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
由等差数列
又因为d=1,故:
那么
对于这一道题:m就相当于Sn,
假设
我们可以枚举
因为我们知道最大的区间长度为
所以枚举
的所有值
那么根据上边的可以得到
只要
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;
}