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poj2109 Power of Cryptography

分享于 2015-05-01

推荐:POJ2109:Power of Cryptography

点击打开题目链接 Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16059   Accepted: 8103 Description Current work in

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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<10 101 and there exists an integer k, 1<=k<=10 9 such that k n = 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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234
这题可以用double型做,

类型            长度 (bit)           有效数字                   绝对值范围

推荐:POJ 2109 Power of Cryptography

不知道那里贪心了。。。学长说贪心。 注意:数据类型范围,还是不太了解; 原题链接:http://poj.org/problemid=2109 Power of Cryptography Time Limit: 1000M

    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-8 int 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; }

推荐:Power of Cryptography POJ2109

Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 11348   Accepted: 5847 Description Current work in cryptography inv

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

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