poj 1730Perfect Pth Powers（分解质因数）

                                                         Perfect Pth Powers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16746 Accepted: 3799

Description

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.

Input

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.

Output

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

Sample Input

171073741824250

Sample Output

1302

题意：给出一个整数x，把x写成x=a^p，求p最大是多少？

分析:把x分解质因数，x = a1^b1 * a2^b2 … ak^bk，则最终结果为b1,b2,…bk的最大公约数。注意x有可能是负数。

如果x是负数，则要把求得的答案一直除以2，知道结果一个奇数，因为一个数的偶数次方不可能是负数。

#include <cstdio>#include <cmath>#include <cstring>const int N = 66700;int is[N];int prime[7000], prime_cnt;void get_prime() {   //筛法预处理出素数    for(int i = 0; i < N; i++) is[i] = 1;    is[0] = is[1] = 0;    prime_cnt = 0;    int m = (int)sqrt(N + 0.5);    for(int i = 2; i < N; i++) {        if(is[i]) {            prime[prime_cnt++] = i;            if(i <= m) {                for(int j = i * i; j < N; j += i)                    is[j] = 0;            }        }    }}int gcd(int a, int b) { //求最大公约数    if(a < b) return gcd(b, a);    if(b == 0) return a;    return gcd(b, a % b);}int main() {    get_prime();    long long n;  //不知道当n为int时为什么会TLE    while(~scanf("%lld", &n) && n) {        int flag = 0;        if(n < 0) {            flag = 1;            n = -n;        }        int ans = 0;        for(int i = 0; i < prime_cnt && n > 1; i++) {            if(n % prime[i] == 0) {                int cnt = 0;                while(n % prime[i] == 0) {                    n /= prime[i];                    cnt++;                }                ans = gcd(ans, cnt);            }        }        if(n > 1) ans = gcd(ans, 1); //如果n不为1，则此时的n必为是一个素数        if(flag) {            while(ans % 2 == 0) ans /= 2;        }        printf("%d\n", ans);    }    return 0;}