hdu 5505 GT and numbers(分解质因子)

解题思路

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
= pri[i];
cnt
= 0;
while (k % pri[i] == 0) {
cnt
++;
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;
}