UVA 10780 Again Prime No Time.(数学)
给定两个整数m和n,求最大的k使得m^k是n!的约数
对m质因子分解,然后使用勒让德定理求得n!包含的质数p的阶数,min(b[i] / a[i])即为结果k, 若为0无解
#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include<utility> using namespace std; typedef long long LL; const int N = 5008, INF = 0x3F3F3F3F; #define MS(a, num) memset(a, num, sizeof(a)) #define PB(A) push_back(A) #define FOR(i, n) for(int i = 0; i < n; i++) int p[N], a[N], b[N]; int lp(int n, int p){ int ans = 0; for(int i = p; i <= n; i*= p){ ans += n / i; } return ans; } int solve(int m, int n){ memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); int tot = 0; for(int i = 2; i * i <= m; i++){ if(m % i == 0){ p[tot] = i; while(m % i == 0){ a[tot]++; m /= i; } tot++; } } if(m > 1){ p[tot] = m; a[tot++] = 1; } int ans = INF; for(int i = 0; i < tot; i++){ b[i] = lp(n, p[i]); if(b[i] < a[i]){ return -1; } ans = min(ans, b[i] / a[i]); } return ans; } int main(){ int t, m, n; cin >>t; for(int cas = 1; cas <= t; cas++){ scanf("%d %d", &m, &n); int ans = solve(m, n); printf("Case %d:\n", cas); if(ans == -1){ printf("Impossible to divide\n"); }else{ printf("%d\n", ans); } } return 0; }