二分答案，每次確定mid是第幾大的數，然後調整。

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn=50000;
long long n,k;
long long A[maxn],B[maxn];


long long Ca(long long x){
    long long f;
    long long ret1=0;
    for(int i=0;i<n;i++){
        f=x/A[i];
        if
 
(x%A[i]!=0)
            f++;
        if(B[n-1]>=f)
            ret1+=(n-(long long)(lower_bound(B,B+n,f)-B));
    }
    return ret1;
}

int main()
{
    cin>>n>>k;
    for(int i=0;i<n;i++){
        cin>>A[i]>>B[i];
    }
    sort(A,A+n);
    sort(B,B+n);
    long long l=A[0
 
]*B[0],r=A[n-1]*B[n-1],mid;
    while(l<=r){
        mid=(l+r)/2;
        long long x=Ca(mid);
        if(x>=k){
            l=mid+1;
        }
        else{
            r=mid-1;
        }
    }
    cout<<l-1<<endl;
    return 0;
}