zoj3167------------Find 7 Faster Than John Von Neumann 大数乘法

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 smallestn? 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?

Input

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

Output

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

You can assume:

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

Sample Input

3 2
4 2
4 3      

Sample Output

15
21
11      

大数乘法，一个数组存要乘的数，一个数组存结果，一个用来传递中间量

代码：

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    int K,M,n,w;
    int m[1000],m1[3],t[1000];
    while(cin>>K>>M)
    {
        memset(m,0,sizeof(m));
        memset(m1,0,sizeof(m1));
        m[0]=1;
        for(int i=0;M!=0;i++)
        {
            m1[i]=M%10;
            M=M/10;
        }         
        n=0;
        while(1)
        {
            memset(t,0,sizeof(t));
			   for(int i=0;i<=2;i++)
            {
                for(int j=0;j<1000;j++)
                {
                    t[j+i]=t[j+i]+m[j]*m1[i];
                }    
                for(int j=0;j<1000;j++)
                {
                    if(t[j]>9)
                    {
                        t[j+1]=t[j+1]+t[j]/10%10;
                        t[j]=t[j]%10;
						  }
                }
            }
            for(int j=0;j<1000;j++)
            	m[j]=t[j];
            n++;
            if(m[K-1]==7)
            {
                cout<<n<<endl;    
                break;
            }
        }            
    }
    return 0;
}