从0到400000遍历过了，但是我觉得更快的map图应该会更快呀，有点好奇，代码是这样的

TLE代码

#include <bits/stdc++.h>
#define int long long 

using namespace std;

const int N = 310;

void solve() {
	int n;
    cin >> n;
    int a[N];
    for(int i = 0; i < n; i ++ ) cin >> a[i];
    unordered_map<int, int> dp;
    dp[0] = 1;
    for(int i = n - 1; i >= 0; i -- ) {
        unordered_map<int, int> dp2;
        dp2.clear();
        for(auto j : dp) {
            dp2[j.first] = 1;
        }
        dp.clear();
        for(auto j : dp2) {
            dp[j.first - a[i]] = 1;
            dp[j.first + a[i]] = 1;
        }
    }
    int res = 1e9;
    for(auto j : dp) {
        res = min(res, abs(j.first));
    }
    cout << res << endl;
}

signed main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0);
	
	int t = 1;
	//cin >> t;
	while(t -- ) {
		solve();
	}
}

AC代码

#include <bits/stdc++.h>
#define int long long 

using namespace std;

const int N = 310, M = 400010;

void solve() {
	int n;
    cin >> n;
    int a[N];
    for(int i = 0; i < n; i ++ ) cin >> a[i];
    int dp[M];
    memset(dp, 0, sizeof dp);
    dp[200000] = 1;
    for(int i = 0; i < n; i ++ ) {
        int dp2[M];
        memcpy(dp2, dp, sizeof dp);
        memset(dp, 0, sizeof dp);
        for(int j = 0; j < M; j ++ ) {
            if(dp2[j] == 1) {
                dp[j - a[i]] = 1;
                dp[j + a[i]] = 1;
               // cout << j - a[i] + 200000 << " " << j + a[i] + 200000 << endl;
            }
        }
    }
    int res = 1e9;
    for(int i = 0; i < M; i ++ ) {
        if(dp[i]) res = min(res, abs(i - 200000));
    }
    
    cout << res << endl;
}

signed main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0);
	
	int t = 1;
	//cin >> t;
	while(t -- ) {
		solve();
	}
}