HDOJ-2058

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21744    Accepted Submission(s): 6391

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]

 

本题题意：

输入两个数n，m，n代表数列为从1~n的等比为1的等比数列，求其中连续哪几个数的和为m，并输出。

运用等差数列的求和公式就能很好的解决这个问题。下面重申一下等差数列的求和公式：

          

附AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>//包含开根函数sqrt() 

using namespace std;

int main(){
    int n,m;
    while(~scanf("%d %d",&n,&m)&&m||n){
        for(int i=sqrt(2*m);i>=1;i--){//1连加到sqrt(2*m) > m 
            int a=(m-(i*(i-1))/2)/i;//等差求和公式推出 
            if(m==a*i+(i*(i-1))/2)
            printf("[%d,%d]
",a,a+i-1);
        }
        printf("
");//注意每组数据空一行 
    }
    return 0;
}