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[];

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

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

 bool judge(long long x){
     for(int i=;i<=;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,,sizeof(pri));
         int n;
         cin>>n;
         long long a,b;
         while(n--){
             cin>>a>>b;
             getApri(a,b);
         }
         long long low = ,high = 1ll<<;
         long long ans;
         while(low<=high){
             long long mid  = (low+high)/;
             if(judge(mid)){
                 ans = mid;
                 high  = mid-;
             }
             else
                 low = mid +;
         }
         cout<<ans<<endl;
     }

     return ;
 }