Power of Cryptography
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 16238 Accepted: 8195

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 4357186184021382204544Sample Output

4
3
1234Source

México and Central America 2004

题意：给一对数 n(1<=n<=200)和p(1<=p<=10^101)，求k使k^n=p。

分析：1、很自然的，因为觉得数据很大，会去想高精度。然后加二分猜数。

然后不会高精度啊。。

2、于是想到转换数学运算：指对互化。用double存，但是double 精确位只有6—7。而没有logx Y，只有先转化为以e为底的对数。用lognP=logn/logP。用两次函数，

精确度不能满足要求。

3、换思路：k^n=p，则p^(1/n)=k。且函数可以直接用pow(x,y)去求x^y。

收获：巩固了一下基础。启发了一下思维。

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

代码：

#include
#include
#include
using namespace std;

int main()
{
    double n,p;
    while(scanf("%lf%lf",&n,&p)!=EOF)
    {
        printf("%.0lf\n",pow(p,1/n));
    }
    return 0;
}