UVa 113 Power of Cryptography (主意题&泰勒公式与误差分析)

UVa 113 Power of Cryptography (想法题&泰勒公式与误差分析)

113 - 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

Background

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of 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 of only theoretical interest.

This problem involves the efficient computation of integer roots of numbers.

The Problem

Given an integer  and an integer  you are to write a program that determines  , the positive root of p. In this problem, given such integers n and p, p will always be of the form  for an integerk (this integer is what your program must find).

The 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  ,  and there exists an integer k,  such that  .

The Output

For each integer pair n and p the value  should be printed, i.e., the number k such that  .

Sample Input

21632774357186184021382204544

Sample Output

431234

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：

由泰勒公式得

（n有下界是因为 ）

由上式知，当p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（所以有效数字15位还是比较足的）

这样就满足题目要求，所以可以用double过这一题。

完整代码：

/*0.015s*/#include <cstdio>#include <cmath>int main(void){double n, p;while (~scanf("%lf%lf", &n, &p))printf("%.0f\n", pow(p, 1 / n));return 0;}