hdu 2058 The sum problem（简单因式分解，，）

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.

思路：

假设答案为[i,j]，则有：j²+j-i²-i=2M   -->   (j+i+1)(j-i)=2M

对2M进行分解，，然后判断，，搞一搞，，

代码：

int n,m;

int main(){
    while(cin>>n>>m,n||m){
        m*=2;
        int t=(int)sqrt(m+0.5);
        rep2(i,t,1){
            if(m%i==0){
                int a,b;
                a=m/i, b=i;
                int x,y;
                x=(a-b-1)/2;
                y=(a+b-1)/2;
                if(x<0 || y>n || (2*x+1)!=(a-b) || (2*y+1)!=(a+b)) continue;
                printf("[%d,%d]
",x+1,y);
            }
        }
        cout<<endl;
    }
}