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 }