pid=3641"> Portal

The question ：

Minimum ( x! ) = 0 mod (a1^b1*a2^b2...an^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;
memset(vis,0,sizeof(vis));
for(int i=2;i<maxn;i++){
if(!vis[i]){
p[cnt++]=i;
for(int j=i+i;j<maxn;j+=i)
vis[j]=1;
}
}
} 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++){
if(get_num(x,p[i])<num[p[i]])
return false;
}
return true;
} int main()
{
init();
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
memset(num,0,sizeof(num));
for(int i=0;i<n;i++){
scanf("%I64d%I64d",a+i,b+i);
int tmp = a[i];
for(int j=0;p[j]*p[j]<=tmp&&j<cnt;j++){
if(tmp%p[j]==0){
LL tot=0;
while(tmp%p[j]==0) tmp=tmp/p[j],tot++;
num[p[j]]+=tot*b[i];
}
}
if(tmp>1) num[tmp]+=b[i];
}
LL ans = 0;
for(int i=0;i<maxn;i++)
ans=max(ans,(LL)i*num[i]);
LL l=0,r=ans;
while(l<=r){
LL mid=(l+r)>>1;
if(check(mid)) r=mid-1;
else l=mid+1;
}
printf("%I64d\n",l);
}
return 0;
}
/*
111
6
6 1000000000000
15 1000000000000
13 1000000000000
7 1000000000000
2 1000000000000
3 1000000000000
*/

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