hdu2058（数学题，有点坑）

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24916 Accepted Submission(s): 7449

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 1050 300 0

Sample Output

[1,4][10,10][4,8][6,9][9,11][30,30]

Author

8600

Source

校庆杯Warm Up

等差数列求和公式，但是一直超时。。。因为没有剪枝，其实sqrt（m*2)就够了，我一直是从m开始虽然没错但就是超时，数学题还是要多练啊！

#include <iostream> #include <cmath>#include <iomanip>#include <string>using namespace std;int main(void){int n,m;while(cin>>n>>m && n||m){for(int i=sqrt(2*m);i>=1;i--){int a=(m-((i-1)*i)/2)/i;if(m==a*i+(i*(i-1))/2&&a>0)cout<<'['<<a<<','<<a+i-1<<']'<<endl;}cout<<endl;}return 0;}

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