Matrix（二分套二分）

Matrix

http://poj.org/problem?id=3685

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8943		Accepted: 2738

Description

Given a N × N matrix A, whose element in the i-th row and j-th column A_ij is an number that equals i² + 100000 × i + j² - 100000 × j + i × j, you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778 

#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
typedef long long ll;
using namespace std;

ll n,k;
ll cal(ll i,ll j){
    return i*i+100000*i+j*j-100000*j+i*j;
}

bool Check(ll num){
    ll sum=0;
    for(ll i=1;i<=n;i++){
        ll L=1,R=n,mid;
        while(L<=R){
            mid=L+R>>1;
            if(cal(mid,i)<=num){
                L=mid+1;
            }
            else{   
                R=mid-1;
            }
        }
        sum+=R;
    }
    return sum>=k;
}

int main(){

    int T;
    cin>>T;
    while(T--){
       cin>>n>>k;
       ll L=-1e18,R=1e18,mid;
       while(L<=R){
           mid=L+R>>1;
           if(Check(mid)){
               R=mid-1;
           }
           else{
               L=mid+1;
           }
       }
       cout<<L<<endl;
    }

}