Problem - D - Codeforces

题意：

思路：

这道题需要倒着步骤考虑，就是先去假设已经分为了两部分，这左右两部分的和相等，然后去想上一个步骤

倒着一个步骤后，可以发现这样的性质：

Code：

#include <bits/stdc++.h>

#define int long long

using i64 = long long;

constexpr int N = 1e5 + 10;
constexpr int M = 1e5 + 10;
constexpr int P = 2600;
constexpr i64 Inf = 1e18;
constexpr int mod = 1e9 + 7;
constexpr double eps = 1e-6;

int n;
int a[N];

void solve() {
    std::cin >> n;

    std::map<int,int> mp, mp2;
    int sum = 0;
    for (int i = 1; i <= n; i ++) {
        std::cin >> a[i];
        sum += a[i];
        mp2[a[i]] ++;
    }

    if (sum & 1) {
        std::cout << "NO" << "\n";
        return;
    }

    int s = 0;
    for (int i = 1; i <= n; i ++) {
        s += a[i];
        mp[a[i]] ++;
        mp2[a[i]] --;
        if (s > sum / 2) {
            if (mp[s - sum / 2]) {
                std::cout << "YES" << "\n";
                return;
            }
        }else if (s < sum / 2) {
            if (mp2[sum / 2 - s]) {
                std::cout << "YES" << "\n";
                return;
            }
        }else {
            std::cout << "YES" << "\n";
            return;
        }
    }

    std::cout << "NO" << "\n";
}
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t = 1;

    while (t--) {
        solve();
    }
    
    return 0;
}