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).
代码语言：javascript复制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();
}
}
}
