POJ 3685 二分套二分

Matrix

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
代码：

 #include<stdio.h>
 #include<iostream>
 #include<algorithm>
 #include<string.h>
 #include<math.h>
 #include<stdlib.h>
 #include<ctype.h>
 #include<stack>
 #include<queue>
 #include<map>
 #include<set>
 #include<vector>
 #define ll long long
 #define  db double
 using namespace std;
 ;
 ;
 ll f(ll x,ll y)
 {
     *x+y*y-*y+x*y);
 }
 ll n,m;
 ll cal(ll x){
     ll cnt=,ans=;
     ;i<=n;i++){
         ll sl=,sr=n;
         while (sl<=sr){
             ll mid=(sr+sl)/;
             if(f(mid,i)<=x){
                 ans=mid;
                 sl=mid+;
             }
             ;
         }
         cnt+=ans;
     }
     return  cnt;
 }
 int main(){
     int t;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%lld%lld",&n,&m);
         ll l=-*n,r=n*n+*n+n*n+n*n;
         ll ans=;
         while (l<=r){
             ll mid=(l+r)/;
             ;}
             ;}
         }
         printf("%lld\n",ans);
     }
     ;
 }