uva 10780 - Again Prime? No Time.(数论）

来源：互联网发布：丽升网络评卷系统编辑：程序博客网时间：2024/05/24 00:57

题目链接：uva 10780 - Again Prime? No Time.

题目大意：给定m和n，计算n！中有多少个因子m。

解题思路：将1~n分解质因子，然后在将m分解，对应每个因子计算说可以保证的个数，维护最小值。

#include <stdio.h>#include <string.h>#include <math.h>#define min(a,b) (a)<(b)?(a):(b)const int N = 10005;const int INF = 0x3f3f3f3f;int n, m, c[N];void div (int k) {int t = ceil(sqrt(k));for (int i = 2; i <= k && i <= t; i++) {while (k % i == 0) {c[i]++;k /= i;}}c[k]++;}void solve () {int i;memset(c, 0, sizeof(c));for (i = 2; i <= n; i++)div(i);int ans = INF, t = ceil(sqrt(m));for (i = 2; i <= m && i <= t; i++) {if (m%i == 0) {int cnt = 0;while (m % i == 0) {cnt++;m /= i;}ans = min (ans, c[i]/cnt);}}if (m != 1)ans = min(ans, c[m]);if (ans == INF || ans == 0) printf("Impossible to divide\n");else printf("%d\n", ans);}int main () {int cas;scanf("%d", &cas);for (int i = 1; i <= cas; i++) {scanf("%d%d", &m, &n);printf("Case %d:\n", i);solve ();}return 0;}

1 0