message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arpa is researching the Mexican wave.
There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.
* At time 1, the first spectator stands.
* At time 2, the second spectator stands.
* ...
* At time k, the k-th spectator stands.
* At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
* At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
* ...
* At time n, the n-th spectator stands and the (n - k)-th spectator sits.
* At time n + 1, the (n + 1 - k)-th spectator sits.
* ...
* At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.
Input
The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).
Output
Print single integer: how many spectators are standing at time t.
Examples
Input
10 5 3
Output
3
Input
10 5 7
Output
5
Input
10 5 12
Output
3
Note
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
* At t = 0 ---------- <image> number of standing spectators = 0.
* At t = 1 ^--------- <image> number of standing spectators = 1.
* At t = 2 ^^-------- <image> number of standing spectators = 2.
* At t = 3 ^^^------- <image> number of standing spectators = 3.
* At t = 4 ^^^^------ <image> number of standing spectators = 4.
* At t = 5 ^^^^^----- <image> number of standing spectators = 5.
* At t = 6 -^^^^^---- <image> number of standing spectators = 5.
* At t = 7 --^^^^^--- <image> number of standing spectators = 5.
* At t = 8 ---^^^^^-- <image> number of standing spectators = 5.
* At t = 9 ----^^^^^- <image> number of standing spectators = 5.
* At t = 10 -----^^^^^ <image> number of standing spectators = 5.
* At t = 11 ------^^^^ <image> number of standing spectators = 4.
* At t = 12 -------^^^ <image> number of standing spectators = 3.
* At t = 13 --------^^ <image> number of standing spectators = 2.
* At t = 14 ---------^ <image> number of standing spectators = 1.
* At t = 15 ---------- <image> number of standing spectators = 0.
Submitted Solution:
```
n,k,t = map(int,input().split())
if t <= k:
print(t)
elif t<= n:
print(k)
else:
print(k-(t-n +k))
``` | instruction | 0 | 27,078 | 3 | 54,156 |
No | output | 1 | 27,078 | 3 | 54,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arpa is researching the Mexican wave.
There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.
* At time 1, the first spectator stands.
* At time 2, the second spectator stands.
* ...
* At time k, the k-th spectator stands.
* At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
* At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
* ...
* At time n, the n-th spectator stands and the (n - k)-th spectator sits.
* At time n + 1, the (n + 1 - k)-th spectator sits.
* ...
* At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.
Input
The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).
Output
Print single integer: how many spectators are standing at time t.
Examples
Input
10 5 3
Output
3
Input
10 5 7
Output
5
Input
10 5 12
Output
3
Note
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
* At t = 0 ---------- <image> number of standing spectators = 0.
* At t = 1 ^--------- <image> number of standing spectators = 1.
* At t = 2 ^^-------- <image> number of standing spectators = 2.
* At t = 3 ^^^------- <image> number of standing spectators = 3.
* At t = 4 ^^^^------ <image> number of standing spectators = 4.
* At t = 5 ^^^^^----- <image> number of standing spectators = 5.
* At t = 6 -^^^^^---- <image> number of standing spectators = 5.
* At t = 7 --^^^^^--- <image> number of standing spectators = 5.
* At t = 8 ---^^^^^-- <image> number of standing spectators = 5.
* At t = 9 ----^^^^^- <image> number of standing spectators = 5.
* At t = 10 -----^^^^^ <image> number of standing spectators = 5.
* At t = 11 ------^^^^ <image> number of standing spectators = 4.
* At t = 12 -------^^^ <image> number of standing spectators = 3.
* At t = 13 --------^^ <image> number of standing spectators = 2.
* At t = 14 ---------^ <image> number of standing spectators = 1.
* At t = 15 ---------- <image> number of standing spectators = 0.
Submitted Solution:
```
n,k,t=map(int,input().split())
if t>n:
t=t%n
t=k-t
if t<k:
print(t)
elif t<n and t>=k:
print(k)
``` | instruction | 0 | 27,079 | 3 | 54,158 |
No | output | 1 | 27,079 | 3 | 54,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arpa is researching the Mexican wave.
There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.
* At time 1, the first spectator stands.
* At time 2, the second spectator stands.
* ...
* At time k, the k-th spectator stands.
* At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
* At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
* ...
* At time n, the n-th spectator stands and the (n - k)-th spectator sits.
* At time n + 1, the (n + 1 - k)-th spectator sits.
* ...
* At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.
Input
The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).
Output
Print single integer: how many spectators are standing at time t.
Examples
Input
10 5 3
Output
3
Input
10 5 7
Output
5
Input
10 5 12
Output
3
Note
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
* At t = 0 ---------- <image> number of standing spectators = 0.
* At t = 1 ^--------- <image> number of standing spectators = 1.
* At t = 2 ^^-------- <image> number of standing spectators = 2.
* At t = 3 ^^^------- <image> number of standing spectators = 3.
* At t = 4 ^^^^------ <image> number of standing spectators = 4.
* At t = 5 ^^^^^----- <image> number of standing spectators = 5.
* At t = 6 -^^^^^---- <image> number of standing spectators = 5.
* At t = 7 --^^^^^--- <image> number of standing spectators = 5.
* At t = 8 ---^^^^^-- <image> number of standing spectators = 5.
* At t = 9 ----^^^^^- <image> number of standing spectators = 5.
* At t = 10 -----^^^^^ <image> number of standing spectators = 5.
* At t = 11 ------^^^^ <image> number of standing spectators = 4.
* At t = 12 -------^^^ <image> number of standing spectators = 3.
* At t = 13 --------^^ <image> number of standing spectators = 2.
* At t = 14 ---------^ <image> number of standing spectators = 1.
* At t = 15 ---------- <image> number of standing spectators = 0.
Submitted Solution:
```
n, k, t = (int(x) for x in input().split())
print(min(k, (n + k) % t))
``` | instruction | 0 | 27,080 | 3 | 54,160 |
No | output | 1 | 27,080 | 3 | 54,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.
Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).
Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.
Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.
Input
The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.
The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.
Output
If it is impossible to obtain a tank with exactly V ml of water, print NO.
Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format:
Each line has to contain 3 numbers denoting a compressed operation: "cnt x y" (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y.
The number of these lines must not exceed N + 5.
Examples
Input
2 3 5
2 3
Output
YES
1 2 1
Input
2 3 4
2 3
Output
NO
Input
5 2 0
1 3 5 7 9
Output
YES
2 2 1
3 3 1
4 4 1
5 5 1
Submitted Solution:
```
N = int(5e3 + 10)
n, m, v = map(int, input().split())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append(a[i] % m)
f = [-1] * N
g = [-1] * N
h = [[-1 for j in range(N)] for i in range(N)]
f[0] = 0
for i in range(n):
nf = f.copy()
ng = g.copy()
for j in range(m):
if (~f[j]):
k = (j + b[i]) % m
upd = f[j] + a[i]
if (nf[k] < upd):
h[i][k] = g[j]
nf[k] = upd
ng[k] = i
f = nf.copy()
g = ng.copy()
r = v % m
if (~f[r]):
if (f[r] >= v):
print("YES")
ext = 0
if (g[r] == 0):
ext = n - 1
snk = g[r]
p = snk
while (True):
lp = h[p][r]
if (lp == -1):
break
print((a[lp] - 1) // m + 1, lp + 1, snk + 1)
r = (r - b[p] + m) % m
p = lp
r = v % m
if (f[r] > v):
print((f[r] - v) // m, snk + 1, ext + 1)
else:
s = 0
for i in range(n):
s += a[i]
if (s >= v):
print("YES")
quit()
flag = [1] * N
snk = g[r]
flag[snk] = 0
p = snk
while (True):
lp = h[p][r]
if (lp == -1):
break
print((a[lp] - 1) // m + 1, lp + 1, snk + 1)
r = (r - b[p] + m) % m
p = lp
flag[p] = 0
c = []
for i in range(n):
if (flag[i]):
c.append(i)
for i in c[1:]:
print((a[i] - 1) // m + 1, i + 1, c[0] + 1)
r = v % m
print((v - f[r]) // m, c[0] + 1, snk + 1)
else:
print("NO")
else:
print("NO")
``` | instruction | 0 | 27,092 | 3 | 54,184 |
No | output | 1 | 27,092 | 3 | 54,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.
Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).
Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.
Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.
Input
The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.
The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.
Output
If it is impossible to obtain a tank with exactly V ml of water, print NO.
Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format:
Each line has to contain 3 numbers denoting a compressed operation: "cnt x y" (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y.
The number of these lines must not exceed N + 5.
Examples
Input
2 3 5
2 3
Output
YES
1 2 1
Input
2 3 4
2 3
Output
NO
Input
5 2 0
1 3 5 7 9
Output
YES
2 2 1
3 3 1
4 4 1
5 5 1
Submitted Solution:
```
import sys
n, k, v = map(int, input().split())
a = list(map(int, input().split()))
total = sum(a)
if total < v:
print('NO')
exit()
dp = [[0]*k for _ in range(n+1)]
prev = [[-1]*(k+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(n):
for j in range(k):
if not dp[i][j]:
continue
dp[i+1][j] = 1
prev[i+1][j] = j
dp[i+1][(j+a[i]) % k] = 1
prev[i+1][(j+a[i]) % k] = j
if not dp[n][v % k]:
print('NO')
exit()
group_a, group_b = [], []
j = v % k
for i in range(n, 0, -1):
if prev[i][j] == j:
group_b.append(i-1)
else:
group_a.append(i-1)
j = prev[i][j]
ans = []
if not group_b:
water = total
for i in range(2, n+1):
ans.append(f'5000 {i+1} 1')
assert (total - v) >= 0 and (total - v) % k == 0
rem = (total - v) // k
ans.append(f'{rem} 1 2')
elif not group_a:
for i in range(2, n+1):
ans.append(f'5000 {i+1} 1')
assert v % k == 0
req = v // k
ans.append(f'{req} 1 2')
else:
water_a, water_b = sum(a[i] for i in group_a), sum(a[i] for i in group_b)
for i in range(1, len(group_a)):
ans.append(f'5000 {group_a[i]+1} {group_a[0]+1}')
for i in range(1, len(group_b)):
ans.append(f'5000 {group_b[i]+1} {group_b[0]+1}')
if water_a > v:
assert (water_a - v) % k == 0
rem = (water_a - v) // k
ans.append(f'{rem} {group_a[0]+1} {group_b[0]+1}')
elif water_a < v:
assert (v - water_a) % k == 0
req = (v - water_a) // k
ans.append(f'{req} {group_b[0]+1} {group_a[0]+1}')
sys.stdout.buffer.write('\n'.join(ans).encode('utf-8'))
``` | instruction | 0 | 27,093 | 3 | 54,186 |
No | output | 1 | 27,093 | 3 | 54,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.
Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).
Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.
Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.
Input
The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.
The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.
Output
If it is impossible to obtain a tank with exactly V ml of water, print NO.
Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format:
Each line has to contain 3 numbers denoting a compressed operation: "cnt x y" (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y.
The number of these lines must not exceed N + 5.
Examples
Input
2 3 5
2 3
Output
YES
1 2 1
Input
2 3 4
2 3
Output
NO
Input
5 2 0
1 3 5 7 9
Output
YES
2 2 1
3 3 1
4 4 1
5 5 1
Submitted Solution:
```
import sys
n, k, v = map(int, input().split())
a = list(map(int, input().split()))
total = sum(a)
if total < v:
print('NO')
exit()
dp = [[0]*k for _ in range(n+1)]
prev = [[-1]*(k+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(n):
for j in range(k):
if not dp[i][j]:
continue
dp[i+1][j] = 1
prev[i+1][j] = j
dp[i+1][(j+a[i]) % k] = 1
prev[i+1][(j+a[i]) % k] = j
if not dp[n][v % k]:
print('NO')
exit()
group_a, group_b = [], []
j = v % k
for i in range(n, 0, -1):
if prev[i][j] == j:
group_b.append(i-1)
else:
group_a.append(i-1)
j = prev[i][j]
ans = []
if not group_b:
water = total
for i in range(2, n+1):
ans.append(f'5000 {i+1} 1')
assert (total - v) >= 0 and (total - v) % k == 0
rem = (total - v) // k
if rem > 0:
ans.append(f'{rem} 1 2')
elif not group_a:
for i in range(2, n+1):
ans.append(f'5000 {i+1} 1')
assert v % k == 0
req = v // k
ans.append(f'{req} 1 2')
else:
water_a, water_b = sum(a[i] for i in group_a), sum(a[i] for i in group_b)
for i in range(1, len(group_a)):
ans.append(f'5000 {group_a[i]+1} {group_a[0]+1}')
for i in range(1, len(group_b)):
ans.append(f'5000 {group_b[i]+1} {group_b[0]+1}')
if water_a > v:
assert (water_a - v) % k == 0
rem = (water_a - v) // k
ans.append(f'{rem} {group_a[0]+1} {group_b[0]+1}')
elif water_a < v:
assert (v - water_a) % k == 0
req = (v - water_a) // k
ans.append(f'{req} {group_b[0]+1} {group_a[0]+1}')
sys.stdout.buffer.write(('YES\n'+'\n'.join(ans)).encode('utf-8'))
``` | instruction | 0 | 27,094 | 3 | 54,188 |
No | output | 1 | 27,094 | 3 | 54,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.
Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).
Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.
Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.
Input
The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.
The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.
Output
If it is impossible to obtain a tank with exactly V ml of water, print NO.
Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format:
Each line has to contain 3 numbers denoting a compressed operation: "cnt x y" (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y.
The number of these lines must not exceed N + 5.
Examples
Input
2 3 5
2 3
Output
YES
1 2 1
Input
2 3 4
2 3
Output
NO
Input
5 2 0
1 3 5 7 9
Output
YES
2 2 1
3 3 1
4 4 1
5 5 1
Submitted Solution:
```
N = int(10)
n, m, v = map(int, input().split())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append(a[i] % m)
f = [-1] * N
g = [-1] * N
h = [[-1 for j in range(N)] for i in range(N)]
f[0] = 0
for i in range(n):
nf = f.copy()
ng = g.copy()
for j in range(m):
if (~f[j]):
k = (j + b[i]) % m
upd = f[j] + a[i]
if (nf[k] < upd):
h[i][k] = g[j]
nf[k] = upd
ng[k] = i
f = nf.copy()
g = ng.copy()
r = v % m
if (~f[r]):
if (f[r] >= v):
print("YES")
ext = 0
if (g[r] == 0):
ext = n - 1
snk = g[r]
p = snk
while (True):
lp = h[p][r]
if (lp == -1):
break
print((a[lp] - 1) // m + 1, lp + 1, snk + 1)
r = (r - b[lp] + m) % m
p = lp
r = v % m
if (f[r] > v):
print((f[r] - v) // m, snk + 1, ext + 1)
else:
s = 0
for i in range(n):
s += a[i]
if (s >= v):
print("YES")
flag = [1] * N
snk = g[r]
flag[snk] = 0
p = snk
while (True):
lp = h[p][r]
if (lp == -1):
break
print((a[lp] - 1) // m + 1, lp + 1, snk + 1)
r = (r - b[lp] + m) % m
p = lp
flag[p] = 0
c = []
for i in range(n):
if (flag[i]):
c.append(i)
for i in c[1:]:
print((a[i] - 1) // m + 1, i + 1, c[0] + 1)
r = v % m
print((v - f[r]) // m, c[0] + 1, snk + 1)
else:
print("NO")
else:
print("NO")
``` | instruction | 0 | 27,095 | 3 | 54,190 |
No | output | 1 | 27,095 | 3 | 54,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,128 | 3 | 54,256 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
n=int(input())
l=[]
def d(x,y):
return x**2 + y**2
lv=[0 for i in range(n)]
for i in range(n):
l1=list(map(int,input().strip().split()))
l.append(l1)
vx=l[0][0]
vy=l[0][1]
lv[0]=1
for i in range(1,n-1,2):
vx1,vy1,vx2,vy2=l[i][0],l[i][1],l[i+1][0],l[i+1][1]
_,c1,c2=min((d(vx+vx1+vx2,vy+vy1+vy2),1,1),(d(vx+vx1-vx2,vy+vy1-vy2),1,-1),(d(vx-vx1+vx2,vy-vy1+vy2),-1,1),(d(vx-vx1-vx2,vy-vy1-vy2),-1,-1))
vx=vx+c1*vx1+c2*vx2
vy=vy+vy1*c1+vy2*c2
lv[i]=c1
lv[i+1]=c2
if (lv[n-1]==0):
_,c1=min((d(vx+l[n-1][0],vy+l[n-1][1]),1),(d(vx-l[n-1][0],vy-l[n-1][1]),-1))
lv[n-1]=c1
print (" ".join(map(str,lv)))
``` | output | 1 | 27,128 | 3 | 54,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,129 | 3 | 54,258 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
import random
n = int(input())
V = [tuple(map(int,input().split())) for i in range(n)]
dist = lambda x,y:x*x+y*y
indices = sorted((dist(*v),i) for i,v in enumerate(V))
result = [0]*n
vx,vy = 0,0
for d,i in reversed(indices):
x,y = V[i]
_,c = min(((dist(vx+x,vy+y),1),(dist(vx-x,vy-y),-1)))
vx += c*x
vy += c*y
result[i] = c
LIMIT = 2.25 * 10**12
while dist(vx,vy) > LIMIT:
i = random.randrange(n)
c = result[i]
x,y = V[i]
if dist(vx,vy) > dist(vx-2*c*x,vy-2*c*y):
vx -= 2*c*x
vy -= 2*c*y
result[i] = -c
print(' '.join(map(str,result)))
``` | output | 1 | 27,129 | 3 | 54,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,130 | 3 | 54,260 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
from math import sqrt
import sys
bound = 1.5 * 10 ** 6
def get_dist(pos):
x,y = pos
return sqrt(x**2 + y**2)
l_3 = [(1,1,1),(1,1,-1),(1,-1,1),(1,-1,-1)]
l_2 = [(1,1),(1,-1)]
def solve_2(v1,v2):
signs = None
best = float('inf')
return_pos = None
for a,b in l_2:
x = v1[0] * a + v2[0] * b
y = v1[1] * a + v2[1] * b
new_pos = (x,y)
dist = get_dist(new_pos)
if dist < best:
signs = [a,b]
best = dist
return_pos = new_pos
return signs, return_pos
def solve_3(v1,v2,v3):
signs = None
best = float('inf')
return_pos = None
for a,b,c in l_3:
x = v1[0] * a + v2[0] * b + v3[0] * c
y = v1[1] * a + v2[1] * b + v3[1] * c
new_pos = (x,y)
dist = get_dist(new_pos)
if dist < best:
signs = [a,b,c]
best = dist
return_pos = new_pos
return signs, return_pos
n = int(input())
v_list = [list(map(int,input().split())) for _ in range(n)]
result = []
if n == 1:
print (1)
sys.exit()
if n == 2:
s,current_pos = solve_2(v_list[0],v_list[1])
print (*s)
sys.exit()
s, current_pos = solve_3(v_list[0],v_list[1],v_list[2])
result += s
i = 3
while i<n:
count = min(2,n-i)
if count == 2:
s,current_pos = solve_3(current_pos,v_list[i],v_list[i+1])
result += s[1:]
i += 2
elif count == 1:
s,current_pos = solve_2(current_pos,v_list[i])
result += s[1:]
i += 1
print (*result)
``` | output | 1 | 27,130 | 3 | 54,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,131 | 3 | 54,262 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
import random
n = int(input())
v = []
for i in range(0, n):
x, y = map(int, input().split())
v.append([x, y, i])
while 1>0:
random.shuffle(v)
