HDU 2058 The sum problem——————数学

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31416 Accepted Submission(s): 9409

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]

这道题的答案可以根据等差数列的性质得到：
最大的区间长度：
由 $1+2+3+⋯+2m>m\large \sqrt{1+2+3+\cdots+2m}> m$
$lmax=2×m\large l_{max} =\sqrt{2\times m}$

由等差数列
$Sn=(a1+an)×n2\Large S_n = \frac{(a_1 + a_n)\times n}{2}$
$Sn=(a1+a1+d∗(n−1))×n2\Large S_n = \frac{(a_1 + a_1+d*(n-1))\times n}{2}$
$Sn=(2∗a1+d∗(n−1))×n2\Large S_n = \frac{(2*a_1+d*(n-1))\times n}{2}$

又因为d=1，故：
$Sn=(2∗a1+n−1)×n2\Large S_n = \frac{(2*a_1+n-1)\times n}{2}$
$a1=Snn−n−12\Large a_1 = \frac{S_n}{n}-\frac{n-1}{2}$
那么
$an=a1+n−1\Large a_n = a_1 + n-1$

对于这一道题：m就相当于Sn,
假设 $ m=(ai+aj)×(j−i+1)2\;\Large m = \frac{(a_i+a_j)\times (j-i+1)}{2}$
我们可以枚举 $a_i$
因为我们知道最大的区间长度为 $lmax=2×m\large l_{max} =\sqrt{2\times m}$
所以枚举 $1 到 l$ 的所有值
那么根据上边的可以得到
$a1=ml−l−12\Large a_1 = \frac{m}{l}-\frac{l-1}{2}$
$an=a1+n−1\Large a_n = a_1 + n-1$
只要 $(a1+a2)∗l2==m(a_1+a_2)*\frac{l}{2} == m$
就好了

代码很短，就这几行：

#include<bits/stdc++.h>
using namespace std;
int main()
{
        int n,m;
        while(~scanf("%d %d",&n,&m)&&(n|m))
        {
                int l=sqrt(2*m);//最大区间长度为 L
                int a1,an;
                for(int k=l;k>0;--k)//区间长度为k
                {
                        a1 = m/k-(k-1)/2;
                        an = a1 + k-1;
                        if((a1 + an )*k/2 == m)
                                printf("[%d,%d]\n",a1,an);
                }
                printf("\n");
        }
        return 0;
}