POJ 2260 Error Correction（水）

题目链接

 枚举一下，就可以。。最近超时的次数太多都怕了。。。SB的忘了个换行PE一次。

 1 #include <stdio.h>
 2 #include <string.h>
 3 int p[101][101],sum1[101],sum2[101],n;
 4 int judge()
 5 {
 6     int i;
 7     for(i = 1;i <= n;i ++)
 8     {
 9         if(sum1[i]%2 != 0)
10         break;
11         if(sum2[i]%2 != 0)
12         break;
13     }
14     if(i == n+1)
15     return 1;
16     else
17     return 0;
18 }
19 int main()
20 {
21     int i,j,z,c,r;
22     while(scanf("%d",&n)!=EOF)
23     {
24         if(n == 0) break;
25         z = 0;
26         memset(sum1,0,sizeof(sum1));
27         memset(sum2,0,sizeof(sum2));
28         for(i = 1;i <= n;i ++)
29         for(j = 1;j <= n;j ++)
30         scanf("%d",&p[i][j]);
31         for(i = 1;i <= n;i ++)
32         {
33             for(j = 1;j <= n;j ++)
34             sum1[i] += p[i][j];
35         }
36         for(i = 1;i <= n;i ++)
37         for(j = 1;j <= n;j ++)
38         {
39             sum2[j] += p[i][j];
40         }
41         if(judge())
42         {
43             printf("OK\n");
44         }
45         else
46         {
47             for(i = 1;i <= n&&z == 0;i ++)
48             for(j = 1;j <= n&&z == 0;j ++)
49             {
50                 sum1[i]++;
51                 sum2[j]++;
52                 if(judge())
53                 {
54                     z = 1;
55                     c = i;
56                     r = j;
57                 }
58                 sum1[i]--;
59                 sum2[j]--;
60             }
61             if(z)
62             printf("Change bit (%d,%d)\n",c,r);
63             else
64             printf("Corrupt\n");
65         }
66     }
67     return 0;
68 }

 

 

posted @ 2012-06-17 21:48  Naix_x  阅读(172)  评论(0)    收藏  举报