题解 | 元素方碑

最简单的一集，注意到i永远无法和i+1，i-1发生互动，分别处理偶位和奇位的数字和，判断是否和整体均值相等即可

    #include<iostream>
	#include<vector>
	using namespace std;
	#define vt vector
	#define int long long
	#define endl "\n"
	#define pb push_back
	signed main() {
	int t;cin >> t;
	while (t--) {int n;cin >> n;vt<int> a(n + 1);
	int sum = 0;
    int num_0 = 0;
    int num_1 = 0;

    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum += a[i];
        if (i % 2 == 0) num_0 += a[i];
        else num_1 += a[i];
    }

    if (sum % n != 0) {
        cout << "NO" << endl;
        continue;
    }

    int to = sum / n;
    int evenCnt = n / 2;
    int oddCnt = (n + 1) / 2;

    if (num_0 == to * evenCnt && num_1 == to * oddCnt) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}
}