POJ-1730 perfect PTH powers factorization factor

This question is to find the maximum number of power to which a number can be opened. This includes a negative number and must be read in long data. First, we break down the prime factor of this number and calculate their respective numbers. Then, we find a GCD for their numbers and finally output them. If it is a positive number, it is output directly. If it is a negative number, you need to find the largest Odd Factor.

The Code is as follows:

#include <cstdlib>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;int p[66000], cnt;long long rec[6600], N; int son[6600];int gcd(int a, int b){    return b == 0 ? a : gcd(b, a % b);    }void pre(){    int k;    for (int i = 4; i <= 66000; i += 2) {        p[i] = 1;    }    for (int i = 3; i <= 257; i += 2) {        if (!p[i]) {            k = 2 * i;            for (int j = i * i; j <= 66000; j += k) {                p[j] = 1;            }        }    }    cnt = 1;    rec[1] = 2;    for (int i = 3; i <= 66000; i += 2) {        if (!p[i]) {            rec[++cnt] = i;        }    }//    printf("cnt = %d\n", cnt);}int deal(long long x){    bool flag = true;    int ans = 0;    for (int i = 1; i <= cnt; ++i) {        while (x % rec[i] == 0) {                 son[i]++;            x /= rec[i];        }    }    if (x != 1) {        return 1;    }    for (int i = 1; i <= cnt; ++i) {        if (son[i]) {            ans = gcd(ans, son[i]);        }    }    return ans;}// 1728 = 12 ^ 3  11664 = 108^2int main(){    int ans;    pre();    while (scanf("%I64d", &N), N) {        memset(son, 0, sizeof (son));         ans = deal(N < 0 ? -N : N);        if (N < 0) {            while (!(ans & 1)) {                ans >>= 1;            }            printf("%d\n", ans);        }        else {            printf("%d\n", ans);        }    }    return 0;    }