x = y = 0
ans = [0]*n
for i in range(n):
if (x+v[i][0])**2+(y+v[i][1])**2 < (x-v[i][0])**2+(y-v[i][1])**2:
x += v[i][0]
y += v[i][1]
ans[v[i][2]] = 1
else:
x -= v[i][0]
y -= v[i][1]
ans[v[i][2]] = -1
if x*x+y*y <= 1500000**2:
print(*ans)
break
2
``` | output | 1 | 27,131 | 3 | 54,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,132 | 3 | 54,264 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
def D(V, S):
x = 0
y = 0
for s, v in zip(S, V):
x += s*v[0]
y += s*v[1]
# error_print(math.sqrt(x**2+y**2))
return math.sqrt(x**2+y**2)
def XY(V, S):
x = 0
y = 0
for s, v in zip(S, V):
x += s*v[0]
y += s*v[1]
return x, y
@mt
def slv(N, V):
V1 = [[*v, i, 1] for i, v in enumerate(V) if v[0] * v[1] > 0]
V2 = [[*v, i, 1] for i, v in enumerate(V) if v[0] * v[1] < 0]
V3 = [[*v, i, 1] for i, v in enumerate(V) if v[0] == 0]
V4 = [[*v, i, 1] for i, v in enumerate(V) if v[0] != 0 and v[1] == 0]
V1.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
V2.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
V3.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
V4.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
def f(V):
sx, sy = 0, 0
for i, v in enumerate(V):
if abs(sx + v[0]) + abs(sy + v[1]) < abs(sx - v[0]) + abs(sy - v[1]):
v[3] = 1
else:
v[3] = -1
sx += v[3]*v[0]
sy += v[3]*v[1]
return sx, sy
SV = [f(V_) for V_ in [V1, V2, V3, V4]]
S = min([(D(SV, s), s) for s in product([-1, 1], repeat=4)])
for s, V_ in zip(S[1], [V1, V2, V3, V4]):
for v in V_:
v[3] *= s
V1.extend(V2)
V1.extend(V3)
V1.extend(V4)
V1.sort(key=lambda x: x[2])
ans = [v[3] for v in V1]
error_print(D(V, ans) < 1.5e+6)
return ' '.join(map(str, ans))
def main():
N = read_int()
V = [read_int_n() for _ in range(N)]
print(slv(N, V))
if __name__ == '__main__':
main()
``` | output | 1 | 27,132 | 3 | 54,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1 | instruction | 0 | 27,133 | 3 | 54,266 |
Tags: brute force, data structures, geometry, greedy, math, sortings
Correct Solution:
```
import random
def val(arr):
return arr[0]**2+arr[1]**2
LIMIT = (1.5*(10**6))**2
n = int(input())
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
dp = [[-1 for _ in range(n)] for _ in range(2)]
store = [[0 for _ in range(n)] for _ in range(2)]
dp[0][0] = arr[0]
dp[1][0] = [-1*num for num in arr[0]]
store[0][0]= 1
store[1][0]= -1
for i in range(1,n):
val1 = val([dp[0][i-1][0]+arr[i][0],dp[0][i-1][1]+arr[i][1]])
val2 = val([dp[1][i-1][0]+arr[i][0],dp[1][i-1][1]+arr[i][1]])
if(val1 < val2):
dp[0][i] = [dp[0][i-1][0]+arr[i][0],dp[0][i-1][1]+arr[i][1]]
store[0][i] = 1
else:
dp[0][i] = [dp[1][i-1][0]+arr[i][0],dp[1][i-1][1]+arr[i][1]]
store[0][i] = -1
val3 = val([dp[0][i-1][0] - arr[i][0],dp[0][i-1][1] - arr[i][1]])
val4 = val([dp[1][i-1][0] - arr[i][0],dp[1][i-1][1] - arr[i][1]])
if(val3 < val4):
dp[1][i] = [dp[0][i-1][0] - arr[i][0],dp[0][i-1][1] - arr[i][1]]
store[1][i] = 1
else:
dp[1][i] = [dp[1][i-1][0]-arr[i][0],dp[1][i-1][1]-arr[i][1]]
store[1][i] = -1
arr1 = []
if(dp[0][n-1] < dp[1][n-1]):
arr1.append(1)
else:
arr1.append(-1)
j = 0
for i in range(n-1,0,-1):
if(arr1[j] == 1):
arr1.append(store[0][i])
else:
arr1.append(store[1][i])
j+=1
t = arr1[0]
while(val(dp[t][n-1]) > LIMIT):
i = random.randint(0,n-1)
# print(i)
v = arr1[n-1-i]
# print(v)
if(val(dp[t][n-1])> val([dp[t][n-1][0] - 2*v*arr[i][0],dp[t][n-1][1] - 2*v*arr[i][1]])):
dp[t][n-1] = [dp[t][n-1][0] - 2*v*arr[i][0],dp[t][n-1][1] - 2*v*arr[i][1]]
arr1[n-1-i] = -1*v
if(n == 34):
print("-1 1 -1 1 -1 -1 -1 -1 1 1 -1 1 1 1 -1 1 1 1 1 1 1 1 1 -1 1 1 1 -1 1 -1 1 1 -1 -1")
else:
for i in range(n-1,-1,-1):
print(arr1[i],end = " ")
``` | output | 1 | 27,133 | 3 | 54,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
import bisect
def len2(x, y):
return x*x+y*y
def test(v1, v2):
x1 = v1[0]+v2[0]
y1 = v1[1]+v2[1]
x2 = v1[0]-v2[0]
y2 = v1[1]-v2[1]
t1 = len2(x1, y1)
t2 = len2(x2, y2)
if t1 <= t2:
xnow = x1
ynow = y1
t = t1
flag = 1
else:
xnow = x2
ynow = y2
t = t2
flag = -1
return(([xnow, ynow, t, v1[3]+v2[3]], flag))
n = int(input())
v = []
ans = [1 for i in range(n)]
r = 1000000**2
for i in range(n):
(x, y) = (int(j) for j in input().split())
v.append([x,y, len2(x, y), [i]])
start = time.time()
#v.sort(key=lambda x: x[2], reverse = True)
for i in range(n-2):
(now, flag) = test(v[i], v[i+1])
if now[2] > r:
(now, flag) = test(v[i], v[i+2])
if now[2] > r:
(now, flag) = test(v[i+1], v[i+2])
if flag == -1:
for j in v[i+2][3]:
ans[j] *= -1
v[i+1] = v[i]
v[i+2] = now
else:
if flag == -1:
for j in v[i+2][3]:
ans[j] *= -1
v[i+2] = now
else:
if flag == -1:
for j in v[i+1][3]:
ans[j] *= -1
v[i] = now
(now, flag) = test(v[n-2], v[n-1])
if flag == -1:
for j in v[n-1][3]:
ans[j] *= -1
#print(now)
#print(v)
#print(ans)
#print((v[0][0]**2+v[0][1]**2)**0.5)
for i in range(n):
print(ans[i], end = " ")
print()
finish = time.time()
#print(finish - start)
``` | instruction | 0 | 27,134 | 3 | 54,268 |
No | output | 1 | 27,134 | 3 | 54,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1
Submitted Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
def D(V, S):
x = 0
y = 0
for s, v in zip(S, V):
x += s*v[0]
y += s*v[1]
# error_print(math.sqrt(x**2+y**2))
return math.sqrt(x**2+y**2)
def XY(V, S):
x = 0
y = 0
for s, v in zip(S, V):
x += s*v[0]
y += s*v[1]
return x, y
@mt
def slv(N, V):
S = [1 for _ in range(N)]
V1 = [[*v, i, 1] for i, v in enumerate(V) if v[0] * v[1] >= 0]
V2 = [[*v, i, 1] for i, v in enumerate(V) if v[0] * v[1] < 0]
V1.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
V2.sort(reverse=True, key=lambda x: abs(x[0]) + abs(x[1]))
sx, sy = 0, 0
for i, v in enumerate(V1):
if abs(sx + v[0]) + abs(sy + v[1]) < abs(sx - v[0]) + abs(sy - v[1]):
v[3] = 1
else:
v[3] = -1
sx += v[3]*v[0]
sy += v[3]*v[1]
sx1 = sx
sy1 = sy
sx, sy = 0, 0
for i, v in enumerate(V2):
if abs(sx + v[0]) + abs(sy + v[1]) < abs(sx - v[0]) + abs(sy - v[1]):
v[3] = 1
else:
v[3] = -1
sx += v[3]*v[0]
sy += v[3]*v[1]
sx2 = sx
sy2 = sy
error_print(sx1, sy1)
error_print(sx2, sy2)
if sx1 + sx2 + sy1 + sy2 > sx1 - sx2 + sy1 - sy2:
for v in V1:
v[3] *= -1
V1.extend(V2)
V1.sort(key=lambda x: x[2])
ans = [v[3] for v in V1]
error_print(D(V, ans) < 1.5e+6)
return ' '.join(map(str, ans))
def main():
N = read_int()
V = [read_int_n() for _ in range(N)]
print(slv(N, V))
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,135 | 3 | 54,270 |
No | output | 1 | 27,135 | 3 | 54,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
# = input()
def len2(a):
return a[1]*a[1]+a[2]*a[2]
n = int(input())
#() = (i for i in input().split())
# = [i for i in input().split()]
#() = (int(i) for i in input().split())
# = [int(i) for i in input().split()]
v = []
ans = [0 for i in range(n)]
for i in range(n):
(x, y) = (int(i) for i in input().split())
v.append([i,x,y])
start = time.time()
v.sort(key=len2, reverse = True)
xnow = v[0][1]
ynow = v[0][2]
ans[v[0][0]] = 1
for i in range(1, n):
x1 = xnow+v[i][1]
y1 = ynow+v[i][2]
x2 = xnow-v[i][1]
y2 = ynow-v[i][2]
# print(x1, y1, x2, y2)
if x1*x1+y1*y1 < x2*x2+y2*y2:
xnow = x1
ynow = y1
ans[v[i][0]] = 1
else:
xnow = x2
ynow = y2
ans[v[i][0]] = -1
#print((xnow*xnow+ynow*ynow)**(0.5))
for i in range(n):
print(ans[i], end = " ")
print()
finish = time.time()
#print(finish - start)
``` | instruction | 0 | 27,136 | 3 | 54,272 |
No | output | 1 | 27,136 | 3 | 54,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
Output
Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
It can be shown that a solution always exists under the given constraints.
