 1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000",
"output": "0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 "
},
{
"input": "15\n-1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 0 -1000000000 -1000000000 0 0 1000000000 -1000000000 0 -1000000000",
"output": "6 5 4 3 2 1 0 1 1 0 0 1 1 0 1 "
},
{
"input": "15\n-1000000000 -1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000 1000000000 0 0 0 0",
"output": "11 10 9 8 7 6 5 4 3 2 1 0 0 0 0 "
},
{
"input": "4\n0 0 2 0",
"output": "0 0 1 0 "
},
{
"input": "15\n1 2 3 4 0 1 2 3 -5 -4 -3 -1 0 5 4",
"output": "4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 "
},
{
"input": "2\n0 -1",
"output": "0 1 "
},
{
"input": "5\n0 -1 -1 -1 0",
"output": "0 1 2 1 0 "
},
{
"input": "5\n0 0 0 -1 0",
"output": "0 0 0 1 0 "
},
{
"input": "3\n0 0 -1",
"output": "0 0 1 "
},
{
"input": "3\n0 -1 -1",
"output": "0 1 2 "
},
{
"input": "12\n0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 0",
"output": "0 1 2 3 4 5 5 4 3 2 1 0 "
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1",
"output": "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 "
},
{
"input": "30\n0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 "
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "2\n0 -1000000000",
"output": "0 1 "
},
{
"input": "2\n0 1000000000",
"output": "0 1 "
},
{
"input": "2\n-1000000000 0",
"output": "1 0 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "2\n0 0",
"output": "0 0 "
},
{
"input": "3\n0 -1000000000 -1000000000",
"output": "0 1 2 "
},
{
"input": "3\n0 1000000000 1000000000",
"output": "0 1 2 "
},
{
"input": "3\n1000000000 1000000000 0",
"output": "2 1 0 "
},
{
"input": "3\n0 0 -1000000000",
"output": "0 0 1 "
},
{
"input": "3\n0 1000000000 0",
"output": "0 1 0 "
},
{
"input": "3\n-1000000000 0 0",
"output": "1 0 0 "
},
{
"input": "3\n0 0 0",
"output": "0 0 0 "
},
{
"input": "3\n0 0 0",
"output": "0 0 0 "
},
{
"input": "3\n0 0 0",
"output": "0 0 0 "
},
{
"input": "4\n0 -1000000000 -1000000000 -1000000000",
"output": "0 1 2 3 "
},
{
"input": "4\n1000000000 -1000000000 0 -1000000000",
"output": "2 1 0 1 "
},
{
"input": "4\n1000000000 -1000000000 1000000000 0",
"output": "3 2 1 0 "
},
{
"input": "4\n0 0 -1000000000 1000000000",
"output": "0 0 1 2 "
},
{
"input": "4\n0 0 1000000000 -1000000000",
"output": "0 0 1 2 "
},
{
"input": "4\n-1000000000 1000000000 0 0",
"output": "2 1 0 0 "
},
{
"input": "4\n0 0 0 -1000000000",
"output": "0 0 0 1 "
},
{
"input": "4\n1000000000 0 0 0",
"output": "1 0 0 0 "
},
{
"input": "4\n1000000000 0 0 0",
"output": "1 0 0 0 "
},
{
"input": "4\n0 0 0 0",
"output": "0 0 0 0 "
},
{
"input": "4\n0 0 0 0",
"output": "0 0 0 0 "
},
{
"input": "4\n0 0 0 0",
"output": "0 0 0 0 "
},
{
"input": "5\n0 1000000000 1000000000 1000000000 1000000000",
"output": "0 1 2 3 4 "
},
{
"input": "5\n1000000000 -1000000000 -1000000000 1000000000 0",
"output": "4 3 2 1 0 "
},
