hdu2058java

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21555    Accepted Submission(s): 6320

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]

/*import java.util.*;
class Main{
public static void main(String args[])
{Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{int n=cin.nextInt();
int m=cin.nextInt();
if(n==0&&m==0)
break;
for(int i=1;i<=n;i++)
{int sum=i;
int k=0,s=0,flag=0;
for(int j=i+1;j<=n;j++)
{ sum+=j;
if(sum==m)
{flag=1;
k=i;
s=j;
break;}
else if(sum>m)
{break;
}
}
if(flag==1)
System.out.println("["+k+","+s+"]");
}
if(n>=m)
System.out.println("["+m+","+m+"]");
System.out.println();
}
}
}*/

上面的那个代码会超时，下面的代码不会超时，这是一个数学题，a+(i*(i+1)/2)=m,当a最小的时候是等于1，所以1+(i*(i+1)/2)<=m;

所以i<=sqrt(2*m);因为这是一个等差数列，d为1；
import java.util.*;
class Main{
public static void main(String args[])
{Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{long n=cin.nextInt();
long m=cin.nextInt();
if(n==0&&m==0)
break;
long d=0;
for(int i=(int)Math.sqrt(2*m);i>0;i--)
{
d=m-(i+i*i)/2;
if(d%i==0)
System.out.println("["+(d/i+1)+","+(d/i+i)+"]");
}

System.out.println();
}
}
}