Examples
Input
3
999999 0
0 999999
999999 0
Output
1 1 -1
Input
1
-824590 246031
Output
1
Input
8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899
Output
1 1 1 1 1 1 1 -1
Submitted Solution:
```
import random
def val(arr):
return arr[0]**2+arr[1]**2
LIMIT = (1.5*(10**6))**2
n = int(input())
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
dp = [[-1 for _ in range(n)] for _ in range(2)]
store = [[0 for _ in range(n)] for _ in range(2)]
dp[0][0] = arr[0]
dp[1][0] = [-1*num for num in arr[0]]
store[0][0]= 1
store[1][0]= -1
for i in range(1,n):
val1 = val([dp[0][i-1][0]+arr[i][0],dp[0][i-1][1]+arr[i][1]])
val2 = val([dp[1][i-1][0]+arr[i][0],dp[1][i-1][1]+arr[i][1]])
if(val1 < val2):
dp[0][i] = [dp[0][i-1][0]+arr[i][0],dp[0][i-1][1]+arr[i][1]]
store[0][i] = 1
else:
dp[0][i] = [dp[1][i-1][0]+arr[i][0],dp[1][i-1][1]+arr[i][1]]
store[0][i] = -1
val3 = val([dp[0][i-1][0] - arr[i][0],dp[0][i-1][1] - arr[i][1]])
val4 = val([dp[1][i-1][0] - arr[i][0],dp[1][i-1][1] - arr[i][1]])
if(val3 < val4):
dp[1][i] = [dp[0][i-1][0] - arr[i][0],dp[0][i-1][1] - arr[i][1]]
store[1][i] = 1
else:
dp[1][i] = [dp[1][i-1][0]-arr[i][0],dp[1][i-1][1]-arr[i][1]]
store[1][i] = -1
arr1 = []
if(dp[0][n-1] < dp[1][n-1]):
arr1.append(1)
else:
arr1.append(-1)
j = 0
for i in range(n-1,0,-1):
if(arr1[j] == 1):
arr1.append(store[0][i])
else:
arr1.append(store[1][i])
j+=1
t = arr1[0]
while(val(dp[t][n-1]) > LIMIT):
i = random.randint(0,n-1)
# print(i)
v = arr1[n-1-i]
# print(v)
if(val(dp[t][n-1])> val([dp[t][n-1][0] - 2*v*arr[i][0],dp[t][n-1][1] - 2*v*arr[i][1]])):
dp[arr1[0]][n-1] = [dp[t][n-1][0] - 2*v*arr[i][0],dp[t][n-1][1] - 2*v*arr[i][1]]
arr1[n-1-i] = -1*v
for i in range(n-1,-1,-1):
print(arr1[i],end = " ")
``` | instruction | 0 | 27,137 | 3 | 54,274 |
No | output | 1 | 27,137 | 3 | 54,275 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Ghosts line up in a straight line from left to right with $ N $ people.
At first, the $ i $ th ghost from the left is facing left if $ U_i $ is'L', and facing right if it is'R'.
They are so timid that I don't want to see scary ghosts as much as possible.
You can instruct each ghost to look back.
It's scary to look back, so once the $ i $ th ghost looks back, it creates a fear of $ A_i $.
When the ghost looking to the left looks back, it turns to the right.
When the ghost looking to the right looks back, it turns to the left.
For $ i <j $, if the $ i $ th ghost is pointing to the right and the $ j $ th ghost is pointing to the left, then the $ i $ th ghost and the $ j $ th ghost are pointing to the left. Is defined as facing each other.
The following constraints are given in $ M $.
The $ S_i $ th ghost is scared of the $ T_i $ th ghost.
In the final state, if the $ S_i $ th ghost and the $ T_i $ th ghost face each other, a fear level of $ B_i $ will occur.
Find the minimum value of the total fear that occurs when you give the optimal instructions.
Constraints
The input satisfies the following conditions.
* $ 1 \ le N \ le 500 $
* $ 0 \ le M \ le \ min (N \ times (N-1), 1000) $
* $ | U | = N $
* $ U_i = $'L' or'R'
* $ 1 \ le A_i \ le 1000 $
* $ 1 \ le B_i \ le 1000 $
* $ 1 \ le S_i, T_i \ le N $
* $ (S_i, T_i) \ ne (S_j, T_j), $ if $ i \ ne j $
* $ S_i \ ne T_i $
Input
The input is given in the following format.
$ N $ $ M $
$ U $
$ A_1 $ $ A_2 $ $ ... $ $ A_N $
$ S_1 $ $ T_1 $ $ B_1 $
$ S_2 $ $ T_2 $ $ B_2 $
$ \ vdots $
$ S_M $ $ T_M $ $ B_M $
All inputs are given as integers.
Output
Output the minimum value of the total fear level.
Examples
Input
3 1
RRL
5 5 1
3 1 10
Output
1
Input
5 5
RLRLL
9 5 1 3 10
1 3 5
2 4 9
4 2 16
1 5 2
3 5 10
Output
8 | instruction | 0 | 27,352 | 3 | 54,704 |
"Correct Solution:
```
from collections import deque
N,M = map(int,input().split())
U = input()
A = list(map(int,input().split()))
src = [tuple(map(int,input().split())) for i in range(M)]
edges = {}
for s,t,b in src:
s,t = s-1,t-1
if s>t: s,t = t,s
if (s,t) in edges:
edges[(s,t)] += b
else:
edges[(s,t)] = b
P = N+2
es = [[] for i in range(P)] # [[to1,cap1,rev1], ...]
def add_edge(fr,to,cap):
es[fr].append([to,cap,len(es[to])])
es[to].append([fr,0,len(es[fr])-1])
for i,(u,a) in enumerate(zip(U,A)):
if u == 'L':
add_edge(0,i+2,a)
add_edge(i+2,1,0)
else:
add_edge(0,i+2,0)
add_edge(i+2,1,a)
for (s,t),b in edges.items():
add_edge(t+2,s+2,b)
INF = float('inf')
level = [0] * P
iters = [0] * P
def dinic_max_flow(s,t):
global iters
def _bfs(s):
global level
level = [-1] * P
level[s] = 0
q = deque([s])
while q:
v = q.popleft()
for to,cap,rev in es[v]:
if cap > 0 and level[to] < 0:
level[to] = level[v] + 1
q.append(to)
def _dfs(v,t,f):
if v == t: return f
for i in range(iters[v],len(es[v])):
iters[v] += 1
to,cap,rev = es[v][i]
if es[v][i][1] > 0 and level[v] < level[to]:
d = _dfs(to,t,min(f,es[v][i][1]))
if d > 0:
es[v][i][1] -= d #cap
es[to][rev][1] += d
return d
return 0
flow = 0
while True:
_bfs(s)
if level[t] < 0: return flow
iters = [0] * P
f = 0
while True:
f = _dfs(s,t,INF)
if f <= 0: break
flow += f
print(dinic_max_flow(0,1))
``` | output | 1 | 27,352 | 3 | 54,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree i is described by the esthetic appeal ai — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
* The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible;
* the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same;
* and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart?
Input
The first line contains a single integer n — the initial number of trees in the woodland belt, 2 ≤ n. The second line contains space-separated integers ai — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
* to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem A1);
* to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems A1+A2).
Output
In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees k.
In the next line print k integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to n from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal.
Examples
Input
5
1 2 3 1 2
Output
8 1
1
Input
5
1 -2 3 1 -2
Output
5 2
2 5
Submitted Solution:
```
import math
import sys
from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
n=inp()
l=inlt()
g=Counter(l)
fst=defaultdict(lambda:-1)
lst=defaultdict(lambda:-1)
sm=[max(0,l[0])]
for i in range(1,n):
sm.append(sm[-1]+max(0,l[i]))
for i in range(n):
if fst[l[i]]==-1:
fst[l[i]]=i
for i in range(n-1,-1,-1):
if lst[l[i]]==-1:
lst[l[i]]=i
mx=-sys.maxsize
v=-1
for each in g:
if g[each]>=2:
if each<0:
val=2*each-sm[fst[each]]+sm[lst[each]]
else:
val=sm[lst[each]]-sm[fst[each]]+each
if val>mx:
mx=val
v=each
cnt=0
rem=[]
for i in range(n):
if i<fst[v]:
rem.append(i+1)
elif i>lst[v]:
rem.append(i+1)
elif l[i]<0 and l[i]!=v:
rem.append(i+1)
print(mx,len(rem))
print(*rem)
``` | instruction | 0 | 27,782 | 3 | 55,564 |
No | output | 1 | 27,782 | 3 | 55,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree i is described by the esthetic appeal ai — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
* The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible;
* the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same;
* and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart?
Input
The first line contains a single integer n — the initial number of trees in the woodland belt, 2 ≤ n. The second line contains space-separated integers ai — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
* to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem A1);
* to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems A1+A2).
Output
In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees k.
In the next line print k integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to n from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal.
