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]

题目的意思:输入两个整数N,M。 N, M( 1 <= N, M <= 1000000000),如果在范围[1,M]内连续整数的和为N,按从小到大次序输出所有这样的连续段,当输入的M,N都为0时结束。
计算的思路:
不考虑子列的终点,而是考虑子列的起点和子列元素的个数,分别记为i,j。由等差数列求和公式,得(i+(i+j-1))*j/2==M ,即(2*i+j-1)*j/2==M(2式),故得i=(2*M/j-j+1)/2,将i,j代回2式,成立则[i,i+j-1]满足条件。注意j最小为1,而由2式,得(j+2*i)*j=2*M,而i>=1,故j*j<=(int)sqrt(2*M).

import java.util.Scanner;

public class Main{

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n = sc.nextInt();
            int m = sc.nextInt();
            if(m==0&&n==0){
                return ;
            }

            int j =(int)Math.pow(2.0*m, 0.5);
            for(j=j;j>0;j--){
                int i;
                i = (2*m/j-j+1)/2;
                if(j*(j+2*i-1)/2==m){
                    System.out.println("["+i+","+(i+j-1)+"]");
                }
            }
            System.out.println();
        }
    }
}
posted on 2016-01-31 21:48  cnxo  阅读(125)  评论(0编辑  收藏  举报