同余模定理 HDOJ 5373 The shortest problem

 

题目传送门

 1 /*
 2     题意：题目讲的很清楚：When n=123 and t=3 then we can get 123->1236->123612->12361215.要求t次操作后，能否被11整除
 3     同余模定理：每次操作将后缀值加到上次操作的值%11后的后面，有点绕，纸上模拟一下就行了
 4 */
 5 /************************************************
 6 * Author        :Running_Time
 7 * Created Time  :2015-8-12 8:50:23
 8 * File Name     :E.cpp
 9  ************************************************/
10 
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29 
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 1e5 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 int a[MAXN];
37 
38 int part(int x) {
39     int ret = 0;
40     while (x)   {
41         ret += x % 10;  x /= 10;
42     }
43     return ret;
44 }
45 
46 int cal(int x)  {
47     int ret = 1;
48     while (x)   {
49         ret *= 10;  x /= 10;
50     }
51     return ret;
52 }
53 
54 int main(void)    {        //HDOJ 5373 The shortest problem
55     int n, t, cas = 0;
56     while (scanf ("%d%d", &n, &t) == 2) {
57         if (n == -1 && t == -1) break;
58 
59         a[1] = n;   int sum = part (n);
60         for (int i=2; i<=t+1; ++i)    {
61             a[i] = a[i-1] % 11 * cal (sum) + sum;
62             sum += part (sum);
63         }
64 
65         printf ("Case #%d: %s\n", ++cas, (a[t+1] % 11 == 0) ? "Yes" : "No");
66     }
67 
68     return 0;
69 }

 

posted @ 2015-08-12 10:39  Running_Time  阅读(208)  评论(0)  编辑  收藏  举报