Search is not available for this dataset
name stringlengths 2 88 | description stringlengths 31 8.62k | public_tests dict | private_tests dict | solution_type stringclasses 2
values | programming_language stringclasses 5
values | solution stringlengths 1 983k |
|---|---|---|---|---|---|---|
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
arr = list(map(int, input().split()))
m = int(input())
q = []
for i in range(m):
a, b = map(int, input().split())
q.append((a, b))
def f(arr, m):
vis = [False] * len(arr)
arr2 = []
for i in range(len(arr)):
arr2.append([arr[i], n - i])
arr2.sort(reverse=True... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const long double error = 2e-6;
const long double PI = acosl(-1);
inline long long int MOD(long long int x, long long int m = mod) {
long long int y = x % m;
return (y >= 0) ? y : y + m;
}
const int inf = 1e9;
const long long int infl ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | import sys
input = sys.stdin.buffer.readline
mx = 2*10**5 + 1
bit = [0]*(mx)
def add(idx):
idx += 1
while idx < mx:
bit[idx] += 1
idx += idx & - idx
# log(n) !!!
def lower_bound(val):
pos = 0
tot = 0
for i in range(20,-1,-1):
if pos + (1 << i) < mx and tot + bit[pos + (1 <... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int m;
cin >> m;
while (m--) {
map<int, pair<int, vector<int>>> mp;
int k, pos;
cin >> k >> pos;
vector<int> temp;
mp[0] = make_pair(0, t... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7;
int a[N];
vector<pair<int, int> > vec;
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first != b.first)
return a.first > b.first;
else
return a.second < b.second;
}
void solve() {
int k, pos;
cin >> k >> pos;
ve... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | import sys
def main():
n = int(input())
al = [[int(x),n-i] for i,x in enumerate(input().split())]
al.sort()
al.reverse()
arr = [[[]for i in range(n)]for i in range(n)]
for i in range(1,n+1):
for j in range(i-1,n):
arr[j][n-al[i-1][1]] = al[i-1][0]
for i in range(n):
arr[i] = list(filter(None, arr[i])) ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
b = list(map(int,input().split()))
a = [[0] * 2 for i in range(n)]
for i in range(n):
a[i][0] = b[i]
a[i][1] = i
for i in range(n-1):
for j in range(n-i-1):
if a[j][0] > a[j+1][0]:
a[j], a[j+1] = a[j+1], a[j]
elif (a[j][0] == a[j + 1][0]) and (a[j][1] < a[j + 1][... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = list(map(int, input().split()))
m = int(input())
for i in range(m):
k, pos = map(int, input().split())
save = []
for j in range(len(a)):
save.append(a[j])
ans = []
for x in range(k):
maximum = 0
pr = -1
for j in range(len(save)):
if sa... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n=int(input())
s=sorted([[v,-i] for i,v in enumerate(map(int,input().split()))])
for _ in range(int(input())):
k,i=map(int,input().split())
ans=sorted(s[-k:],key=lambda x:-x[1])
print(ans[i-1][0]) |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long N = 200000;
vector<bool> visited(N + 1, false);
bool isprime(long long x) {
for (long long i = 2; i <= sqrt(x); i++) {
if (x % i == 0) return false;
}
return true;
}
void findfact(long long x, map<long long, vector<long long>>& m) {
for (long long i = ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using LL = long long;
const LL INF = INT_MAX;
const int N = 1e5 + 7;
const LL MOD = 1e9 + 7;
void ArrayIn(int size, int a[]) {
for (int i = 0; i < size; i++) scanf("%d", &a[i]);
}
void ArrayOut(int size, int a[]) {
for (int i = 0; i < size; i++) printf("%d ", a[i]);
p... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long int no = 3e6 + 5, modulo = 1e9 + 7, inf = 1e18, N = 3e3 + 1;
long long int ar[no], br[no], cr[no];
void solve() {
long long int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, x = 0, y = 0, z = 0,
w = 0, k = 0;
cin >> n;
vector<long long int>... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long double pi = 2 * acos(0.0);
struct arr {
long long index, val;
};
struct query {
long long n, ind, index, ans;
};
bool compare(arr a1, arr a2) {
if (a1.val == a2.val) return (a1.index < a2.index);
return (a1.val > a2.val);
}
bool compare2(arr a1, arr a2) {... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = list(map(int, input().split()))
m = int(input())
for i in range(m):
d = dict()
k, pos = map(int, input().split())
b = a + []
for i in range(k):
d[b.index(max(b))] = max(b)
b[b.index(max(b))] = 0
print(d[sorted(d.keys())[pos - 1]]) |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1e12;
int main() {
int n;
cin >> n;
vector<int> a(n), s;
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
s = a;
sort(s.begin(), s.end());
int m;
cin >> m;
for (int i = 0; i < m; ++i) {