{
"input": "5\n1000000000 -1000000000 1000000000 -1000000000 0",
"output": "4 3 2 1 0 "
},
{
"input": "5\n0 0 -1000000000 -1000000000 -1000000000",
"output": "0 0 1 2 3 "
},
{
"input": "5\n1000000000 0 -1000000000 0 -1000000000",
"output": "1 0 1 0 1 "
},
{
"input": "5\n1000000000 1000000000 1000000000 0 0",
"output": "3 2 1 0 0 "
},
{
"input": "5\n0 0 0 -1000000000 -1000000000",
"output": "0 0 0 1 2 "
},
{
"input": "5\n-1000000000 1000000000 0 0 0",
"output": "2 1 0 0 0 "
},
{
"input": "5\n1000000000 1000000000 0 0 0",
"output": "2 1 0 0 0 "
},
{
"input": "5\n0 0 0 0 -1000000000",
"output": "0 0 0 0 1 "
},
{
"input": "5\n0 0 1000000000 0 0",
"output": "0 0 1 0 0 "
},
{
"input": "5\n1000000000 0 0 0 0",
"output": "1 0 0 0 0 "
},
{
"input": "5\n0 0 0 0 0",
"output": "0 0 0 0 0 "
},
{
"input": "5\n0 0 0 0 0",
"output": "0 0 0 0 0 "
},
{
"input": "5\n0 0 0 0 0",
"output": "0 0 0 0 0 "
},
{
"input": "6\n0 1000000000 -1000000000 1000000000 -1000000000 1000000000",
"output": "0 1 2 3 4 5 "
},
{
"input": "6\n-1000000000 -1000000000 1000000000 1000000000 1000000000 0",
"output": "5 4 3 2 1 0 "
},
{
"input": "6\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 0",
"output": "5 4 3 2 1 0 "
},
{
"input": "6\n0 0 1000000000 1000000000 -1000000000 -1000000000",
"output": "0 0 1 2 3 4 "
},
{
"input": "6\n0 0 1000000000 1000000000 -1000000000 -1000000000",
"output": "0 0 1 2 3 4 "
},
{
"input": "6\n-1000000000 1000000000 -1000000000 -1000000000 0 0",
"output": "4 3 2 1 0 0 "
},
{
"input": "6\n0 0 0 -1000000000 1000000000 1000000000",
"output": "0 0 0 1 2 3 "
},
{
"input": "6\n-1000000000 1000000000 -1000000000 0 0 0",
"output": "3 2 1 0 0 0 "
},
{
"input": "6\n-1000000000 -1000000000 1000000000 0 0 0",
"output": "3 2 1 0 0 0 "
},
{
"input": "6\n0 0 0 0 -1000000000 1000000000",
"output": "0 0 0 0 1 2 "
},
{
"input": "6\n0 0 0 -1000000000 1000000000 0",
"output": "0 0 0 1 1 0 "
},
{
"input": "6\n1000000000 1000000000 0 0 0 0",
"output": "2 1 0 0 0 0 "
},
{
"input": "6\n0 0 0 0 0 -1000000000",
"output": "0 0 0 0 0 1 "
},
{
"input": "6\n0 0 0 1000000000 0 0",
"output": "0 0 0 1 0 0 "
},
{
"input": "6\n1000000000 0 0 0 0 0",
"output": "1 0 0 0 0 0 "
},
{
"input": "6\n0 0 0 0 0 0",
"output": "0 0 0 0 0 0 "
},
{
"input": "6\n0 0 0 0 0 0",
"output": "0 0 0 0 0 0 "
},
{
"input": "6\n0 0 0 0 0 0",
"output": "0 0 0 0 0 0 "
},
{
"input": "7\n0 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 -1000000000",
"output": "0 1 2 3 4 5 6 "
},
{
"input": "7\n1000000000 1000000000 -1000000000 0 -1000000000 1000000000 -1000000000",
"output": "3 2 1 0 1 2 3 "
},
{
"input": "7\n1000000000 1000000000 -1000000000 1000000000 -1000000000 -1000000000 0",
"output": "6 5 4 3 2 1 0 "
},
{
"input": "7\n0 0 1000000000 1000000000 1000000000 1000000000 -1000000000",
"output": "0 0 1 2 3 4 5 "
},
{
"input": "7\n0 1000000000 1000000000 -1000000000 1000000000 1000000000 0",
"output": "0 1 2 3 2 1 0 "
