Power of Cryptography（用double的泰勒公式可行分析）

Power of Cryptography

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=49

http://poj.org/problem?id=2109

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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

题意：给出n和p，求出 ，但是p可以很大（）

如何存储p？不用大数可不可以？

先看看double行不行：指数范围在-307~308之间(以10为基数)，有效数字为15位。

误差分析：

令f(p)=p^(1/n)，Δ=f(p+Δp)-f(p)

则由泰勒公式得

（Δp的上界是因为double的精度最多是15位，n有下界是因为 ）

由上式知，当Δp最大，n最小的时候误差最大。

根据题目中的范围，带入误差公式得Δ<9.0e-7，说明double完全够用（这从一方面说明有效数字15位还是比较足的(相对于float)）

这样就满足题目要求，所以可以用double过这一题。

 1 #include<stdio.h>
 2 #include<math.h>
 3 using namespace std;
 4 double n;
 5 double p ;
 6
 7 int main ()
 8 {
 9    // freopen ("a.txt" , "r" , stdin);
10     while (~ scanf ("%lf%lf" , &n , &p)) {
11         printf ("%.0f\n" , pow (p , 1.0 / n)) ;
12     }
13     return 0 ;
14 }

转载：http://blog.csdn.net/synapse7/article/details/11672691