hdu 5373 模拟

简单模拟题，可以利用一下能被11整除的数的特点：奇数位的数字和与偶数位的数字和之差能被11整除。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cstdio>
 5 using namespace std;
 6 
 7 const int N = 1000000;
 8 int s[N];
 9 int mid[N];
10 
11 int main ()
12 {
13     int n, t, _case = 0;
14     while ( scanf("%d%d", &n, &t) != EOF )
15     {
16         if ( n == -1 && t == -1 ) break;
17         int sum = 0, p = 0, cnt = 0;
18         while ( n )
19         {
20             sum += n % 10;
21             mid[p++] = n % 10;
22             n /= 10;
23         }
24         for ( int i = p - 1; i >= 0; i-- )
25         {
26             s[cnt++] = mid[i];
27         }
28         while ( t-- )
29         {
30             int tmp = sum;
31             p = 0;
32             while ( tmp )
33             {
34                 sum += tmp % 10;
35                 mid[p++] = tmp % 10;
36                 tmp /= 10;
37             }
38             for ( int i = p - 1; i >= 0; i-- )
39             {
40                 s[cnt++] = mid[i];
41             }
42         }
43         int ssum = 0;
44         for ( int i = 0; i < cnt; i++ )
45         {
46             if ( i & 1 ) ssum += s[i];
47             else ssum -= s[i];
48         }
49         _case++;
50         if ( ssum % 11 == 0 )
51         {
52             printf("Case #%d: Yes\n", _case);
53         }
54         else
55         {
56             printf("Case #%d: No\n", _case);
57         }
58     }
59     return 0;
60 }

 

posted @ 2015-08-11 19:49  hxy_has_been_used  阅读(230)  评论(0)  编辑  收藏  举报