poj1730

好吧，这次被负数给卡死了。

以及一开始思想的错误，找到最大的p使b^p = x；

先对x进行质因数分解，因为x < 2 ^ 32，所以质因子最大为sqrt(2 ^ 32)，把大约小于100000的质数扫一下，然后进行质因子分解。

不一定要所有因子的次数都相同，只要求出所以因子的最大公约数就可以得到p。但是当x时负数的时候，只能是奇数次方，所以要将p一直除到为奇数为止。

AC代码：

#include <cstdio>#include <string.h>#include <cmath>const int MAX_NUMBER = 100000;bool vis[MAX_NUMBER + 1];int prime[MAX_NUMBER], factor[MAX_NUMBER];int prime_number;void getAllPrime() {memset(vis, 0, sizeof(vis));int m = (int)sqrt(MAX_NUMBER + 0.5);for (int i = 2; i <= m; i++) {if (!vis[i]) {for (int j = i * i; j <= MAX_NUMBER; j += i) {vis[j] = 1;}}}prime_number = 0;for (int i = 2; i <= MAX_NUMBER; i++) {if (!vis[i]) {prime[++prime_number] = i;}}}int gcd(int a, int b) {if (b == 0) {return a;}else {return gcd(b, a % b);}}int main() {getAllPrime();int number;while (scanf("%d", &number) != EOF) {memset(factor, 0, sizeof(factor));if (!number) {break;}int ans = 0;for (int i = 1; i <= prime_number; i++) {while (number % prime[i] == 0) {number /= prime[i];factor[i]++;}}int first = 0;int flag = 0;int first_factor;for (int i = 1; i <= prime_number; i++) {if (factor[i] != 0) {if (first == 0) {first_factor = factor[i];first = 1;}else {first_factor = gcd(first_factor, factor[i]);}}}if (first == 0) {first_factor = 1;}if (number < 0) {while (first_factor % 2 == 0) {first_factor = first_factor / 2;}printf("%d\n", first_factor);}else {printf("%d\n", first_factor);}}return 0;}