},
{
"input": "7\n1000000000 -1000000000 -1000000000 1000000000 -1000000000 0 0",
"output": "5 4 3 2 1 0 0 "
},
{
"input": "7\n0 0 0 1000000000 -1000000000 -1000000000 1000000000",
"output": "0 0 0 1 2 3 4 "
},
{
"input": "7\n-1000000000 0 0 -1000000000 0 -1000000000 1000000000",
"output": "1 0 0 1 0 1 2 "
},
{
"input": "7\n1000000000 1000000000 1000000000 -1000000000 0 0 0",
"output": "4 3 2 1 0 0 0 "
},
{
"input": "7\n0 0 0 0 -1000000000 -1000000000 1000000000",
"output": "0 0 0 0 1 2 3 "
},
{
"input": "7\n0 -1000000000 0 0 0 -1000000000 1000000000",
"output": "0 1 0 0 0 1 2 "
},
{
"input": "7\n1000000000 1000000000 1000000000 0 0 0 0",
"output": "3 2 1 0 0 0 0 "
},
{
"input": "7\n0 0 0 0 0 -1000000000 1000000000",
"output": "0 0 0 0 0 1 2 "
},
{
"input": "7\n0 -1000000000 0 0 0 0 -1000000000",
"output": "0 1 0 0 0 0 1 "
},
{
"input": "7\n-1000000000 1000000000 0 0 0 0 0",
"output": "2 1 0 0 0 0 0 "
},
{
"input": "7\n0 0 0 0 0 0 -1000000000",
"output": "0 0 0 0 0 0 1 "
},
{
"input": "7\n0 0 0 0 0 1000000000 0",
"output": "0 0 0 0 0 1 0 "
},
{
"input": "7\n1000000000 0 0 0 0 0 0",
"output": "1 0 0 0 0 0 0 "
},
{
"input": "7\n0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 "
},
{
"input": "7\n0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 "
},
{
"input": "7\n0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 "
},
{
"input": "8\n0 -1000000000 -1000000000 1000000000 1000000000 1000000000 1000000000 -1000000000",
"output": "0 1 2 3 4 5 6 7 "
},
{
"input": "8\n0 -1000000000 1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000",
"output": "0 1 2 3 4 5 6 7 "
},
{
"input": "8\n1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 0",
"output": "7 6 5 4 3 2 1 0 "
},
{
"input": "8\n0 0 -1000000000 -1000000000 1000000000 1000000000 1000000000 -1000000000",
"output": "0 0 1 2 3 4 5 6 "
},
{
"input": "8\n1000000000 0 0 -1000000000 -1000000000 1000000000 -1000000000 -1000000000",
"output": "1 0 0 1 2 3 4 5 "
},
{
"input": "8\n1000000000 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 0 0",
"output": "6 5 4 3 2 1 0 0 "
},
{
"input": "8\n0 0 0 1000000000 1000000000 -1000000000 -1000000000 -1000000000",
"output": "0 0 0 1 2 3 4 5 "
},
{
"input": "8\n-1000000000 0 0 1000000000 1000000000 0 -1000000000 1000000000",
"output": "1 0 0 1 1 0 1 2 "
},
{
"input": "8\n1000000000 1000000000 1000000000 -1000000000 -1000000000 0 0 0",
"output": "5 4 3 2 1 0 0 0 "
},
{
"input": "8\n0 0 0 0 1000000000 1000000000 1000000000 -1000000000",
"output": "0 0 0 0 1 2 3 4 "
},
{
"input": "8\n1000000000 0 1000000000 -1000000000 0 -1000000000 0 0",
"output": "1 0 1 1 0 1 0 0 "
},
{
"input": "8\n-1000000000 -1000000000 -1000000000 -1000000000 0 0 0 0",
"output": "4 3 2 1 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 1000000000 1000000000 -1000000000",
"output": "0 0 0 0 0 1 2 3 "
},
{
"input": "8\n-1000000000 0 -1000000000 0 0 1000000000 0 0",
