hdu 5373 - The shortest problem 解题报告

The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1141 Accepted Submission(s): 554



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：给你n,t。例1：n=35,t=2。第一次3+5=8,变为358，第二次，3+5+8=16,变为35816.发现35816可以被11整除。输出yes。

题解：这里最重要的是要知道怎么样的数可以被十一整除，可以通过百度或学习数论的知识了解到：

当一个数的奇数位之和与偶数位之和的差的绝对值能被11整除，那么该数就可以被11整除。如35816，奇数位之和为17，偶数位之和是6，差的绝对值为11，所以能被11整除。

参考代码：

#include<stdio.h>
#include<stdlib.h>
#define ll __int64
int s[30],k;
ll luwherehandsome(ll x)
{
int tem=0;
while(x)
{
s[k++]=x%10;
tem+=x%10;
x/=10;
}
return tem;
}
int main()
{
int n,t,v,q=0;
ll tem,ans,m;
while(~scanf("%d%d",&n,&t))
{
ans=0;
q++;
if(n==-1&&t==-1)
break;
k=0;
v=1;
m=luwherehandsome(n);
while(k--)
{
if(v)
{
ans+=s[k];
v=0;
}
else
{
ans-=s[k];
v=1;
}
}
while(t--)
{
k=0;
m+=luwherehandsome(m);
while(k--)
{
if(v)
{
ans+=s[k];
v=0;
}
else
{
ans-=s[k];
v=1;
}
}
}
printf("Case #%d: ",q);
if(abs(ans)%11)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}