2015 Multi-University Training Contest 7 1005

The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 380 Accepted Submission(s): 186


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each input , if the final number are divisible by 11, output “Yes”, else output ”No”. without quote.

[align=left]Sample Input[/align]

35 2
35 1

-1 -1

[align=left]Sample Output[/align]

Case #1: Yes
Case #2: No

[align=left]Source[/align]
2015 Multi-University Training Contest 7

题意：将前面的每位数相加得到的数加到最后，如此往复执行n次，问最后得到的数字能否被11整除
分析：模拟一次就行，注意答案可能计算中的结果可能会超int

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
LL num,n,ans,sum;

LL calc(LL x)
{
LL res=0;
while(x)
{
res+=x%10;
x=x/10;
}
return res;
}

LL count_digit(LL x)
{
if(x==0) return 10;
LL ok=1;
while(x)
{
x=x/10;
ok=ok*10;
}
return ok;
}
int main()
{
//freopen("in.txt","r",stdin);
int Case=0;
while(scanf("%I64d %I64d",&num,&n)!=EOF)
{
if(num==-1 && n==-1) break;

ans=calc(num);//ans表示当前要加在后排的数
sum=num;//sum表示累加的和
for(int i=1;i<=n;i++)
{
sum=sum*count_digit(ans)+ans;
sum=sum%11;
ans=ans+calc(ans);
}

printf("Case #%d: ",++Case);
if(sum%11==0) printf("Yes\n");
else printf("No\n");
}
return 0;
}

View Code