Examples
Input
5
1 2 3 1 2
Output
8 1
1
Input
5
1 -2 3 1 -2
Output
5 2
2 5
Submitted Solution:
```
n, t = int(input()), list(map(int, input().split()))
a, b = {}, {}
for i, j in enumerate(t):
if not j in a: a[j] = i
else: b[j] = i
p = [(a[j], b[j] - 1) for j in b]
s = [j if j > 0 else 0 for j in t]
u = v = 2 * t[p[0][0]] - 1
for i in range(n - 1): s[i + 1] += s[i]
for i, j in p:
u = 2 * t[i] + s[j] - s[i]
if u > v: a, b, v = i, j, u
s = list(range(1, a + 1)) + [i for i, j in enumerate(t[a + 1: b], a + 2) if j < 0] + list(range(b + 3, n + 1))
print(v, len(s))
print(' '.join(map(str, s)))
``` | instruction | 0 | 27,783 | 3 | 55,566 |
No | output | 1 | 27,783 | 3 | 55,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 11/2/18
"""
def steps(start, target):
ans = 0
for i, v in enumerate(start):
u = target[i]
if v != u:
for j in range(i+1, len(start)):
a, b = start[j], target[j]
if a != b and a == u:
start[i], start[j] = start[j], start[i]
break
ans += 1
return ans
def solve(seq):
hc = seq.count('H')
tc = len(seq) - hc
ans = float('inf')
for i in range(tc+1):
s = ['T'] * i + ['H'] * hc + ['T'] * (tc-i)
ans = min(steps(seq.copy(), s), ans)
for i in range(hc+1):
s = ['H'] * i + ['T'] * tc + ['H'] * (hc-i)
ans = min(steps(seq.copy(), s), ans)
return ans
N = int(input())
line = list(input())
print(solve(line))
``` | instruction | 0 | 27,843 | 3 | 55,686 |
Yes | output | 1 | 27,843 | 3 | 55,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
n, s = int(input()), input() * 2
h = s.count('H') // 2
print(h - max(s[i:i + h].count('H') for i in range(n)))
# Made By Mostafa_Khaled
``` | instruction | 0 | 27,844 | 3 | 55,688 |
Yes | output | 1 | 27,844 | 3 | 55,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
n, t = int(input()), input()
k, x = t.count('T'), 'T'
if 2 * k > n: k, x = n - k, 'H'
p = t[: k]
y = d = p.count(x)
t += p
for i in range(n):
if t[i] == x:
if t[i + k] != x: d -= 1
elif t[i + k] == x:
d += 1
if d > y: y = d
print(k - y)
``` | instruction | 0 | 27,845 | 3 | 55,690 |
Yes | output | 1 | 27,845 | 3 | 55,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
n = int(input())
s = input() * 2
count = s.count("H") // 2
print(count - max(s[i:i + count].count('H') for i in range(n)))
``` | instruction | 0 | 27,846 | 3 | 55,692 |
Yes | output | 1 | 27,846 | 3 | 55,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
n = int(input())
s = input()
H = s.count("H")
mn = int(10e3)
for i in range(n-H):
cnt = s[i:i+H-1].count("T")
mn = min(cnt,mn)
print(mn)
``` | instruction | 0 | 27,847 | 3 | 55,694 |
No | output | 1 | 27,847 | 3 | 55,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a, b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if (k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
if a[mid] < x:
lo = mid + 1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while (n % 2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n // 2
for i in range(3, int(n ** 0.5) + 2, 2):
while (n % i == 0):
primes[i] = primes.get(i, 0) + 1
n = n // i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a, b):
temp = a
a = b
b = temp
return a, b
# find function
def find(x, link):
while (x != link[x]):
x = link[x]
return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x, y = swap(x, y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
'''
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i * i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x // spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
# defining a couple constants
MOD = int(1e9) + 7;
CMOD = 998244353;
INF = float('inf');
NINF = -float('inf');
'''
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
from sys import stdin
import math
from collections import Counter
n=int(input())
s=input()
t=s.count("H")
#print(s)
print(min(s[i:(i+t)%n].count("T") for i in range(n)))
``` | instruction | 0 | 27,848 | 3 | 55,696 |
No | output | 1 | 27,848 | 3 | 55,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
a = int(input())
s = input()
d = s.count('H')
p = []
for i in range(len(s)):
if i+d > len(s):
n = d+i - len(s)
m = d - n
h = s[:m] + s[-n:]
k = h.count("T")
p.append(k)
else:
h = s[i:d+i]
k = h.count("T")
mi = a
for i in range(len(p)):
if p[i] < mi:
mi = p[i]
if s.count("H") == 1 or s.count("T") == 0:
print(0)
else:
print(mi)
``` | instruction | 0 | 27,849 | 3 | 55,698 |
No | output | 1 | 27,849 | 3 | 55,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him.
Input
The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one.
Output
Print the single number which is the minimal number of swaps that let the trainer to achieve his goal.
Examples
Input
3
HTH
Output
0
Input
9
HTHTHTHHT
Output
2
Note
In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7.
Submitted Solution:
```
R = lambda : map(int, input().split())
n = int(input())
s = input()
c = 'H' if s.count('H') < s.count('T') else 'T'
oc = 'H' if c == 'T' else 'T'
l = s.find(c)
cc = s.count(c)
print(s[l:l + cc].count(oc))
``` | instruction | 0 | 27,850 | 3 | 55,700 |
No | output | 1 | 27,850 | 3 | 55,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between the n-th tree and the first tree is dn. The height of the i-th tree is hi.
Drazil starts each day with the morning run. The morning run consists of the following steps:
* Drazil chooses two different trees
* He starts with climbing up the first tree
* Then he climbs down the first tree, runs around the park (in one of two possible directions) to the second tree, and climbs on it
* Then he finally climbs down the second tree.
But there are always children playing around some consecutive trees. Drazil can't stand children, so he can't choose the trees close to children. He even can't stay close to those trees.
If the two trees Drazil chooses are x-th and y-th, we can estimate the energy the morning run takes to him as 2(hx + hy) + dist(x, y). Since there are children on exactly one of two arcs connecting x and y, the distance dist(x, y) between trees x and y is uniquely defined.
Now, you know that on the i-th day children play between ai-th tree and bi-th tree. More formally, if ai ≤ bi, children play around the trees with indices from range [ai, bi], otherwise they play around the trees with indices from <image>.
Please help Drazil to determine which two trees he should choose in order to consume the most energy (since he wants to become fit and cool-looking monkey) and report the resulting amount of energy for each day.
Input
The first line contains two integer n and m (3 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting number of trees and number of days, respectively.
The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 109), the distances between consecutive trees.
The third line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 109), the heights of trees.
Each of following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) describing each new day. There are always at least two different trees Drazil can choose that are not affected by children.
Output
For each day print the answer in a separate line.
Examples
Input
5 3
2 2 2 2 2
3 5 2 1 4
1 3
2 2
4 5
Output
12
16
18
Input
3 3
5 1 4
5 1 4
3 3
2 2
1 1
Output
17
22
11
Submitted Solution:
```
import itertools
def ans():
def dis(com1, com2, a):
com1, com2= min(com1, com2), max(com1, com2)
if a< com1 or a> com2:
return [com1, com2]
else:
return [com2, com1]
n, m= [int(i) for i in input().split()]
d= [int(i) for i in input().split()]
h= [int(i) for i in input().split()]
for i in range(m):
safet= []
a, b= [int(p)- 1 for p in input().split()]
#print(a, b)
for j in range(n-(b- a+ 1)):
if b+1< n:
safet.append(b+1)
b+= 1
else:
safet.append(0)
b= 0
#print(safet)
temp= 0
for com in itertools.combinations(safet, 2):
com= list(com)
com.sort()
[x, y]= dis(com[0], com[1], a)
if x< y:
ans= max(temp, 2*(h[com[0]]+ h[com[1]])+ sum(d[x: y]))
else:
ans= max(temp, 2*(h[com[0]]+ h[com[1]])+ sum(d)- sum(d[y: x]))
#print(com, ans, "p", 2*(h[com[0]]+ h[com[1]]), sum(d), sum(d[y: x]))
temp= ans
#print(com, ans)
print(ans)
return
ans()
``` | instruction | 0 | 27,867 | 3 | 55,734 |
No | output | 1 | 27,867 | 3 | 55,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between the n-th tree and the first tree is dn. The height of the i-th tree is hi.
Drazil starts each day with the morning run. The morning run consists of the following steps:
* Drazil chooses two different trees
* He starts with climbing up the first tree
* Then he climbs down the first tree, runs around the park (in one of two possible directions) to the second tree, and climbs on it
* Then he finally climbs down the second tree.
But there are always children playing around some consecutive trees. Drazil can't stand children, so he can't choose the trees close to children. He even can't stay close to those trees.
If the two trees Drazil chooses are x-th and y-th, we can estimate the energy the morning run takes to him as 2(hx + hy) + dist(x, y). Since there are children on exactly one of two arcs connecting x and y, the distance dist(x, y) between trees x and y is uniquely defined.
Now, you know that on the i-th day children play between ai-th tree and bi-th tree. More formally, if ai ≤ bi, children play around the trees with indices from range [ai, bi], otherwise they play around the trees with indices from <image>.
Please help Drazil to determine which two trees he should choose in order to consume the most energy (since he wants to become fit and cool-looking monkey) and report the resulting amount of energy for each day.
Input
The first line contains two integer n and m (3 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting number of trees and number of days, respectively.
The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 109), the distances between consecutive trees.
The third line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 109), the heights of trees.
Each of following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) describing each new day. There are always at least two different trees Drazil can choose that are not affected by children.
Output
For each day print the answer in a separate line.
Examples
Input
5 3
2 2 2 2 2
3 5 2 1 4
1 3
2 2
4 5
Output
12
16
18
Input
3 3
5 1 4
5 1 4
3 3
2 2
1 1
Output
17
22
11
Submitted Solution:
```
import itertools
def ans():
def dis(com1, com2, a):
com1, com2= min(com1, com2), max(com1, com2)
if a< com1 or a> com2:
return [com1, com2]
else:
return [com2, com1]
n, m= [int(i) for i in input().split()]
d= [int(i) for i in input().split()]
h= [int(i) for i in input().split()]
for i in range(m):
safet= []
a, b= [int(p)- 1 for p in input().split()]
#print(a, b)
for j in range(n-(b- a+ 1)):
if b+1< n:
safet.append(b+1)
b+= 1
else:
safet.append(0)
b= 0
#print(safet)
temp= 0
for com in itertools.combinations(safet, 2):
com= list(com)
[x, y]= dis(com[0], com[1], a)
if x< y:
ans= max(temp, 2*(h[com[0]]+ h[com[1]])+ sum(d[x: y]))
else:
ans= max(temp, 2*(h[com[0]]+ h[com[1]])+ sum(d)- sum(d[y: x]))
temp= ans
#print(ans, temp)
print(ans)
return
ans()
``` | instruction | 0 | 27,868 | 3 | 55,736 |
No | output | 1 | 27,868 | 3 | 55,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between the n-th tree and the first tree is dn. The height of the i-th tree is hi.
Drazil starts each day with the morning run. The morning run consists of the following steps:
* Drazil chooses two different trees
* He starts with climbing up the first tree
* Then he climbs down the first tree, runs around the park (in one of two possible directions) to the second tree, and climbs on it
* Then he finally climbs down the second tree.
But there are always children playing around some consecutive trees. Drazil can't stand children, so he can't choose the trees close to children. He even can't stay close to those trees.
If the two trees Drazil chooses are x-th and y-th, we can estimate the energy the morning run takes to him as 2(hx + hy) + dist(x, y). Since there are children on exactly one of two arcs connecting x and y, the distance dist(x, y) between trees x and y is uniquely defined.
Now, you know that on the i-th day children play between ai-th tree and bi-th tree. More formally, if ai ≤ bi, children play around the trees with indices from range [ai, bi], otherwise they play around the trees with indices from <image>.
Please help Drazil to determine which two trees he should choose in order to consume the most energy (since he wants to become fit and cool-looking monkey) and report the resulting amount of energy for each day.
Input
The first line contains two integer n and m (3 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting number of trees and number of days, respectively.
The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 109), the distances between consecutive trees.
The third line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 109), the heights of trees.
Each of following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) describing each new day. There are always at least two different trees Drazil can choose that are not affected by children.
Output
For each day print the answer in a separate line.
