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  • POJ 2234

     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<algorithm>
     4 #define MAXN 100
     5 using namespace std;
     6 //把所有堆的石子数目用二进制数表示出来,当全部这些数按位异或结果为0时当前局面为必败局面,否则为必胜局面;
     7 int a[MAXN];
     8 void op(int & num);
     9 int trans_10_to_2(int num);
    10 int main()
    11 {
    12     //freopen("acm.acm","r",stdin);
    13     int num;
    14     int i;
    15     int tem;
    16     int sum;
    17     while(cin>>num)
    18     {
    19         sum = 0;
    20         memset(a,0,sizeof(a));
    21         for(i = 0; i < num; ++ i)
    22         {
    23             cin >> tem;
    24             trans_10_to_2(tem);
    25         }
    26         for_each(a,a+MAXN,op);
    27         for(i = 0; i < MAXN; ++ i)
    28         {
    29             if(a[i] == 1)
    30             {
    31                 cout<<"Yes"<<endl;
    32                 break;
    33             }
    34         }
    35         if(i == MAXN)
    36             cout<<"No"<<endl;
    37     }
    38 }
    39 
    40 
    41 int trans_10_to_2(int num)
    42 {
    43     int mod;
    44     int i;
    45     i = 0;
    46     while(num)
    47     {
    48         mod = num%2;
    49         num /= 2;
    50         a[i ++] += mod;
    51     }
    52     return i;
    53 }
    54 
    55 void op(int & num)
    56 {
    57     if(num % 2 == 0)
    58         num = 0;
    59     else
    60         num = 1;
    61 }

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  • 原文地址:https://www.cnblogs.com/gavinsp/p/4566713.html
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