HD2058The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12422 Accepted Submission(s): 3757

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]
 
这也是一道水题，不过一开始没看懂题意，后面读了几遍就懂了。
题目给两个数N,M，要求找出从1开始到N里的子序列的和是M，并全部输出，一开始可能会想到求和公式，首项加末项乘以项数除以二，公式里有两个变量，题目告知M可以是一个很大的数，如果利用公式直接穷举去遍历所有的数的话的话会溢出，所以要换个角度去想。
 
考虑子串的长度为j，首项为i的数列满足要求，则(i+i+j-1)*j/2=m,==>(2*i-1+j)*j=2m，其中2*i-1>0;
所以j*j<2*m;j<sqrt(2*m);这样确定子串的长度了；
利用j和m，求出i的表达式，接下来就很简单了：
 
#include<iostream>
#include<cmath>
using namespace std;
int main(){
    int n,m,i;
    while(cin>>n>>m && !(n==0 && m==0)){
       int j = sqrt((double)2*m);
        for(j;j>=1;j--){   //去掉douubl会在提交时显示编译错误，要声明函数重载时的类型
             i =(2*m/j+1-j)/2;
            if((2*i-1+j)*j/2==m)
                cout<<"["<<i<<","<<i+j-1<<"]"<<endl;
        }
        cout<<endl;
    }
    return 0;
}