hdu 5505 GT and numbers(分解质因子)

题目链接：hdu 5505 GT and numbers

解题思路

将N，M分解质因，然后对单一因子进行考虑， 2x∗n>m ，注意如果N不是M的因子时是-1.

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef unsigned long long ll;
const int maxn = 1000005;

int N;
ll M;
int P, pri[maxn + 5], vis[maxn + 5];
int n, fac[maxn + 5], cnt[maxn + 5];

void init () {
    P = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i <= maxn; i++) {
        if (vis[i]) continue;
        pri[P++] = i;
        for (int j = i + i; j <= maxn; j += i)
            vis[j] = 1;
    }
}

void solve (int k) {
    n = 0;
    for (int i = 0; i < P; i++) {
        if (k % pri[i] == 0) {
            fac[n] = pri[i];
            cnt[n] = 0;
            while (k % pri[i] == 0) {
                cnt[n]++;
                k /= pri[i];
            }
            n++;
        }
    }
}

int get(int n, int x) {
    int ret = 0;
    while (x < n) {
        x = x + x;
        ret++;
    }
    return ret;
}

int main () {
    init();

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%llu", &N, &M);
        solve(N);

        int ans = 0;
        bool flag = true;
        for (int i = 0; i < n; i++) {
            int c = 0;
            while (M % fac[i] == 0) {
                c++;
                M /= fac[i];
            }
            if (c < cnt[i]) flag = false;
            ans = max(ans, get(c, cnt[i]));
        }
        printf("%d\n", M == 1 && flag ? ans : -1);
    }
    return 0;
}