題意:輸入兩個數m,n求最大的整數K使得m^k是n!的約數

題解:將m分解,m = p1^a1*p2^a2*p3^a3.... n!也分解，一個一個分解太慢，素數篩可以快一點，二分K就可以了

#include <bits/stdc++.h>
#define N 11100
#define ll long long
using namespace std;
bool isprime[N];
ll prime[N], num, c[N], temp[N], c1[N], k;
void doprime(ll n){
    ll i,j;
    num = 0;
    memset(isprime, 
 
true,sizeof(isprime));
    isprime[1] = 0;
    for(i=2;i<=n;i++){
        if(isprime[i]){
            prime[num++] = i;
            for(j=i*i;j<=n;j+=i) isprime[j] = false;
        }
    }
}
ll check(ll k){
    ll sum = 0;
    for(ll i=0; i<num; i++)
        if(c[i]<k*c1[i]) return
 
 0;
    return 1;
}
int main(){
    ll n = 5,m = 24, T, nn=1;
    doprime(10100);
    scanf("%lld", &T);
    while(T--){
        memset(c1, 0 ,sizeof(c1));
        memset(c, 0 ,sizeof(c));
        scanf("%lld%lld", &m, &n);
        for(ll i=1;i<=n;i++) temp[i] = i;
        for(ll i=0
 
;i<num&&prime[i]<=n;i++){
            ll fi = prime[i];
            for(ll j=fi;j<=n;j+=prime[i])
                while(temp[j]%prime[i] == 0) temp[j] /= prime[i],c[i]++;
        }
        for(ll i=1;i<=n;i++) if(temp[i] != 1) c[i]++;
        for(ll i=0;i<num;i++)
            while(m%prime[i] == 0) c1[i]++,m /= prime[i];
        ll l = 0, r = 1e9, ans = 0;
        while(l<=r){
            ll mid = (l+r)>>1;
            if(check(mid)) ans = mid, l = mid+1;
            else r = mid-1;
        }
        cout<<"Case "<<nn++<<":\n";
        if(ans == 0) cout<<"Impossible to divide\n";
        else cout<<ans<<"\n";
    }

    return 0;
}

UVA10780 Again Prime? No Time.