TOJ 1438.Perfect Pth Powers

题目链接：http://acm.tju.edu.cn/toj/showp1438.html

1438.   Perfect Pth Powers

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 2341   Accepted Runs: 466

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We say that x is a perfect square if, for some integer b, x = b². Similarly, x is a perfect cube if, for some integer b, x = b³. More generally, x is a perfect pth power if, for some integer b, x = b^p. Given an integer x you are to determine the largest p such that x is a perfect pth power.

Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.

For each test case, output a line giving the largest integer p such that x is a perfect pth power.

Sample Input

171073741824250

Output for Sample Input

1302

Source: Waterloo Local Contest Jan. 31, 2004

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水题，直接用pow就好，注意n的范围。

#include <stdio.h>#include <cmath>using namespace std;int main(){long long n;while(~scanf("%lld",&n),n){if(n>0){for(int i=31;i>0;i--){double p=pow(1.0*n,1.0/i);for(long long j=p-10;j<=p+10;j++)if( (long long)pow(1.0*j,i)==n){printf("%d\n",i);i=0;break;}}}else {n=-n;for(int i=31;i>0;i-=2){double p=pow(1.0*n,1.0/i);for(long long j=p-10;j<=p+10;j++)if( (long long)pow(1.0*j,i)==n){printf("%d\n",i);i=0;break;}}}}}