Find 7 Faster Than John Von Neumann

描述：

It was said that when testing the first computer designed by von Neumann, people gave the following problem to both the legendary professor and the new computer: If the 4th digit of 2^n is 7, what is the smallest n? The machine and von Neumann began computing at the same moment, and von Neumann gave the answer first. Now you are challenged with a similar but more complicated problem: If the K-th digit of M^n is 7, what is the smallest n?

輸入：

Each case is given in a line with 2 numbers: K and M (< 1,000).

輸出：

For each test case, please output in a line the smallest n.

You can assume: 

The answer always exist. 
The answer is no more than 100.

樣例輸入：

3 2
4 2
4 3

樣例輸出：

15
21
11

題目大意：

給出k和m求出最小的n，使得m的n次方的第k位爲7。

代碼如下：

#include<stdio.h>
#include<string.h>
#define N 1005
int n;
int k,m;
int asd[1005];
void check(int q)
{
	while(asd[q]!=7)                         //判斷第k位是否已經等於7 
	{
		for(int i=0;i<N;i++)				//利用數組來存入大數一位一位的處理			
		{
			asd[i]=asd[i]*m;
		}
		for(int i=0;i<N;i++)				
		{
			asd[i+1]=asd[i]/10+asd[i+1];	//不斷地將低位的數字除以10後加到高位去  
			asd[i]=asd[i]%10;				 
		}	
		n++;	
	}
}
int main()
{
	while(scanf("%d %d",&k,&m)!=EOF)
	{
		memset(asd,0,sizeof(asd));
		n=0;
		asd[0]=1;
		check(k-1);
		printf("%d\n",n); 
	}
	return 0;
}