[CF383D] Antimatter

[CF383D] Antimatter - dp

Description

给定长度为 n 的序列，可以给任意数加上正号或者负号，求有多少种连续子段和为 0 的情况。

Solution

设 (f[i][j]) 表示以第 i 个数结尾的子串，有多少个和为 j

#include <bits/stdc++.h>
using namespace std;
#define int long long

const int N = 1005;
const int M = 20005;
const int B = 10002;
const int mod = 1e9 + 7;

int n, f[N][M], a[N], ans;

signed main()
{
    ios::sync_with_stdio(false);

    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    for (int i = 1; i <= n; i++)
    {
        f[i][B + a[i]] = 1;
        f[i][B - a[i]] = 1;
        for (int j = 0; j < M; j++)
        {
            if (j + a[i] < M)
                (f[i][j] += f[i - 1][j + a[i]]) %= mod;
            if (j - a[i] >= 0)
                (f[i][j] += f[i - 1][j - a[i]]) %= mod;
        }
        (ans += f[i][B]) %= mod;
    }

    cout << ans;
}