int k, pos;
cin >> k >> pos;
int c = 0;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 123, MAXN = 5e5 + 47, MEGAINF = 1e18 + 228;
template <class T>
inline istream& operator>>(istream& in, vector<T>& a) {
for (auto& i : a) in >> i;
return in;
}
template <class T>
inline ostream& operator<<(ostream& out, vector<T>& a) {
for (... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100100;
using namespace std;
int ans[110][110];
struct node {
int nub, pos;
} a[110];
bool cmp(node c, node b) {
if (c.nub == b.nu... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | def mergesort(l, r, arr, pos):
if r - l == 1:
return arr, pos
m = (l + r) // 2
arr, pos = mergesort(l, m, arr, pos)
arr, pos = mergesort(m, r, arr, pos)
c = [0 for i in range(r)]
d = [0 for i in range(r)]
poi_a = l
poi_b = m
for i in range(l, r):
if poi_a == m:
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
namespace kotespace {
template <class T>
class duplet {
private:
public:
T x, y;
duplet(){};
duplet(T a, T b) : x(a), y(b){};
bool operator<(const duplet P) const {
return (x < P.x || (x == P.x && y < P.y));
}
bool operator>(const duplet P) const {
ret... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 2, mod = 1e9 + 7;
void solve() {
int n;
cin >> n;
vector<pair<int, int> > a(n);
for (int i = 0; i < n; i++) {
cin >> a[i].first;
a[i].second = i;
}
sort(a.begin(), a.end(), [](pair<int, int> i, pair<int, int> j) {
if (i.first == j... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using ld = long double;
void solve();
bool comp(const pair<int64_t, int64_t> &a, const pair<int64_t, int64_t> &b) {
if (a.first != b.first) return (a.first < b.first);
return (a.second > b.second);
}
int32_t main() {
ios_base::sync_with_stdio(0);
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
bool choice_first(std::vector<size_t> a, std::vector<size_t> b) {
uint64_t sum_a, sum_b;
sum_a = sum_b = 0;
for (auto elem : a) {
sum_a += elem;
}
for (auto elem : b) {
sum_b += elem;
}
return sum_a > sum_b || (sum_a == sum_b && a < b);
}
int main() {
std::ios_base::sync... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | /*
*
* @Author Ajudiya_13(Bhargav Girdharbhai Ajudiya)
* Dhirubhai Ambani Institute of Information And Communication Technology
*
*/
import java.util.*;
import java.io.*;
import java.lang.*;
public class Code85
{
public static void main(String[] args)
{
InputReader in = new InputReader(System.... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | /*
TO LEARN
1-segment trees
2-euler tour
3-fenwick tree and interval tree
*/
/*
TO SOLVE
uva 1103
*/
/*
bit manipulation shit
1-Computer Systems: A Programmer's Perspective
2-hacker's delight
3-(02-03-bits-ints)
4-machine-basics
5-Bits Manipulation tutorialspoint
*/
import java.util.*;
import java.ma... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int Z = (int)3e3 + 228;
const int N = (int)3e5 + 228;
const int INF = (int)1e9 + 228;
const int MOD = (int)1e9 + 7;
int a[N];
map<int, int> cnt;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
vector<int> b;
for (int ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, q, arr[101], sr[101];
vector<int> dp[101];
multiset<int> s;
int main() {
cin >> n;
for (long long i = 0; i < n; i++) {
cin >> arr[i];
sr[i] = arr[i];
}
sort(sr, sr + n, [](int a, int b) { return a > b; });
for (long long i = 1; i <= n; i++) {
fo... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void func() {
int n;
cin >> n;
vector<int> arr(n), acop(n);
for (int i = 0; i < n; ++i) {
cin >> arr[i];
acop[i] = arr[i];
}
sort(acop.begin(), acop.end(), greater<int>());
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int k, pos, cnt = 0;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const long long INF = 1e9 + 5;
const double eps = 1e-7;
const double PI = acos(-1.0);
inline void debug_vi(vector<int> a) {
for (long long i = (long long)(0); i < (long long)(a.size()); i++)
cout << a[i] << " ";
}
inline void debug_vl... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
public class OptimalSubsequences {
InputStream is;
PrintWriter pw;
String INPUT = "";
long L_INF = (1L << 60L);
void solve() {
int k,pos,n=ni(), m;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2, typename T3>
struct pair3 {
T1 first;
T2 second;
T3 third;
};
template <typename T1, typename T2, typename T3, typename T4>
struct pair4 {
T1 first;
T2 second;
T3 third;
T4 fourth;
};
const long long MOD = 1000000007;
const ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct eq {
long long si, f, s, r;
};
bool compare(eq a, eq b) {
if (a.f == b.f) return a.s < b.s;
return a.f < b.f;
}
bool comp(eq a, eq b) { return a.si < b.si; }
void solve() {
long long n;
cin >> n;
