UVA - 10622 Perfect P-th Powers

Description

Problem E: Perfect Pth Powers

We say that x is a perfect square if, for some integer b,x = b^2. Similarly,x is a perfect cube if, for some integerb, x = b^3. More generally,x is a perfect pthpower if, for some integer b, x = b^p. Given anintegerx you are to determine the largest p such that xis 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 therange of a (32-bit) int inC, 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 thatx isa perfect pth power.

Sample Input

171073741824250

Output for Sample Input

1302

题意：求x=b^p,的 最大的p

思路：首先将n分解质因数，然后找因数个数的gcd就是了，负数的话，一直除以2，直到答案是奇数的时候

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>typedef long long ll;using namespace std;const int maxn = 333333;ll n;int prime[maxn], vis[maxn], cnt = 0;int gcd(int a, int b) {return b==0?a:gcd(b, a%b);}int solve() {ll tmp = n;if (n < 0)n = -n;int ans = 0;for (int i = 0; i < cnt && prime[i] <= n; i++) {int count = 0;while (n % prime[i] == 0) {count++;n /= prime[i];}ans = gcd(ans, count);}if (ans == 0)ans = 1;if (tmp < 0) {while (ans % 2 == 0)ans /= 2;}return ans;}int main() {for (int i = 2; i < maxn; i++) {if (vis[i])continue;prime[cnt++] = i;for (int j = i; j < maxn; j += i)vis[j] = 1;}while (scanf("%lld", &n) != EOF && n) {printf("%d\n", solve());}return 0;}