HDU 5373 (模拟 水~)

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2129    Accepted Submission(s): 868

Problem Description

In this problem, we should solve an interesting game. At first, we have an integer n, then we begin to make some funny change. We sum up every digit of the n, then insert it to the tail of the number n, then let the new number be the interesting number n. repeat it for t times. When n=123 and t=3 then we can get 123->1236->123612->12361215.

 

Input

Multiple input.
We have two integer n (0<=n<=104 ) , t(0<=t<=105) in each row.
When n==-1 and t==-1 mean the end of input.

 

Output

For each input , if the final number are divisible by 11, output “Yes”, else output ”No”. without quote.

 

Sample Input

35 2
35 1
-1 -1

 

Sample Output

Case #1: Yes
Case #2: No 

 

题意:每次把每一位的数字加起来放在数字的最后面,最后结果模11是不是0.

直接模拟,维护模的结果.

#include <bits/stdc++.h>
using namespace std;
#define maxn 1111111

int pos, bit;
int n, t;
long long sum;
long long gg[maxn];

long long get (long long num) {
    long long ans = 0;
    while (num) {
        ans += num%10;
        num /= 10;
        bit++;
    }
    return ans;
}

int main () {
    gg[0] = 1;
    for (int i = 1; i <= 1000000; i++) gg[i] = gg[i-1]*10%11;
    int kase = 0;
    while (scanf ("%d%d", &n, &t) == 2) {
        if (n == -1 && t == -1)
            break;
        printf ("Case #%d: ", ++kase);
        long long mod = n%11;
        bit = 0;
        long long sum = 0, pre; pre = get (n);
        for (int i = 1; i <= t; i++) {
            sum += pre;
            bit = 0;
            pre = get (sum);
            mod = mod*gg[bit]%11+sum;
            mod %= 11;
        }
        printf ("%s\n", mod ? "No" : "Yes");
    }
    return 0;
}