CodeForces 1006C Three Parts of the Array

题目链接：CodeForces 1006C Three Parts of the Array

题目大意：

题解：
两个指针分别指向头尾，向中间移动，计算\(sum1\)和\(sum2\)，哪边小移动哪边的指针，记录\(sum1 = sum2\)时\(sum1\)的值，

#include <iostream>
using namespace std;

long long num[1000010];
int n;

int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> num[i];
    }
    int p = 1, q = n;
    long long sum1 = 0, sum2 = 0, ans = 0;
    while (p <= q) {
        if (sum1 > sum2) {
            sum2 += num[q--];
        } else if (sum2 > sum1) {
            sum1 += num[p++];
        }
        if (sum1 == sum2) {
            ans = max(ans, sum1);
            sum1 += num[p++];
        }
    }
    cout << ans << endl;
    return 0;
}