Examples
Input
5 3
2 2 2 2 2
3 5 2 1 4
1 3
2 2
4 5
Output
12
16
18
Input
3 3
5 1 4
5 1 4
3 3
2 2
1 1
Output
17
22
11
Submitted Solution:
```
print("I cannot do it QAQ")
``` | instruction | 0 | 27,869 | 3 | 55,738 |
No | output | 1 | 27,869 | 3 | 55,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between the n-th tree and the first tree is dn. The height of the i-th tree is hi.
Drazil starts each day with the morning run. The morning run consists of the following steps:
* Drazil chooses two different trees
* He starts with climbing up the first tree
* Then he climbs down the first tree, runs around the park (in one of two possible directions) to the second tree, and climbs on it
* Then he finally climbs down the second tree.
But there are always children playing around some consecutive trees. Drazil can't stand children, so he can't choose the trees close to children. He even can't stay close to those trees.
If the two trees Drazil chooses are x-th and y-th, we can estimate the energy the morning run takes to him as 2(hx + hy) + dist(x, y). Since there are children on exactly one of two arcs connecting x and y, the distance dist(x, y) between trees x and y is uniquely defined.
Now, you know that on the i-th day children play between ai-th tree and bi-th tree. More formally, if ai ≤ bi, children play around the trees with indices from range [ai, bi], otherwise they play around the trees with indices from <image>.
Please help Drazil to determine which two trees he should choose in order to consume the most energy (since he wants to become fit and cool-looking monkey) and report the resulting amount of energy for each day.
Input
The first line contains two integer n and m (3 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting number of trees and number of days, respectively.
The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 109), the distances between consecutive trees.
The third line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 109), the heights of trees.
Each of following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) describing each new day. There are always at least two different trees Drazil can choose that are not affected by children.
Output
For each day print the answer in a separate line.
Examples
Input
5 3
2 2 2 2 2
3 5 2 1 4
1 3
2 2
4 5
Output
12
16
18
Input
3 3
5 1 4
5 1 4
3 3
2 2
1 1
Output
17
22
11
Submitted Solution:
```
_, days = [int(char) for char in input().split()]
distances = [int(char) for char in input().split()]
heights = [int(char) for char in input().split()]
trees = list(zip(distances, heights))
for _ in range(days):
kid_a, kid_b = [int(char) for char in input().split()]
if kid_a <= kid_b:
path = trees[kid_b:] + trees[0:kid_a - 1]
else:
path = trees[kid_b:kid_a - 1]
mostEnergy = 0
for firstTree in range(len(path) - 1):
distance = 0
for secondTree in range(firstTree + 1, len(path)):
distance += path[firstTree][0]
e = distance + 2 * (path[firstTree][1] + path[secondTree][1])
if e > mostEnergy:
mostEnergy = e
print(mostEnergy)
``` | instruction | 0 | 27,870 | 3 | 55,740 |
No | output | 1 | 27,870 | 3 | 55,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
import math
def dis(a0, b0, x, y):
return ((a0-x) ** 2 + (b0-y) ** 2) ** 0.5
# distance from point to segment
def minDistance(A, B, E) :
AB = [B[0] - A[0], B[1] - A[1]]
BE = [E[0] - B[0], E[1] - B[1]]
AE = [E[0] - A[0], E[1] - A[1]]
AB_BE = AB[0] * BE[0] + AB[1] * BE[1];
AB_AE = AB[0] * AE[0] + AB[1] * AE[1];
reqAns = 0;
# Case 1
if AB_BE > 0:
# Finding the magnitude
y = E[1] - B[1]
x = E[0] - B[0]
return (x * x + y * y) ** 0.5
# Case 2
elif (AB_AE < 0) :
y = E[1] - A[1]
x = E[0] - A[0]
return (x * x + y * y) ** 0.5
# Case 3
else:
# Finding the perpendicular distance
x1, y1, x2, y2 = AB[0], AB[1], AE[0], AE[1]
mod = (x1 * x1 + y1 * y1) ** 0.5
return abs(x1 * y2 - y1 * x2) / mod
n, x0, y0 = map(int, input().split())
P = [list(map(int, input().split())) for _ in range(n)]
max_ = 0
min_ = float("inf")
for i in range(n):
p = P[i]
min_ = min(min_, minDistance(P[i], P[(i+1)%n], [x0, y0]))
max_ = max(max_, dis(x0, y0, p[0], p[1]))
print(math.pi * (max_ * max_ - min_ * min_))
``` | instruction | 0 | 27,901 | 3 | 55,802 |
Yes | output | 1 | 27,901 | 3 | 55,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
from math import *
def dist(a,b,p):
xu = b[0]-a[0]
yu = b[1]-a[1]
xv = p[0]-a[0]
yv = p[1]-a[1]
c1 = xu*xv + yu*yv
c2 = xu*xu + yu*yu
x,y = 0,0
if c1<=0:
x = a[0]-p[0]
y = a[1]-p[1]
elif c2<=c1:
x = b[0]-p[0]
y = b[1]-p[1]
else:
x = a[0] + xu*(c1/c2)-p[0]
y = a[1] + yu*(c1/c2)-p[1]
return x*x + y*y
n,cx,cy = map(int,input().split())
pts = [ list(map(int,input().split())) for _ in range(n) ]
mini,maxi = float('inf'),0
for i in range(n):
px,py = pts[i]
maxi = max((px-cx)**2+(py-cy)**2,maxi)
mini = min(dist(pts[i-1],pts[i],[cx,cy]),mini)
print((maxi-mini)*pi)
``` | instruction | 0 | 27,902 | 3 | 55,804 |
Yes | output | 1 | 27,902 | 3 | 55,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
s = input().split()
n = int(s[0])
px = int(s[1])
py = int(s[2])
y = []
x = []
for i in range(n):
s=input().split()
x.append(int(s[0]) - px)
y.append(int(s[1]) - py)
#print(x[i],y[i])
l = []
x0, y0 = x[-1], y[-1]
for i in range(n):
l.append(x[i]*x[i] + y[i]*y[i])
dx, dy = x[i] - x0, y[i] - y0
if (x0*dx+y0*dy)*(x[i]*dx + y[i]*dy) < 0:
x0 = x0*y[i]-x[i]*y0
l.append(x0*x0/(dx*dx+dy*dy))
x0,y0 = x[i],y[i]
a = 3.141592653589793 * (max(l) -min(l))
#print (px,py)
#print (x[imax],y[imax])
#print (x[imin],y[imin])
#print (((px - x[imax])*(px - x[imax]) + (py - y[imax])*(py - y[imax])))
#print (((px - x[imin])*(px - x[imin]) + (py - y[imin])*(py - y[imin])))
print (a)
``` | instruction | 0 | 27,903 | 3 | 55,806 |
Yes | output | 1 | 27,903 | 3 | 55,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
#!/usr/bin/python
# -*- coding: utf-8 -*-
import sys
import math
def dist2(x1, y1, x2, y2):
return (x1 - x2) ** 2 + (y1 - y2) ** 2
def find_dist(x1, y1, x2, y2, X, Y):
if x2 == x1:
if min(y1, y2) <= Y <= max(y1, y2):
return (x1 - X) ** 2
else:
return min(dist2(x1, y1, X, Y), dist2(x2, y2, X, Y))
a = (y2 - y1) / (x2 - x1)
b = y1 - a * x1
x = (X + a * (Y - b)) / (1 + a * a)
if min(x1, x2) <= x <= max(x1, x2):
return dist2(x, a * x + b, X, Y)
else:
return min(dist2(x1, y1, X, Y), dist2(x2, y2, X, Y))
def main():
N, X, Y = map(int, sys.stdin.readline().split())
maxr = 0
minr = 10**15
pts = [tuple(map(int, sys.stdin.readline().split())) for _ in range(N)]
pts.append(pts[0])
for i in range(0, len(pts)-1):
x, y = pts[i]
r = dist2(x, y, X, Y)
if r > maxr:
maxr = r
r = find_dist(x, y, pts[i+1][0], pts[i+1][1], X, Y)
if r < minr:
minr = r
print ("%.10f" % (math.pi * (maxr - minr)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,904 | 3 | 55,808 |
Yes | output | 1 | 27,904 | 3 | 55,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
import math
def distance(x1,y1,x2,y2):
return math.sqrt(math.pow(x1-x2,2)+math.pow(y1-y2,2))
m = input().split()
n = int(m[0])
px,py = int(m[1]), int(m[2])
min_r = math.pow(10,7)
max_r = 0
for i in range(n):
x,y = input().split()
r = distance(px,py,int(x),int(y))
min_r = min(r,min_r)
max_r = max(r,max_r)
print(math.pi*(math.pow(max_r,2)-math.pow(min_r,2)))
``` | instruction | 0 | 27,905 | 3 | 55,810 |
No | output | 1 | 27,905 | 3 | 55,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
import math
def dot(v1, v2):
return v1[0]*v2[0] + v1[1]*v2[1]
def cross(v1, v2):
return v1[0]*v2[1] - v1[1]*v2[0]
def mag(x,y):
return math.sqrt(x**2 + y**2)
def main():
PI = 3.141592653589793
mags = []
n, x0, y0 = map(int, input().split())
x, y = x0, y0
for i in range(n):
x_, y_ = map(int, input().split())
if i == 0:
x1, y1 = x_, y_
x_, y_ = x_ - x0, y_ - y0
if x == x0 and y == y0:
x, y = x_, y_
mags.append(mag(x_,y_))
if dot([x,y],[x-x_, y-y_])*dot([x_,y_],[x_-x,y_-y]) > 0:
mags.append(cross([x,y],[x_-x,y_-y]) / mag(x-x_,y-y_))
x, y = x_, y_
if dot([x,y],[x-x1,y-y1])*dot([x1,y1],[x1-x,y1-y]) > 0:
mags.append(cross([x,y],[x1-x,y1-y]) / mag(x-x1,y-y1))
print((max(mags)**2 - min(mags)**2) * PI)
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,906 | 3 | 55,812 |
No | output | 1 | 27,906 | 3 | 55,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
class vector:
def __init__(self, x, y):
self.x = x
self.y = y
def length(self):
return (self.x ** 2 + self.y ** 2) ** 0.5
n, x, y = map(int, input().split())
min_r = 2000000
max_r = 0
for i in range(n):
x1, y1 = map(int, input().split())
k = vector(x1 - x, y1 - y).length()
max_r = max(max_r, k)
min_r = min(min_r, k)
pi = 3.141592653589
print(pi * max_r ** 2 - pi * min_r ** 2)
``` | instruction | 0 | 27,907 | 3 | 55,814 |
No | output | 1 | 27,907 | 3 | 55,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point P and now he is wondering what is the area of the region that will be cleared from snow. Help him.
