前言:一开始我想错了,一开始我想的是直接统计每一项模完后的和,我们只要能够取到一半,那么就有解,但是时间复杂度太大了
我们做推导
x
+
y
=
s
u
m
x + y = sum
x+y=sum
x
−
y
=
k
∗
n
x - y = k * n
x−y=k∗n
那么我们可以得到
2
∗
x
−
s
u
m
=
k
∗
n
2*x - sum = k*n
2∗x−sum=k∗n
注意:由于取模相当于绕圈圈,所以我们要交替开两个数组记录
题目地址
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = (int)5e3 + 10;
int n, m;
int a[N];
signed main() {
cin >> n >> m;
int cnt = 0;
for (int i = 1; i <= m; i++) {
int u; cin >> u;
a[i] = u % n;
cnt += a[i];
}
cnt %= n;
vector<int> dp(n+10, 0);
dp[0] = 1;
for (int i = 1; i <= m; i++) {
vector<int> temp = dp;
for (int j = 0; j <= n; j++) {
if (dp[j]) {
temp[(j + a[i]) % n] = 1;
}
}
dp = temp;
}
// x + y = s , x - y = k*n
// 2*x - x = k*n
int ok = 0;
for (int i = 0; i <= n; i++) {
if (dp[i] && ((i * 2 - cnt) % n == 0)) {
//cout << " i " << i << endl;
ok = 1; break;
}
}
if (ok) {
cout << "YES";
}
else cout << "NO";
return 0;
}