zoj 3167 Find 7 Faster Than John Von Neumann

原创于 2017-04-04 19:45:35 发布 · 258 阅读

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Find 7 Faster Than John Von Neumann

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?

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:

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

Sample Input

3 2
4 2
4 3

Sample Output

15
21
11

【分析】已知K，M，求使得_Mⁿ的第K位是7的最小n

大数乘法，用数组模拟

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[300],b[300];
void mul(int i){
    int j,k;
    for(j=0;j<300;j++)
        b[j]*=i;
    for(i=k=0;i<300;i++){
        j=(b[i]+k)/10;
        b[i]=(b[i]+k)%10;
        k=j;
    }
}
int main(){
    int k,m;
    while(~scanf("%d%d",&k,&m)){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int p=m;
        int ans=1;
        for(int i=0;p;i++){
            a[i]=b[i]=p%10;
            p/=10;
        }
        while(a[k-1]!=7){
            mul(m);
            for(int i=0;i<300;i++)
                a[i]=b[i];
            ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}