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?

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 1005int 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;}