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): 15122 Accepted Submission(s): 4529



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]

Author

8600

Source

校庆杯Warm Up

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#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
//枚举字串的长度
int main()
{
int n,m,i;
while(scanf("%d%d",&n,&m)!=EOF&&(n!=0||m!=0))
{
for(int j=sqrt(2*m);j>=1;j--)
{
i=((2*m)/j+1-j)/2;
if((i+i+j-1)*j==2*m)
printf("[%d,%d]\n",i,i+j-1);
}
printf("\n");
}
return 0;
}