/*这是尼姆博弈题要让先手赢的话,先手从第i堆牌中取出x张牌后剩下的n-1堆异或的结果和第i剩下的异或为0,那先手一定能赢那么只有当第i堆牌数大于剩下n-1堆牌的异或结果时,先手一定赢*/#include<iostream>using namespace std;int main(){int n, card[102];while(cin >> n && n){int t = 0, con
Being a Good Boy in Spring Festival Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3055 Accepted Submission(s): 1771 Problem Descript
题意:存在下面的编码方式: a - 1 b - 2 ... z - 26 ab - 27 ... az - 51 bc - 52 ... vwxyz - 83681 其中字符串的长度逐渐增加,并且每一个字符串的字符只能是升序。例如b不能排在a的前面。 #include<cstdio>#include<cstring>using namespace std;#de