摘要:
类似多对石头的博弈判断哪个先取时,再第i个数与总的S取异或,如果a[i]>=(s^a[i])就是先取它、View Code #include<iostream>#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<string.h>#include<math.h>using namespace std;int a[1000009];int b[1000009];int prim[100009];int padd=0;int pp(int a){ 阅读全文
posted @ 2011-09-17 21:05 huhuuu 阅读(359) 评论(0) 推荐(0) 编辑