求和公式的应用

来源：互联网发布：预算报价软件编辑：程序博客网时间：2024/05/16 01:20

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21414 Accepted Submission(s): 6285

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]

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

利用多项式求和公式：（a+b)*i/2=m

0 0