题目链接
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
2015 Multi-University Training Contest 7
题意:给你一个数N,现在给你t次操作,每次操作就是把这个数每位数的和加在这个数后面,比如n=123,t=3时,操作流程为:123->1236->123612->12361215,现在问你进行t次操作后的数能否被11整除。
由于操作后数据过大,不可能直接暴力找,所以我们要进行模拟。首先我们要知道,能被11整除的数的特征就是:奇数位上的数的和与偶数位上的和只差能被11整除,大减小。根据这一点我们就可以很方便的模拟这个过程,每次操作维护几个值就行,维护奇数位上的和,偶数位上的和,奇数的个数,偶数的个数,以及这个大数每位数的和,具体看代码。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define ll long long ll ji,ou,sum;//ji记录奇数位的和,ou记录偶数位的和,sum记录奇数位和偶数位的和,即全部 ll numji,numou;//numji记录奇数的个数,numou记录偶数的个数 ll a[100000]; void solve(ll x) { ll k=0; while(x) { k++; a[k]=x%10; x/=10; }//把要加到后面的那个数拆开 for(ll i=k;i>=1;i--) { if(numji==numou)//如果以前操作结束时奇数个数等于偶数个数,那么此时从奇数开始 { numji++; ji+=a[i]; } else if(numji>numou)//此时从偶数开始,因为第一个数是从奇数位开始的,所以奇数的个数一直大于等于偶数的个数 { numou++; ou+=a[i]; } sum+=a[i]; } } int main() { ll n,t,ca=0; while(scanf("%lld%lld",&n,&t)!=EOF) { if(n==-1&&t==-1) break; ca++; ji=0,ou=0; numji=0,numou=0; sum=0; solve(n); for(ll i=1;i<=t;i++) { solve(sum); } ll ans; if(ji>=ou) ans=ji-ou; else ans=ou-ji; printf("Case #%lld: ",ca); if(ans%11==0) puts("Yes"); else puts("No"); } return 0; }