HDU 2058 The sum problem

The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 12104 Accepted Submission (s): 3666

 

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]
Summary:

Observe a + 1, a + 2... A + d
Add all equal to M
That is, (a + 1 + a + d) * d/2 = M,
Here d is the square. We can start with the length d so that we can convert the range from M to M ^ 1/2;

Here we will explain ① and ② In the Code:
①: When a + 1, a + 2... When the sum of a + 3 is M, that is
(A + 1 + a + d) * d/2 = M
The minimum a value is 0, so (d + 1) * d/2 = M, d goes to the maximum value, which reduces the time complexity in this step.
D is sqrt (2 * M)

② :( A + 1 + a + d) * d/2 = M
So a * d + (d + 1)/2 = M
Therefore, to establish the equation, M-(d + 1)/2 must be a multiple of d.

 

 

import java.util.*;import java.io.*;import java.math.BigInteger;public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNextInt()){ int n=sc.nextInt();int m=sc.nextInt();if(n==0&&m==0)System.exit(0);int t1=(int)Math.sqrt(2*m);for(int i=t1;i>=1;i--){long temp=m-(i*i+i)/2;if(temp%i==0)System.out.println("["+(temp/i+1)+","+(temp/i+i)+"]");}System.out.println();}}}