题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3199
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 11AC代码:
#include <stdio.h> #include <string.h> 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 n,k,m,i,p; while(~scanf("%d%d",&k,&m)){ memset(a,sizeof(a)); memset(b,sizeof(b)); p=m; n=1; for(i=0;p;i++){ a[i]=b[i]=p%10; p/=10; } while(a[k-1]-7){ mul(m); for(i=0;i<300;i++) a[i]=b[i]; n++; } printf("%d\n",n); } return 0; }