vector<pair<long long, long long> > v;
for (long long i =... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
bool comp(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
if (a.first == b.first) {
return a.second < b.second;
}
return a.first > b.first;
}
int main() {
int n;
cin >> n;
vector<long long int> arr(n);
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
vector<long long> a(n), b(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < n; i++) {
b[i] = a[i];
}
sort(a.begin(), a.end(), greater<long long>());
long long m;
cin >> m;
vector<long long>... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
long long ar[n];
long long br[n];
for (auto x = 0; x < n; x++) {
cin >> ar[x];
br[x] = ar[x];
}
sort(br, br + n);
long long m;
cin >> m;
while (m--) {
long long pos, k;
set<long long> s;
cin >> k >>... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
signed main() {
long long n;
cin >> n;
vector<long long> v(n);
map<long long, vector<long long>> def;
for (long long i = 0; i < n; i++) {
cin >> v[i];
def[v[i]].push_back(i);
}
vector<long long> g;
for (auto p : def) {
long long a = p.first;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n=int(input())
a=[int(x) for x in input().split()]
z=a[:]
z.sort(reverse=True)
an=[[] for i in range(n+1)]
an[0]=a[:]
for i in range(n):
for j in range(len(a)-1,-1,-1) :
# print(j,z)
if a[j]==z[-1]:
del a[j]
z.pop()
break
an[i+1]=a[:]
an=an[::-1]
for... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long bin_pow(long long a, long long b) {
if (b == 0) return 1;
if (b % 2 == 0) {
long long t = bin_pow(a, b / 2);
return t * t % 1000000007;
} else
return a * bin_pow(a, b - 1) % 1000000007;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<int> srr;
map<int, int> occ;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int arr[200], n, m, pj, k, fast = 1, last;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> arr[i];
srr.push_back(arr[i]);
}
sort(srr.begin(), srr.end(), greater<i... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, n, m, k;
cin >> n;
vector<int> v[n + 1];
vector<long long int> s;
for (int i = 0; i < n; i++) {
cin >> t;
v[n].push_back(t);
s.push_back(t);
}
sort(s.begin(), s.end());
int l = n - 1, x = 0, y;
while (l > 0) {
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java |
import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
public class competetive {
static class pair implements Comparable<pair>{
long f;
int s;
pair(long a,int b){
this.f=a;
this.s=b;
}
long getF(){
return f;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> arr(n);
set<pair<int, int>> left;
for (int i = 0; i < arr.size(); i++) {
cin >> arr[i];
left.insert(make_pair(arr[i], i * -1));
}
int m;
cin >> m;
while (m--) {
int k, j;
cin >> k >> j;
j--;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 1234567891, M = 31;
long long n, m, t, k, f, mi = 2e9, x, y, z, ma, x3, x2, y2, l, r, i, j, ans, vv,
a2[410000], d[410000], dp[810000];
strin... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n=int(input())
a=[int(i) for i in input().split()]
copy1=a[:]
a.sort()
m=int(input())
for i in range(m):
k,pos=map(int,input().split())
ans=[-1]
copy=a[-k:]
for i in copy1:
if i in copy:
copy.pop(copy.index(i))
ans.append(i)
print(ans[pos])
|
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = [int(i) for i in input().split()]
b = sorted(a)
b.reverse()
m = int(input())
for _ in range(m):
k, pos = map(int, input().split())
x = 0
j = 0
d = dict()
ans = 0
for i in b[:k]:
d[i] = d.get(i, 0) + 1
while x != pos:
u = a[j]
if u in d:
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n=int(input())
a=list(map(int,input().split()))
d=[]
for i in range(n):
d.append([a[i],-i])
d.sort(reverse=True)
e=[]
for i in range(n):
e.append([-d[i][1],d[i][0]])
m=int(input())
f=[]
for i in range(m):
b,c=map(int,input().split())
f=e[:b]
f.sort()
print(f[c-1][1]) |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
void ct(vector<int> &a) {
for (auto &i : a) cout << i << ' ';
cout << '\n';
}
void ct(vector<pair<int, int>> &a) {
for (auto &i : a) cout << i.first << ":" << i.second << ' ';
cout <... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.*;
import java.util.*;
public class OptimalSubsequences {
public static void main(String[] args) {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
HashMap<Integer,ArrayList<Integer>> map = new HashMap<>();
int n = in.nextInt();
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m;
pair<int, int> a[112];
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
for (int i = 0; i < n; i++) {
int num;
cin >> num;
a[i] = {-num, i};
}
sort(a, a + n);
cin >> m;
for (int i = 1; i <= m; i++) {
int k, pos;
