求序列的和，杭电0j-2058-优快云博客

本文介绍了如何解决HDU 2058问题，即在一个给定序列中找到所有和为特定数值M的子序列。通过使用数学公式和循环迭代，实现高效求解。

原题地址：http://acm.hdu.edu.cn/showproblem.php?pid=2058

【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]

【AC代码】

 1 #include <iostream>
 2 #include <math.h>
 3 using namespace std;
 4 int main()
 5 {
 6     __int64 m, n;
 7     while(cin >> n >> m, n||m)
 8     {
 9         __int64 a, k;
10         for(k=(int)sqrt(2*m); k>0; k--)
11         {
12             a = m/k-(k-1)/2;
13             if((2*a-1+k)*k == 2*m)
14                 cout << "[" << a << "," << k+a-1 << "]" << endl;
15         }
16     cout << endl;
17     }
18     return 0;
19 }