hdu 3641 Treasure Hunting 强大的二分

/**
大意：给定一组ai,bi .    m = a1^b1 *a2^b2 * a3^ b3 * a4^b4*...*ai^bi 
求最小的x!%m =0
思路: 将ai 质因子分解，若是x!%m=0 那么x！ 质因子分解之后 质因子的个数一定大于等于m的个数。二分求解可得
注意： 二分时，需要将，上下限 设定好，low =0； high = 1ll<<60;
**/

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;

long long pri[150];

void getApri(long long a,long long b){
    int cnt;
    for(int i=2;i*i<=a;i++)if(a%i==0){
        cnt =0;
        while(a%i==0){
            cnt++;
            a =a/i;
        }
        pri[i] += cnt*b;
        if(a==1)
            break;
    }
    if(a>1)
        pri[a] += b;
}

long long cal(long long i,long long x){   x!中有多少个i因子
    long long ret =0;
    while(x){
        x = x/i;
        ret += x;
    }
    return ret;
}

bool judge(long long x){
    for(int i=1;i<=100;i++)if(pri[i]){
        long long tmp = cal(i,x);
        if(tmp<pri[i])
            return false;
    }
    return true;
}

int main(){
    int t;
    cin>>t;
    while(t--){
        memset(pri,0,sizeof(pri));
        int n;
        cin>>n;
        long long a,b;
        while(n--){
            cin>>a>>b;
            getApri(a,b);
        }
        long long low = 0,high = 1ll<<60;
        long long ans;
        while(low<=high){
            long long mid  = (low+high)/2;
            if(judge(mid)){
                ans = mid;
                high  = mid-1;
            }
            else
                low = mid +1;
        }
        cout<<ans<<endl;
    }

    return 0;
}