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
題意：給一對數 n(1<=n<=200)和p(1<=p<=10^101)，求k使k^n=p。
分析：1、很自然的，因為覺得數據很大，會去想高精度。然後加二分猜數。
然後不會高精度啊。。
2、於是想到轉換數學運算：指對互化。用double存，但是double 精確位只有6—7。而沒有logx Y，只有先轉化為以e為底的對數。用lognP=logn/logP。用兩次函數，
精確度不能滿足要求。
3、換思路：k^n=p，則p^(1/n)=k。且函數可以直接用pow(x,y)去求x^y。
收獲：鞏固了一下基礎。啟發了一下思維。
類型 長度 (bit) 有效數字 絕對值范圍
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
代碼：
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
double n,p;
while(scanf("%lf%lf",&n,&p)!=EOF)
{
printf("%.0lf\n",pow(p,1/n));
}
return 0;
}