cin >> k >> po... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n=int(input())
import copy
mins=[0]*n
a=list(map(int,input().split()))
m=int(input())
for i in range(m):
f=a.copy()
k,pos=map(int,input().split())
for l in range(n-k):
if mins[l]==0:
mins[l]=min(f)
for j in range(len(f)-1,-1,-1):
if f[j]==mins[l]:
f.po... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 9;
struct node {
int x, y;
} a[maxn];
int b[maxn];
bool cmp(node a, node b) { return a.x == b.x ? a.y < b.y : a.x > b.x; }
map<int, int> A;
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i].x);
b[i] ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 100;
int arr[N];
multimap<int, int, greater<int>> l;
map<int, int> mp;
vector<int> ans[N];
int main() {
ios_base::sync_with_stdio();
cin.tie(0);
cout.tie(0);
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", arr ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
b = list(map(int, input().split()))
a = []
for i in range(n):
a.append([b[i], n - i])
a.sort()
a.reverse()
p = []
t = []
for i in range(n):
t.append(a[i])
d = t.copy()
d.sort(key=lambda x: -x[1])
p.append(d)
m = int(input())
#print(p)
for i in range(m):
k, pos = map(int, input()... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int bsl(vector<pair<int, int>> &A, int val) {
int left = -1;
int right = (int)A.size();
while (right - left > 1) {
int mid = (left + right) / 2;
if (A[mid].first >= val) {
right = mid;
} else {
left = mid;
}
}
return right;
}
int bsr(ve... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = [int(x) for x in input().split()]
a_sorted = sorted(a, reverse=True)
m = int(input())
for i in range(m):
k, pos = [int(x) for x in input().split()]
vals = a_sorted[0:k]
result = []
for val in a:
if val in vals:
result.append(val)
vals.remove(val)
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, m, x, i, ma = 0;
cin >> n;
vector<pair<long long int, long long int>> v1;
for (i = 1; i <= n; i++) cin >> x, v1.push_back(make_pair(-x, i));
sort(v1.begin(), v1.end());
cin >> m;
while (m--) {
long long int k, pos, j = 0;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int inf = 1e9 + 7;
bool compA(pair<int, int> a, pair<int, int> b) {
if (a.first == b.first) return a.second < b.second;
return a.first > b.first;
}
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void run() ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.*;
import java.util.*;
public class D {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int n = sc.nextInt();
int[] arr = new int[n];
PriorityQueue<Pair> p = new Pr... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long MOD = 1e9 + 7;
const long long mxN = 2e5 + 3;
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
if (p1.first > p2.first)
return true;
else if (p1.first == p2.firs... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
struct node {
int id;
int num;
};
node s[maxn];
int b[maxn];
int num[maxn];
bool cmp(node a, node b) {
if (a.num != b.num)
return a.num > b.num;
else
return a.id < b.id;
}
struct ask {
int id;
int L = 0;
int pos;
};
ask A[ma... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rnd(time(0));
const long long INF = 1e9;
struct Point {
Point() {
cin >> x;
cin >> y;
}
Point(long double x, long double y) : x(x), y(y) {}
long double x, y;
};
void solve() {
int n, m, k, pos;
cin >> n;
vector<int... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
const long long num = 1000000000;
bool check1(pair<long long, long long> a, pair<long long, long long> b) {
return a.first == b.first ? (a.second > b.second) : (a.first < b.first);
}
bool check2(pair<long long, long long> a, pair<long long, lon... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long int powMod(long long int x, long long int y) {
long long int p = 1;
while (y) {
if (y % 2) {
p = (p * x) % 1000000007;
}
y /= 2;
x = (x * x) % 1000000007;
}
return p;
}
long long int invMod(long long int x) { return powMod(x, 10000000... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cer... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = list(map(int,input().split()))
for i in range(n):
a[i] = (a[i],-i)
a.sort()
m = int(input())
for j in range(m):
t = list(map(int, input().split()))
k = t[0]
p = t[1]
print(sorted(a[-k:], key = lambda x: -x[1])[p-1][0]) |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e5 + 7;
const int inf = INT_MAX / 2;
const long long INF = LLONG_MAX / 3;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const string cars[] = {"🚗", "🚕", "🚙"};
signed main() {
cout <... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
l = [int(j) for j in input().split()]
m = int(input())
d = dict()
for i in range(n):
if l[i] in d:
d[l[i]].append(i)
else:
d[l[i]]=[i]
d = sorted(d.items(), reverse =True)
