poj2109 Power of Cryptography

Description

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest. 
This problem involves the efficient computation of integer roots of numbers. 
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n^th. power, for an integer k (this integer is what your program must find).

Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10¹⁰¹ and there exists an integer k, 1<=k<=10⁹ such that kⁿ = p.

Output

For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.

Sample Input

2 163 277 4357186184021382204544

Sample Output

431234

这题可以用double型做，
类型            长度 (bit)           有效数字                   绝对值范围
    float            32                     6~7                 10^(-37) ~ 10^38
    double         64                    15~16              10^(-307) ~10^308
    long double  128                  18~19               10^(-4931) ~ 10 ^ 4932
因为k^n=p,所以k=p^(1.0/n);
#include<stdio.h>#include<math.h>int main(){double n,m;while(scanf("%lf%lf",&n,&m)!=EOF)    {    printf("%.0f\n",pow(m,(1.0)/n));    }    return 0;}
也可以用二分法做
#include<stdio.h>#include<string.h>#include<math.h>#define eps 1e-8int main(){double n,m,i,j,h,temp;int l,r,mid;while(scanf("%lf%lf",&n,&m)!=EOF){r=1000000000;l=1;while(l<=r){mid=(l+r)/2;temp=pow(mid,n);//printf("%lf       %lf\n",temp,pow(mid,n));if(fabs(pow(mid,n)-m)<eps){printf("%d\n",mid);break;       //因为这里求的是一个确定的整数，所以可以直接输出}else if(pow(mid,n)<m){l=mid+1;}else if(pow(mid,n)>m){r=mid-1;}}}return 0;}