空間限制:C/C++ 131072K,其他語言262144K
64bit IO Format: %lld
題目描述
輸入描述:
The first line contains an positive integer T(1≤T≤60), represents there are T test cases.For each test case:The first line contains an integer n(1≤n≤105), indicating there are n songs.The second line contains n integers a1,a2…an (1≤ai≤109 ), the ith integer ai indicates the ith song lasts ai seconds.
輸出描述:
For each test case, output one line "YES" (without quotes) if HH is possible to stop listen at an integral point, and "NO" (without quotes) otherwise.
輸入
3 3 2000 1000 3000 3 2000 3000 1600 2 5400 1800
輸出
NO YES YES
說明
In the first example it's impossible to stop at an integral point.
In the second example if we choose the first and the third songs, they cost 3600 seconds in total, so HH can stop at 13:00:00
In the third example if we choose the first and the second songs, they cost 7200 seconds in total, so HH can stop at 14:00:00
實現代碼
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 1e5 + 5;
int vis[3600], a[100005], dp[3600];
bool judge(int len) {
for (int i = 1; i <= len; i++) {
dp[a[i]] = 1;
if (dp[0]) return true;
for (int j = 1; j < 3600; j++) {
if (vis[j]) {
dp[(a[i] + j) % 3600] += 1;
if (dp[0]) return true;
}
}
for (int j = 0; j < 3600; j++) if (dp[j]) vis[j] = 1; // 避免自己修改自己
}
return false;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int n, len, num;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> len;
memset(vis, 0, sizeof(vis));
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= len; i++) {
cin >> a[i];
a[i] %= 3600;
}
if (judge(len)) cout << "YES\n";
else cout << "NO\n";
}
return 0;
}