# print(d)
for que in range(m):
k, pos = [int(j) for j in input().split()]
min_ = []
i = 0
j = 0
while(k>0):
k-=1
mi... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m;
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first != b.first) return a.first > b.first;
return a.second < b.second;
}
bool cmp2(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
ios_base::sync_with_stdio(0);
cin.tie(... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long int, long long int> a,
pair<long long int, long long int> b) {
if (a.first != b.first)
return a.first < b.first;
else
return a.second > b.second;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<long long int> graph[200007];
long long int visited[200007] = {0};
long long int mx = 0, cnt = 0;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, m, i, j;
cin >> n;
long long int a[n + 10];
vector<pair<long long int, long lo... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = list(map(int, input().split()))
s_a = sorted(a)
m = int(input())
req = []
for _ in range(m):
req.append(list(map(int, input().split())))
d = dict()
for r in req:
if r[0] in d.keys():
print(d[r[0]][r[1] - 1])
else:
del_indx = []
for e in s_a[:n-r[0]]:
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first != b.first) return a.first < b.first;
return a.second > b.second;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector<pair<int, int>> a;
for (int i = 0; i < n; ++i) {
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
using namespace std;
int a[1000], b[1000];
map<int, bool> mk;
int main() {
iostream::sync_with_stdio(0);
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >>... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void solve();
void fast() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
}
template <typename T>
istream &operator>>(istream &in, vector<T> &ans) {
for (auto &d : ans) in >> d;
return in;
}
template <typename T>
ostream &operator<<(ostrea... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int a[101];
bool comp(pair<int, int> x, pair<int, int> y) {
if (x.first == y.first) {
return x.second < y.second;
}
return x.first > y.first;
}
bool comp2(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
int d[101][101];
int main() {
cin.sync_... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append([a[i], -i])
b.sort(reverse=True)
for i in range(int(input())):
k, pos = map(int, input().split())
ans = 0
z = []
for i in range(k):
ans += b[i][0]
z.append([-b[i][1], b[i][0]])
z.sort()
pr... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline void fastio() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
}
void solve() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) cin >> a[i];
map<int, vector<int> > mp;
for (int i = 0; i < n; ++i) {
mp[a[i]]... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
using namespace std;
bool compare(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first != b.first) {
return a.first > b.first;
} else {
return a.second < b.second;
}
}
int32_t main() {
long long n;
cin >> n;
long long arr[n... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
data = input().split()
data1 = []
for i in range(n):
data1.append((int(data[i]), i))
data1.sort(key=lambda x: (x[0], -x[1]))
for i in range(int(input())):
k, pos = map(int, input().split())
temp = sorted(data1[len(data1) - k:], key=lambda x: x[1])
print(temp[pos - 1][0]) |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
vector<pair<int, int>> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i].first;
v[i].second = i + 1;
v[i].first *= -1;
}
sort(v.begin(), v.end());
int t;
cin >> t... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int dv(pair<int, int> p1, pair<int, int> p2) {
if (p1.first < p2.first) return 0;
if (p1.first > p2.first) return 1;
if (p1.second > p2.second) return 0;
return 1;
}
int main() {
int t, n, m, p, k, a, b, c, l, r;
cin >> n;
int A[n];
for (int i = 0; i < n; i+... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<vector<long long>> adj;
map<long, bool> vis, viss;
vector<long long> rnk, parent, sz;
int spf[1000000 + 1];
long long binpow(long long a, long long b, long long m) {
a %= m;
long long res = 1;
while (b > 0) {
if (b & 1) res = res * a % m;
a = a * a % m;... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
bool kmp(int a, int b) { return a < b; }
int main() {
int n, m;
int x = 0;
int y = 0;
int z = 0;
cin >> n;
int a[n];
int d[n][n];
int c[n];
int an[100003];
int k = 0;
int p = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
c[i] = 1;
}
cin >... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int getrnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <typename T1, typename T2>