Input
The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.
Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.
All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.
Output
Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 0 0
0 1
-1 2
1 2
Output
12.566370614359172464
Input
4 1 -1
0 0
1 2
2 0
1 1
Output
21.991148575128551812
Note
In the first sample snow will be removed from that area:
<image>
Submitted Solution:
```
#!/usr/bin/python
# -*- coding: utf-8 -*-
import sys
import math
def dist2(x1, y1, x2, y2):
return (x1 - x2) ** 2 + (y1 - y2) ** 2
def main():
N, X, Y = map(int, sys.stdin.readline().split())
maxr = 0
minr = 10**15
for i in range(N):
x, y = map(int, sys.stdin.readline().split())
r = dist2(x, y, X, Y)
if r < minr:
minr = r
if r > maxr:
maxr = r
print ("%.10f" % (math.pi * (maxr - minr)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,908 | 3 | 55,816 |
No | output | 1 | 27,908 | 3 | 55,817 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,189 | 3 | 56,378 |
"Correct Solution:
```
def solve():
m, n = map(int,input().split())
k = int(input())
mp = [input() for i in range(m)]
dp = [[[0] * 3 for j in range(n + 1)] for i in range(m + 1)]
dic = {"J":0, "O":1, "I":2}
for i in range(1, m + 1):
dpi = dp[i]
dpi_1 = dp[i - 1]
for j in range(1, n + 1):
dpij = dpi[j]
dpij_1 = dpi[j - 1]
dpi_1j = dpi_1[j]
dpi_1j_1 = dpi_1[j - 1]
for p in range(3):
#dp[i][j][p] = dp[i][j - 1][p] + dp[i - 1][j][p] - dp[i - 1][j - 1][p]
dpij[p] = dpij_1[p] + dpi_1j[p] - dpi_1j_1[p]
#dp[i][j][mp[i - 1][j - 1]] += 1
dpij[dic[mp[i - 1][j - 1]]] += 1
for i in range(k):
x1,y1,x2,y2 = map(int,input().split())
dpx2 = dp[x2]
dpx1 = dp[x1 - 1]
dpx2y2 = dpx2[y2]
dpx2y1 = dpx2[y1 - 1]
dpx1y2 = dpx1[y2]
dpx1y1 = dpx1[y1 - 1]
#print(dp[x2][y2][0] - dp[x1 - 1][y2][0] - dp[x2][y1 - 1][0] + dp[x1 - 1][y1 - 1][0], dp[x2][y2][1] - dp[x1 - 1][y2][1] - dp[x2][y1 - 1][1] + dp[x1 - 1][y1 - 1][1], dp[x2][y2][2] - dp[x1 - 1][y2][2] - dp[x2][y1 - 1][2] + dp[x1 - 1][y1 - 1][2])
print(dpx2y2[0] - dpx1y2[0] - dpx2y1[0] + dpx1y1[0], dpx2y2[1] - dpx1y2[1] - dpx2y1[1] + dpx1y1[1], dpx2y2[2] - dpx1y2[2] - dpx2y1[2] + dpx1y1[2])
solve()
``` | output | 1 | 28,189 | 3 | 56,379 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,190 | 3 | 56,380 |
"Correct Solution:
```
# coding: utf-8
M, N = list(map(int, input().split()))
q = int(input())
Map = []
for i in range(M):
Map.append(input())
def get_slide_sum(Map, kw):
l = []
for r in range(M):
tmp = []
for c in range(N):
ll = 0 if c == 0 else tmp[c-1]
ul = 0 if r == 0 or c == 0 else l[r-1][c-1]
ur = 0 if r == 0 else l[r-1][c]
count = ll + ur - ul
if Map[r][c] == kw:
count+=1
tmp.append(count)
l.append(tmp)
return l
J = get_slide_sum(Map, 'J')
O = get_slide_sum(Map, 'O')
I = get_slide_sum(Map, 'I')
def get_sum(inp,L):
a, b, c, d = list(map(int, inp.split()))
ll = 0 if a == 1 else L[a-2][d-1]
ul = 0 if a == 1 or b == 1 else L[a-2][b-2]
ur = 0 if b == 1 else L[c-1][b-2]
return L[c-1][d-1] - ll - ur + ul
for i in range(q):
inp = input()
print(get_sum(inp, J), get_sum(inp, O), get_sum(inp, I))
``` | output | 1 | 28,190 | 3 | 56,381 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,191 | 3 | 56,382 |
"Correct Solution:
```
def solve():
m, n = map(int,input().split())
k = int(input())
mp = [input() for i in range(m)]
dp = [[[0] * 3 for j in range(n + 1)] for i in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
for p in range(3):
dp[i][j][p] = dp[i][j - 1][p] + dp[i - 1][j][p] - dp[i - 1][j - 1][p]
if mp[i - 1][j - 1] == "J":
dp[i][j][0] += 1
elif mp[i - 1][j - 1] == "O":
dp[i][j][1] += 1
else:
dp[i][j][2] += 1
for i in range(k):
x1,y1,x2,y2 = map(int,input().split())
print(dp[x2][y2][0] - dp[x1 - 1][y2][0] - dp[x2][y1 - 1][0] + dp[x1 - 1][y1 - 1][0], dp[x2][y2][1] - dp[x1 - 1][y2][1] - dp[x2][y1 - 1][1] + dp[x1 - 1][y1 - 1][1], dp[x2][y2][2] - dp[x1 - 1][y2][2] - dp[x2][y1 - 1][2] + dp[x1 - 1][y1 - 1][2])
solve()
``` | output | 1 | 28,191 | 3 | 56,383 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,192 | 3 | 56,384 |
"Correct Solution:
```
n,m=map(int,input().split())
j=[[0 for i in range(m+1)] for j in range(n+1)]
o=[[0 for i in range(m+1)] for j in range(n+1)]
i=[[0 for i in range(m+1)] for j in range(n+1)]
field=[]
roop=int(input())
for k in range(n):
field.append(input())
for k in range(n):
j_c,o_c,i_c=0,0,0
for l in range(m):
if field[k][l]=="J":
j_c+=1
if field[k][l]=="O":
o_c+=1
if field[k][l]=="I":
i_c+=1
j[k+1][l+1]=j[k][l+1]+j_c
o[k+1][l+1]=o[k][l+1]+o_c
i[k+1][l+1]=i[k][l+1]+i_c
for k in range(roop):
x1,y1,x2,y2=map(int,input().split())
j_c=j[x2][y2]-j[x2][y1-1]-j[x1-1][y2]+j[x1-1][y1-1]
o_c=o[x2][y2]-o[x2][y1-1]-o[x1-1][y2]+o[x1-1][y1-1]
i_c=i[x2][y2]-i[x2][y1-1]-i[x1-1][y2]+i[x1-1][y1-1]
print(j_c,o_c,i_c)
``` | output | 1 | 28,192 | 3 | 56,385 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,193 | 3 | 56,386 |
"Correct Solution:
```
m, n = map(int,input().split())
k = int(input())
mp = [input() for i in range(m)]
dp = [[[0] * 3 for j in range(n + 1)] for i in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
for p in range(3):
dp[i][j][p] = dp[i][j - 1][p] + dp[i - 1][j][p] - dp[i - 1][j - 1][p]
if mp[i - 1][j - 1] == "J":
dp[i][j][0] += 1
elif mp[i - 1][j - 1] == "O":
dp[i][j][1] += 1
else:
dp[i][j][2] += 1
# print(dp[i][j])
for i in range(k):
x1,y1,x2,y2 = map(int,input().split())
print(dp[x2][y2][0] - dp[x1 - 1][y2][0] - dp[x2][y1 - 1][0] + dp[x1 - 1][y1 - 1][0], dp[x2][y2][1] - dp[x1 - 1][y2][1] - dp[x2][y1 - 1][1] + dp[x1 - 1][y1 - 1][1], dp[x2][y2][2] - dp[x1 - 1][y2][2] - dp[x2][y1 - 1][2] + dp[x1 - 1][y1 - 1][2])
``` | output | 1 | 28,193 | 3 | 56,387 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,194 | 3 | 56,388 |
"Correct Solution:
```
m, n = (int(x) for x in input().split())
k = int(input())
joi_list = [0]*m
for i in range(m):
joi_list[i] = list(input())
c = ['J', 'O', 'I']
t = [[[0]*(n+1) for i in range(m+1)] for j in range(len(c))]
for h in range(len(c)):
for i in range(1,m+1):
co = 0
for j in range(n):
if joi_list[i-1][j] == c[h]:
co += 1
t[h][i][j+1] = t[h][i-1][j+1] + co
ans = [0]*len(c)
for i in range(k):
p1, q1, p2, q2 = (int (x) for x in input().split())
for h in range(len(c)):
ans[h] = t[h][p2][q2]-t[h][p2][q1-1]-t[h][p1-1][q2]+t[h][p1-1][q1-1]
print(*ans)
``` | output | 1 | 28,194 | 3 | 56,389 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,195 | 3 | 56,390 |
"Correct Solution:
```
m,n = map(int,input().split())
k = int(input())
c = [[] for i in range(m)]
for i in range(m):
c[i] = input()
x = ["J","O","I"]
da = [[[0]*n for _ in range(m)] for _ in range(3)]
for l in range(3):
if c[0][0] == x[l]:
da[l][0][0] = 1
else:
da[l][0][0] = 0
for i in range(1,n):
if c[0][i] == x[l]:
da[l][0][i] = da[l][0][i-1] + 1
else:
da[l][0][i] = da[l][0][i-1]
for i in range(1,m):
co = 0
for j in range(n):
if c[i][j] == x[l]:
co += 1
da[l][i][j] = da[l][i-1][j] + co
for i in range(k):
p,q,x,y = map(int,input().split())
p -= 1
q -= 1
x -= 1
y -= 1
ans = [0] * 3
for j in range(3):
if p == 0 and q == 0:
ans[j] = da[j][x][y]
elif p == 0:
ans[j] = da[j][x][y] - da[j][x][q-1]
elif q == 0:
ans[j] = da[j][x][y] - da[j][p-1][y]
else:
ans[j] = da[j][x][y] - da[j][p-1][y] - da[j][x][q-1] + da[j][p-1][q-1]
print(*ans)
``` | output | 1 | 28,195 | 3 | 56,391 |
Provide a correct Python 3 solution for this coding contest problem.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7 | instruction | 0 | 28,196 | 3 | 56,392 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0560
"""
import sys
from sys import stdin
input = stdin.readline
def solve(M, N, K, field):
dp_J = [[0] * (N+1) for _ in range(M+1)]
dp_O = [[0] * (N+1) for _ in range(M+1)]
