HDU 2058 The sum problem Time(等差と推理のような数学の問題)

The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 18728 Accepted Submission(s): 5561
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]
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    //  ：                           Sn=(a1+an)*n/2       2       
   
   
   
   
   
   
   
   

    
    
    
        Sn=na1+n(n-1)d/2
   
   
   
   
   
   
   
   

    
    
    
     a1=1          N   a1=1   2Sn=n(n+1)=2*m    n<=sqrt(2*m)  
   
   
   
   
   
   
   
   

    
    
    
      n=[1,sqrt(2*m)]    a1     an   n               m                
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>
#include <cstdio>
#include <math.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <iomanip>
#include <stdlib.h>
using namespace std;
int main(){
    int n,m,i,a1,an;
    while(cin>>n>>m){
    	if(n==0&&m==0)
    	break;
    	for(i=sqrt(2*m);i>=1;i--){
    		a1=(2*m-i*i+i)/(2*i);
    		an=a1+(i-1);
    		if(an<=n&&(a1+an)*i==2*m)
    		  cout<<"["<<a1<<","<<an<<"]"<<endl;
		}
		cout<<endl;
	}
	return 0;
}