bool relax(T1... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
const int maxn = 1e5 + 5;
using namespace std;
int n, q;
struct node {
int x, y, z;
} a[maxn], b[maxn], c[maxn];
int tot = 1, ans[maxn];
bool tmp(const node &u, const node &v) {
if (u.x != v.x) return u.x > v.x;
return u.y < v.y;
}
bool cc(const node &u, const node &v) {
if (u.x != v.x)... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java |
import java.io.*;
import java.util.*;
import java.math.BigInteger;
import java.lang.Object;
public class Main {
static class sort implements Comparator<int[]>
{
public int compare(int[] a,int[] b)
{
if(a[1] == b[1]) return -b[0]+a[0];
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.Big... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
arr = [int(i) for i in input().split()]
sor = [[arr[i], n - i] for i in range(n)]
sor.sort()
m = int(input())
for i in range(m):
op = []
[q, index] = [int(i) for i in input().split()]
for j in range(q):
op.append(n - sor[-1-j][1])
op.sort()
print(arr[op[index - 1]])
|
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | from math import *
import os, sys
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append((-a[i], i))
b.sort()
#print(b)
for i in range(int(input())):
k, pos = map(int, input().split())
tmp = [... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void Update(const long &);
long FindSum(const long &);
long N, M, K = 0;
vector<long> BIT;
int main() {
cin >> N;
vector<pair<long, long>> A(N + 1);
vector<long> X(N + 1);
for (long i = 1; i <= N; ++i) {
cin >> A[i].first;
A[i].second = i;
X[i] = A[i].fi... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
arr = list(map(int, input().split()))
arr_sorted = [(arr[i], i) for i in range(n)]
arr_sorted.sort(key = lambda el: (el[0], -el[1]), reverse = True)
m = int(input())
for req_i in range(m):
k, pos = map(int, input().split())
pos -= 1
cur_arr = arr_sorted[:k]
cur_arr.sort(key = lambda el:... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
public class Template {
static int mod = 1000000007;
public static void main(String[] args) {
FastScanner sc = new Fa... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<bool> sieve(long long n) {
vector<bool> prime(n + 1, true);
prime[0] = prime[1] = false;
for (long long i = 2; i <= n; ++i) {
if (prime[i] && (i * i) <= n)
for (long long j = i * i; j <= n; j += i) prime[j] = false;
}
return prime;
}
long long pow... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | def f(a):
return -a[1]
import sys
fin = sys.stdin
n = int(input())
arr = list(map(int, fin.readline().split()))
arr_ind = list()
for i in range(n):
arr_ind.append([arr[i], -i])
arr_ind.sort()
arr_ind.reverse()
ans_array = list()
for k in range(1, n + 1):
ans = arr_ind[:k]
ans.sort(key=f)
ans_array.... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
template <class c>
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(20);
long long n = 0;
cin >> n;
deque<long long> posl(n);
for (auto& p : posl) cin >> p;
map<long long, long long> al;
for (auto& p : posl) al[p]++;
long long m = 0;
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | n = int(input())
nar = list(map(int, input().split()))
m = int(input())
for q in range(m):
k, p = list(map(int, input().split()))
mar = []
for i in range(k):
mar.append(nar[i])
for i in range(k, n):
minItem = 0
for j in range(1, k):
if mar[j] <= mar[minItem]:
... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
FastReader input=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int T=1;
while(T-->0)
{
int n=input.nextInt();
ArrayList<Integer> list=ne... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | python3 | import sys
# from math
import bisect
import heapq
# from collections import deque
# from types import GeneratorType
# def bootstrap(func, stack=[]):
# def wrapped_function(*args, **kwargs):
# if stack:
# return func(*args, **kwargs)
# else:
# call = func(*args, **kwargs)
# ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int i, j, k, l, m, n;
cin >> n;
long long int a[n];
vector<long long int> v;
for (i = 0; i < n; i++) {
cin >> a[i];
v.push_back(a[i]);
}
sort(v.begin(), v.end());
long long int var;
cin >> var;
while (var--) {
cin >> k ... |
1227_D1. Optimal Subsequences (Easy Version) | This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily... | {
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
} | {
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10... | CORRECT | java | import java.io.PrintWriter;
import java.util.*;
public class OptimalSubsequences {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = 1;
for (int i = 0; i < t; i++) {
solve(sc, pw);
... |
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