摘要：
题目链接这是一道博弈的题，准确说是尼姆博弈，只要判断各项的异或值即可。代码#include const int maxn = 5000;int x[maxn];int main(){ int t, n, tmp; scanf("%d", &t); while (t--) { scanf("%d", &n); int cnt = 0; for (int i = 1; i <= n; i++) { scanf("%d",&x[i]); if (x[i] == 1) ... 阅读全文