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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]\n",x+1,y); } } cout<<endl; }}
hdu 2058 The sum problem（简单因式分解，，）
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