【二分答案】POJ-2109 Power of Cryptography

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000K   

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

2 163 277 4357186184021382204544

Sample Output

431234

————————————————————ゆっくりとの分割線————————————————————

思路：你知道double的数据范围吗？

1.7E-308～1.7E+308

嗯。

pow(p,1/n)

但是这样精度可能不够。

P.S. 不告诉你cin cout可以A，但是scanf printf就会WA

所以不能容忍 1 / n 这样的不精确数了。

那就二分答案，然后判断 k == pow(m, n) ?

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <climits>#include <iostream>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;/****************************************/int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifdouble n, p;while(~scanf("%lf%lf", &n, &p)) {LL l, r, m;l = 0; r = 1e9+1;while(l < r) {m = (l+r) >> 1;double x = pow(m, n);if(x > p) r = m;else if(x < p) l = m;else {printf("%lld\n", m);break;}}}return 0;}