dp_I = [[0] * (N+1) for _ in range(M+1)]
for y in range(M):
t_J = [0] * (N+1)
t_O = [0] * (N+1)
t_I = [0] * (N+1)
for x in range(N):
land = field[y][x]
if land == 'J':
t_J[x+1] = t_J[x] + 1
t_O[x+1] = t_O[x]
t_I[x+1] = t_I[x]
elif land == 'O':
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x] + 1
t_I[x+1] = t_I[x]
else:
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x]
t_I[x+1] = t_I[x] + 1
for x in range(1, N+1):
dp_J[y+1][x] = dp_J[y][x] + t_J[x]
dp_O[y+1][x] = dp_O[y][x] + t_O[x]
dp_I[y+1][x] = dp_I[y][x] + t_I[x]
for _ in range(K):
result = [0, 0, 0]
y1, x1, y2, x2 = map(int, input().split())
result[0] = dp_J[y2][x2] - dp_J[y1-1][x2] - dp_J[y2][x1-1] + dp_J[y1-1][x1-1]
result[1] = dp_O[y2][x2] - dp_O[y1-1][x2] - dp_O[y2][x1-1] + dp_O[y1-1][x1-1]
result[2] = dp_I[y2][x2] - dp_I[y1-1][x2] - dp_I[y2][x1-1] + dp_I[y1-1][x1-1]
print(' '.join(map(str, result)))
def solve4(M, N, K, field):
dp = [[[0, 0] for _ in range(N+1)] for _ in range(M+1)]
for y in range(1, M+1):
for x in range(1, N+1):
land = field[y-1][x-1]
if land == 'J':
dp[y][x][0] = dp[y-1][x][0] + dp[y][x-1][0] - dp[y-1][x-1][0] + 1
dp[y][x][1] = dp[y-1][x][1] + dp[y][x-1][1] - dp[y-1][x-1][1]
elif land == 'O':
dp[y][x][0] = dp[y-1][x][0] + dp[y][x-1][0] - dp[y-1][x-1][0]
dp[y][x][1] = dp[y-1][x][1] + dp[y][x-1][1] - dp[y-1][x-1][1] + 1
else:
dp[y][x][0] = dp[y-1][x][0] + dp[y][x-1][0] - dp[y-1][x-1][0]
dp[y][x][1] = dp[y-1][x][1] + dp[y][x-1][1] - dp[y-1][x-1][1]
for _ in range(K):
result = [0, 0, 0]
y1, x1, y2, x2 = map(int, input().split())
result[0] = dp[y2][x2][0] - dp[y1-1][x2][0] - dp[y2][x1-1][0] + dp[y1-1][x1-1][0]
result[1] = dp[y2][x2][1] - dp[y1-1][x2][1] - dp[y2][x1-1][1] + dp[y1-1][x1-1][1]
result[2] = (x2 - x1 + 1) * (y2 - y1 + 1) - result[0] - result[1]
print(' '.join(map(str, result)))
def main(args):
M, N = map(int, input().split())
K = int(input())
field = [input().strip() for _ in range(M)]
solve4(M, N, K, field)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 28,196 | 3 | 56,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7
Submitted Solution:
```
# use dp
# python 3.
m, n = map(int, input().split())
k = int(input())
# answer matrix
dp_j = [[0 for i in range(n+1)]for j in range(m+1)]
dp_o = [[0 for i in range(n+1)]for j in range(m+1)]
# input row and calc dp
for i in range(1, m+1):
row = list(input())
# print(row)
for j in range(1, n+1):
dp_j[i][j] = dp_j[i-1][j] + dp_j[i][j-1] - dp_j[i-1][j-1] + (1 if row[j-1] == 'J' else 0)
dp_o[i][j] = dp_o[i-1][j] + dp_o[i][j-1] - dp_o[i-1][j-1] + (1 if row[j-1] == 'O' else 0)
for i in range(k):
a, b, c, d = map(int, input().split())
j_cnt = dp_j[c][d] - dp_j[a-1][d] - dp_j[c][b-1] + dp_j[a-1][b-1]
o_cnt = dp_o[c][d] - dp_o[a-1][d] - dp_o[c][b-1] + dp_o[a-1][b-1]
i_cnt = (abs(c-a)+1) * (abs(d-b)+1) - j_cnt - o_cnt
print(str(j_cnt) + ' ' + str(o_cnt) + ' ' + str(i_cnt))
``` | instruction | 0 | 28,197 | 3 | 56,394 |
Yes | output | 1 | 28,197 | 3 | 56,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long journey, the super-space-time immigrant ship carrying you finally discovered a planet that seems to be habitable. The planet, named JOI, is a harsh planet with three types of terrain, "Jungle," "Ocean," and "Ice," as the name implies. A simple survey created a map of the area around the planned residence. The planned place of residence has a rectangular shape of M km north-south and N km east-west, and is divided into square sections of 1 km square. There are MN compartments in total, and the compartments in the p-th row from the north and the q-th column from the west are represented by (p, q). The northwest corner section is (1, 1) and the southeast corner section is (M, N). The terrain of each section is one of "jungle", "sea", and "ice". "Jungle" is represented by J, "sea" is represented by O, and "ice" is represented by one letter I.
Now, in making a detailed migration plan, I decided to investigate how many sections of "jungle," "sea," and "ice" are included in the rectangular area at K.
input
Read the following input from standard input.
* The integers M and N are written on the first line, separated by blanks, indicating that the planned residence is M km north-south and N km east-west.
* The integer K is written on the second line, which indicates the number of regions to be investigated.
* The following M line contains information on the planned residence. The second line of i + (1 ≤ i ≤ M) contains an N-character string consisting of J, O, and I that represents the information of the N section located on the i-th line from the north of the planned residence. ..
* The following K line describes the area to be investigated. On the second line of j + M + (1 ≤ j ≤ K), the positive integers aj, bj, cj, and dj representing the jth region are written with blanks as delimiters. (aj, bj) represents the northwest corner section of the survey area, and (cj, dj) represents the southeast corner section of the survey area. However, aj, bj, cj, and dj satisfy 1 ≤ aj ≤ cj ≤ M, 1 ≤ bj ≤ dj ≤ N.
output
Output K lines representing the results of the survey to standard output. Line j of the output contains three integers representing the number of "jungle" (J) compartments, the "sea" (O) compartment, and the "ice" (I) compartment in the jth survey area. , In this order, separated by blanks.
Example
Input
4 7
4
JIOJOIJ
IOJOIJO
JOIJOOI
OOJJIJO
3 5 4 7
2 2 3 6
2 2 2 2
1 1 4 7
Output
1 3 2
3 5 2
0 1 0
10 11 7
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0560
"""
import sys
from sys import stdin
input = stdin.readline
def solve(M, N, K, field):
dp_J = [[0] * (N+1) for _ in range(M+1)]
dp_O = [[0] * (N+1) for _ in range(M+1)]
dp_I = [[0] * (N+1) for _ in range(M+1)]
for y in range(M):
t_J = [0] * (N+1)
t_O = [0] * (N+1)
t_I = [0] * (N+1)
for x in range(N):
land = field[y][x]
if land == 'J':
t_J[x+1] = t_J[x] + 1
t_O[x+1] = t_O[x]
t_I[x+1] = t_I[x]
elif land == 'O':
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x] + 1
t_I[x+1] = t_I[x]
else:
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x]
t_I[x+1] = t_I[x] + 1
for x in range(1, N+1):
dp_J[y+1][x] = dp_J[y][x] + t_J[x]
dp_O[y+1][x] = dp_O[y][x] + t_O[x]
dp_I[y+1][x] = dp_I[y][x] + t_I[x]
for _ in range(K):
result = [0, 0, 0]
y1, x1, y2, x2 = map(int, input().split())
result[0] = dp_J[y2][x2] - dp_J[y1-1][x2] - dp_J[y2][x1-1] + dp_J[y1-1][x1-1]
result[1] = dp_O[y2][x2] - dp_O[y1-1][x2] - dp_O[y2][x1-1] + dp_O[y1-1][x1-1]
result[2] = dp_I[y2][x2] - dp_I[y1-1][x2] - dp_I[y2][x1-1] + dp_I[y1-1][x1-1]
print(' '.join(map(str, result)))
def solve2(M, N, K, field):
dp_J = [[0] * (N+1) for _ in range(M+1)]
dp_O = [[0] * (N+1) for _ in range(M+1)]
for y in range(M):
t_J = [0] * (N+1)
t_O = [0] * (N+1)
for x in range(N):
land = field[y][x]
if land == 'J':
t_J[x+1] = t_J[x] + 1
t_O[x+1] = t_O[x]
elif land == 'O':
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x] + 1
else:
t_J[x+1] = t_J[x]
t_O[x+1] = t_O[x]
for x in range(1, N+1):
dp_J[y+1][x] = dp_J[y][x] + t_J[x]
dp_O[y+1][x] = dp_O[y][x] + t_O[x]
for _ in range(K):
result = [0, 0, 0]
y1, x1, y2, x2 = map(int, input().split())
result[0] = dp_J[y2][x2] - dp_J[y1-1][x2] - dp_J[y2][x1-1] + dp_J[y1-1][x1-1]
result[1] = dp_O[y2][x2] - dp_O[y1-1][x2] - dp_O[y2][x1-1] + dp_O[y1-1][x1-1]
result[2] = (x2 - x1 + 1) * (y2 - y1 + 1) - result[0] - result[1]
print(' '.join(map(str, result)))
def main(args):
M, N = map(int, input().split())
K = int(input())
field = [input().strip() for _ in range(M)]
solve2(M, N, K, field)
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 28,198 | 3 | 56,396 |
Yes | output | 1 | 28,198 | 3 | 56,397 |
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