hdu 1850 - Being a Good Boy in Spring Festival(简单博弈)
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思路:
尼姆博弈的变形, 对于每个堆都遍历一下,若起始状态已是T态,则ans = 0; 否则,就枚举第一步需要变动的堆i,求其他n-1个数a[]的异或值temp,
如果temp<a[i],则ans加一,否则就不加,代表这一堆目前无法处理。
代码如下:
const int M = 105;int a[M];int main(){ int n; while(~scanf("%d", &n) && n) { for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); int ans = 0, pre = 0, cur = 0; for(int i = 1; i <= n; ++i) { cur = pre; for(int j = i+1; j <= n; ++j) cur ^= a[j]; pre ^= a[i]; ans += (cur<a[i]); } printf("%d\n", ans); } return 0;}
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