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+*i+j*j-*j+i*j;
 }

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

 }