poj3685（嵌套二分）

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.

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.

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

#include <iostream>

using namespace std;
long long N, M;
const long long INF = 1LL << 50;
long long mtr ( long long  i, long long j )
{
	return i * i + 100000 * i + j * j - 100000 * j + i * j;
}

bool b_s ( long long X )
{
	long long res = 0;

	for ( int i = 1; i <= N; i++ )
		{
			int cnt = N ;
			int l = 1, r = N;

			while ( l <= r )
				{
					int mid = ( r + l ) >> 1;

					if ( mtr ( mid, i ) >= X )
						{
							r = mid - 1;
							cnt = mid - 1;
						}
					else
						{
							l = mid + 1;
						}
				}

			res += cnt ;
		}
	return res >= M;
}
int main()
{
	int n;
	cin >> n ;
	while ( n-- )
		{
			cin >> N >> M;
			long long l = -INF, r = INF;
			long long ans=-1;

			while ( l <= r )
				{
					long long mid = ( r + l )>>1;
					if ( b_s ( mid ) )
						{
							r = mid - 1;
							ans = mid - 1;
						}
					else
						{
							l = mid + 1;
						}
				}

			cout <<ans << endl;
		}

	return 0;
}