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
", ++cas, (a[t+1] % 11 == 0) ? "Yes" : "No");
66 }
67
68 return 0;
69 }