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  • The shortest problem(hdu,多校

    The shortest problem

    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 0    Accepted Submission(s): 0


    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
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <algorithm>
     4 
     5 using namespace std;
     6 
     7 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
     8 //int main(int argc, char** argv) 
     9 
    10 #define N 100005
    11 
    12 int main()
    13 {
    14     int n, m, d, a, sum, r, nu, t = 1;
    15     
    16     while(scanf("%d%d", &n, &m), n != -1 || m != -1)
    17     {
    18         r = n % 11;
    19         sum = 0;
    20         
    21         while(n)   // 不知道为什么非得这么写才对~
    22         {
    23             sum += n % 10;
    24             n /= 10;
    25         }
    26         
    27         while(m--)
    28         {
    29             a = 1;
    30             nu = n = sum;  // nu,存当前次数之前出现的根的总和
    31             sum = 0;
    32             
    33             while(n)
    34             {
    35                 sum += n % 10;  // sum存目前的根总和的根
    36                 n /= 10;
    37                 a *= 10;
    38             }
    39             
    40             r = (r * a + nu) % 11;
    41             sum += nu;
    42         }
    43         
    44         if(r == 0)
    45             printf("Case #%d: Yes
    ", t++);
    46         else
    47             printf("Case #%d: No
    ", t++);
    48     }
    49     return 0;
    50 }
     
    让未来到来 让过去过去
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  • 原文地址:https://www.cnblogs.com/Tinamei/p/4721335.html
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