poj 2109 Power of Cryptography

Power of Cryptography
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 17912   Accepted: 9034

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 nth. 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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

Source

 
正常算法是二分+高精度。
 
可是由下表:

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

可以一步算出结果,如果多步会丢失精度:

 1 //180K    0MS    C++    195B    2014-05-08 13:47:15
 2 #include<stdio.h>
 3 #include<math.h>
 4 int main(void)
 5 {
 6     int n;
 7     double p;
 8     while(scanf("%d%lf",&n,&p)!=EOF)
 9     {
10         printf("%.0lf
",pow(p,1.0/n));
11     }
12     return 0;
13 } 

写了二分法的不过应该是处理不够好,没过就不贴了。

原文地址:https://www.cnblogs.com/GO-NO-1/p/3716048.html