Poj3685 (nested binary)

Matrix

Time limit:6000 ms		Memory limit:65536 K
Total submissions:4658		Accepted:1189

Description

GivenN×NMatrix A, whose element inI-Th row andJ-Th ColumnAIJIs an number that equalsI2 + 100000 ×I+J2-100000 ×J+I×J, You are to findM-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) andM(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

121 12 12 22 32 43 13 23 83 95 15 255 10

Sample output

3-99993312100007-199987-99993100019200013-399969400031-99939

Source

Poj founder monthly contest-2008.08.31, windy7926778 first create a table, and initially think it will increase progressively from bottom left to top right. Taking it seriously, it is not a special rule. When n is relatively large, it will increase. However, the monotonicity of each column can be maintained. It is easy to know how to take I and j as constants to find the derivative. This two points are interesting. In the long range, a number X is divided into two parts, and the number> is the number that satisfies the condition that X is greater than or equal to m and greater than M. It can be determined by dividing each column into two parts. Time Complexity log (10 ^ 18) * n * log (n ).

#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;}

Poj3685 (nested binary)