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  • 2015 Multi-University Training Contest 7 hdu 5373 The shortest problem

    The shortest problem

    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 346    Accepted Submission(s): 167


    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
     
    Source
     
    解题:由于能被11整除的数的特点是奇数位与偶数位的和的绝对值可以被11整除,所以,搞一下就可以了
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 int cur = 0,n,m,d[2];
     4 int solve(int x) {
     5     int a[2] = {0};
     6     cur = 0;
     7     while(x) {
     8         a[cur^1] += x%10;
     9         x /= 10;
    10         cur ^= 1;
    11     }
    12     if(cur&1) swap(d[0],d[1]);
    13     d[0] += a[0];
    14     d[1] += a[1];
    15     return d[0] + d[1];
    16 }
    17 int main() {
    18     int cs = 1;
    19     while(scanf("%d%d",&n,&m),(~n) && (~m)) {
    20         d[0] = d[1] = 0;
    21         for(int i = 0; i <= m; ++i) n = solve(n);
    22         printf("Case #%d: %s
    ",cs++,abs(d[0]-d[1])%11?"No":"Yes");
    23     }
    24     return 0;
    25 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4722245.html
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