"output": "1 0 1 0 0 1 0 0 "
},
{
"input": "8\n1000000000 1000000000 1000000000 0 0 0 0 0",
"output": "3 2 1 0 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 0 -1000000000 -1000000000",
"output": "0 0 0 0 0 0 1 2 "
},
{
"input": "8\n0 0 0 1000000000 -1000000000 0 0 0",
"output": "0 0 0 1 1 0 0 0 "
},
{
"input": "8\n1000000000 1000000000 0 0 0 0 0 0",
"output": "2 1 0 0 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 0 0 -1000000000",
"output": "0 0 0 0 0 0 0 1 "
},
{
"input": "8\n0 1000000000 0 0 0 0 0 0",
"output": "0 1 0 0 0 0 0 0 "
},
{
"input": "8\n1000000000 0 0 0 0 0 0 0",
"output": "1 0 0 0 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 0 "
},
{
"input": "8\n0 0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 0 "
}
] | 1,639,507,467
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 164
| 483
| 21,401,600
|
import sys, math
input=sys.stdin.readline
INF=int(1e9)+7
def solve():
n=int(input())
data=list(map(int, input().split()))
result=[INF]*n
for i in range(n):
if data[i]==0:
result[i]=0
idx=i-1
while idx>=0 and data[idx]!=0:
result[idx]=min(result[idx],i-idx)
idx-=1
idx=i+1
while idx<n and data[idx]!=0:
result[idx]=min(result[idx],idx-i)
idx+=1
print(*result,sep=' ')
t=1
while t:
t-=1
solve()
|
Title: Distances to Zero
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*.
Demo Input:
['9\n2 1 0 3 0 0 3 2 4\n', '5\n0 1 2 3 4\n', '7\n5 6 0 1 -2 3 4\n']
Demo Output:
['2 1 0 1 0 0 1 2 3 ', '0 1 2 3 4 ', '2 1 0 1 2 3 4 ']
Note:
none
|
```python
import sys, math
input=sys.stdin.readline
INF=int(1e9)+7
def solve():
n=int(input())
data=list(map(int, input().split()))
result=[INF]*n
for i in range(n):
if data[i]==0:
result[i]=0
idx=i-1
while idx>=0 and data[idx]!=0:
result[idx]=min(result[idx],i-idx)
idx-=1
idx=i+1
while idx<n and data[idx]!=0:
result[idx]=min(result[idx],idx-i)
idx+=1
print(*result,sep=' ')
t=1
while t:
t-=1
solve()
```
| 3
|
|
765
|
A
|
Neverending competitions
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
- this list contains all Jinotega's flights in this year (in arbitrary order), - Jinotega has only flown from his hometown to a snooker contest and back, - after each competition Jinotega flies back home (though they may attend a competition in one place several times), - and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
|
In the first line of input there is a single integer *n*: the number of Jinotega's flights (1<=≤<=*n*<=≤<=100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next *n* lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
|
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
|
[
"4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO\n",
"3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP\n"
] |
[
"home\n",
"contest\n"
] |
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
| 500
|
[
{
"input": "4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO",
"output": "home"
