UVA10780 - Again Prime? No Time.(分解质因子)

题目链接

题意：输入两个整数n和m，求最大的整数k使得m^k是n!的约数。

思路：m^k等于m的所有质因子的k次方的和，所以只要找到m中的质因子在n!中所能得到的最小的次方，就是k的值。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;int n, m;int main() {    int cas;    int t = 1;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d", &m, &n);        printf("Case %d:\n", t++);        int k = 2;        int ans = INF;        while (m != 1) {            int cnt = 0;             while (m % k == 0) {                m /= k;                 cnt++;            }             if (cnt) {                int a = 0;                for (int i = 0; i <= n; i += k) {                    if (i % k == 0 && i != 0) {                        int temp = i;                        while (temp % k == 0) {                            a++;                             temp /= k;                         }                      }                 }                a = a / cnt;                ans = min(ans, a);            }            k++;        }        if (ans)            printf("%d\n", ans);         else            printf("Impossible to divide\n");    }     return 0;}