题目描述给定n，a求最大的k，使n！可以被a^k整除但不能被a^(k+1)整除。输入描述:两个整数n(2<=n<=1000)，a(2<=a<=1000)输出描述:一个整数.示例1输入复制6 10输出复制1分解质因数取min即可；#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include//#include//#pragma GCC optimize(2)using namespace std;#define maxn 200005#define inf 0x7fffffff//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-11typedef pair pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair pii;inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}int sqr(int x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/int pme[maxn];int tot;int vis[maxn];int cnt[maxn];void init() { for (int i = 2; i < maxn; i++) { if (!vis[i]) { pme[++tot] = i; for (int j = i; j < maxn; j += i) { vis[j] = 1; } } }}int num[maxn];int main() { int n, a; scanf("%d%d", &n, &a); init(); for (int i = 1; i <= tot; i++) { int tmp = n; while (tmp) { cnt[pme[i]] += tmp / pme[i]; tmp /= pme[i]; } } int minn = 999999;/* for(int i=1;i<=tot;i++){ if(cnt[pme[i]]){ cout< } } */ for (int i = 1; i <= tot; i++) { while (a%pme[i] == 0) { num[i]++; a /= pme[i]; } } for (int i = 1; i <= tot; i++) { if(num[i]) minn = min(minn, cnt[pme[i]] / num[i]); } cout << minn << endl;} EPFL - Fighting
