Find 7 Faster Than John Von Neumann

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挑战解决一道数学编程题，找出使M的n次方的第K位为7的最小n值。通过数组模拟大数运算，逐步迭代计算直至找到答案。

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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?

输入：

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

输出：

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.

样例输入：

3 2
4 2
4 3

样例输出：

15
21
11

题目大意：

给出k和m求出最小的n，使得m的n次方的第k位为7。

代码如下：

#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)                         //判断第k位是否已经等于7 
	{
		for(int i=0;i<N;i++)				//利用数组来存入大数一位一位的处理			
		{
			asd[i]=asd[i]*m;
		}
		for(int i=0;i<N;i++)				
		{
			asd[i+1]=asd[i]/10+asd[i+1];	//不断地将低位的数字除以10后加到高位去  
			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;
}