OJ-HDU The sum problem

                                 The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30951    Accepted Submission(s): 9261

Problem Description

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

Input

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

Output

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

Sample Input

20 10

50 30

0 0

Sample Output

[1,4]

[10,10]

[4,8]

[6,9]

[9,11]

[30,30]

思路:

原本是直接两个循环再加上求和公式做,果不其然TLE了

后来查了一下网上的做法,优化了许多

设i为子序列起点,j为起点到终点的距离,可以推出(i+i+j-1)*j/2=M

i=(2*M/j+1-j)/2

所以只需要知道距离就可以推出起点的位置

(2*i+j-1)j=2M

根据放缩法则,可以得出j<=sqrt(2*M),只需要遍历一下就行

AcCode:

package hdu经典100题;

import java.util.Scanner;

public class P2058 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		while(in.hasNext()) {
			long N = in.nextLong();
			long M = in.nextLong();
			if(N==0 && M==0) {
				break;
			}			
			for (long i = (long) Math.sqrt(2*M); i>0; i--) {
				N = ((2*M)/i+1-i)/2;
				if((2*N+i-1)*i/2==M) {
					System.out.println("["+N+","+(N+i-1)+"]");
				}
			}
			System.out.println();
		}
	}

}