Topic link :

pid=3641"> Portal

The question :

Minimum ( x! ) = 0 mod (a1^b1*a2^^bn)

analysis :

First of all a1~an Prime factorization , Then count the index of each qualitative factor . Because with x The increase of , The number of prime factors is gradually added

So we were able to split x. Yes x! Prime factorization is used to infer whether the condition is satisfied . And then find the smallest one .

The code is as follows :

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 110;
typedef long long LL;
bool vis[maxn];
int p[maxn],cnt;
LL a[maxn];
LL b[maxn];
LL num[maxn]; void init(){
cnt = 0;
for(int i=2;i<maxn;i++){
for(int j=i+i;j<maxn;j+=i)
} LL get_num(LL x,int pri){
if(x<pri) return 0;
return get_num(x/pri,pri)+(LL)x/pri;
} bool check(LL x){
for(int i=0;i<cnt;i++){
return false;
return true;
} int main()
int t,n;
for(int i=0;i<n;i++){
int tmp = a[i];
for(int j=0;p[j]*p[j]<=tmp&&j<cnt;j++){
LL tot=0;
while(tmp%p[j]==0) tmp=tmp/p[j],tot++;
if(tmp>1) num[tmp]+=b[i];
LL ans = 0;
for(int i=0;i<maxn;i++)
LL l=0,r=ans;
LL mid=(l+r)>>1;
if(check(mid)) r=mid-1;
else l=mid+1;
return 0;
6 1000000000000
15 1000000000000
13 1000000000000
7 1000000000000
2 1000000000000
3 1000000000000

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