},
{
"input": "3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP",
"output": "contest"
},
{
"input": "1\nESJ\nESJ->TSJ",
"output": "contest"
},
{
"input": "2\nXMR\nFAJ->XMR\nXMR->FAJ",
"output": "home"
},
{
"input": "3\nZIZ\nDWJ->ZIZ\nZIZ->DWJ\nZIZ->DWJ",
"output": "contest"
},
{
"input": "10\nPVO\nDMN->PVO\nDMN->PVO\nPVO->DMN\nDMN->PVO\nPVO->DMN\nPVO->DMN\nPVO->DMN\nDMN->PVO\nPVO->DMN\nDMN->PVO",
"output": "home"
},
{
"input": "11\nIAU\nIAU->RUQ\nIAU->RUQ\nRUQ->IAU\nRUQ->IAU\nIAU->RUQ\nRUQ->IAU\nIAU->RUQ\nRUQ->IAU\nIAU->RUQ\nIAU->RUQ\nRUQ->IAU",
"output": "contest"
},
{
"input": "10\nHPN\nDFI->HPN\nHPN->KAB\nHPN->DFI\nVSO->HPN\nHPN->KZX\nHPN->VSO\nKZX->HPN\nLDW->HPN\nKAB->HPN\nHPN->LDW",
"output": "home"
},
{
"input": "11\nFGH\nFGH->BRZ\nUBK->FGH\nQRE->FGH\nFGH->KQK\nFGH->QRE\nKQK->FGH\nFGH->UBK\nBRZ->FGH\nFGH->ALX\nALX->FGH\nFGH->KQK",
"output": "contest"
},
{
"input": "50\nPFH\nJFV->PFH\nBVP->PFH\nPFH->BVP\nPFH->JFV\nPFH->ETQ\nPFH->LQJ\nZTO->PFH\nPFH->BVP\nPFH->RXO\nPFH->ZTO\nHWL->PFH\nPFH->HIV\nPFH->AFP\nPFH->HWL\nOBB->PFH\nHIV->PFH\nPFH->LSR\nAFP->PFH\nLQJ->PFH\nHWL->PFH\nETQ->PFH\nPFH->HWL\nLSR->PFH\nWBR->PFH\nBNZ->PFH\nHQR->PFH\nZTO->PFH\nPFH->WBR\nPFH->BYJ\nRXO->PFH\nFHZ->PFH\nFHZ->PFH\nPFN->PFH\nPFH->GMB\nPFH->JFV\nJFV->PFH\nGNZ->PFH\nPFH->BNZ\nPFH->GNZ\nPFH->HQR\nBYJ->PFH\nGMB->PFH\nPFH->FHZ\nPFH->FHZ\nPFH->ZTO\nPFH->UGD\nBVP->PFH\nUGD->PFH\nPFH->PFN\nPFH->OBB",
"output": "home"
},
{
"input": "1\nAAK\nAAK->ABA",
"output": "contest"
},
{
"input": "1\nXYZ\nXYZ->XYR",
"output": "contest"
}
] | 1,594,606,980
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 93
| 6,656,000
|
n = int(input())
home = input()
for num in range(n):
x = input()
if num == n-1:
if x[:3] == home:
print('contest')
else:
print('home')
|
Title: Neverending competitions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
- this list contains all Jinotega's flights in this year (in arbitrary order), - Jinotega has only flown from his hometown to a snooker contest and back, - after each competition Jinotega flies back home (though they may attend a competition in one place several times), - and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input Specification:
In the first line of input there is a single integer *n*: the number of Jinotega's flights (1<=≤<=*n*<=≤<=100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next *n* lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output Specification:
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Demo Input:
['4\nSVO\nSVO->CDG\nLHR->SVO\nSVO->LHR\nCDG->SVO\n', '3\nSVO\nSVO->HKT\nHKT->SVO\nSVO->RAP\n']
Demo Output:
['home\n', 'contest\n']
Note:
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
|
```python
n = int(input())
home = input()
for num in range(n):
x = input()
if num == n-1:
if x[:3] == home:
print('contest')
else:
print('home')
```
| 0
|
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