hdoj 2058 The sum problem 【等差数列求和】

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20890 Accepted Submission(s): 6151

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]

ACcode:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#define cle(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define INF 0x3f3f3f3f
#include<algorithm>
using namespace std;
int main()
{
int n, m;
while(scanf("%d%d", &n,&m), n|m)
{
int i, j, k;
for(k = sqrt(2*m); k > 0; --k)
{
i = (2*m/k-k+1)/2;
j = (i+k-1);
if((i+j)*k == 2*m)
printf("[%d,%d]\n",i, j);
}
printf("\n");
}
return 0;
}