HDU2058 The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35412    Accepted Submission(s): 10612

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]
 
这题有个坑，开始我的i从0遍历到n-1,次次都TLE,后来把n-1换成sqrt(2*m)就AC了；
原因很简单：
假设符合条件的子串为[a,a+b],有(b+2a)*(b+1)=2m,则b不需要从0遍历到n-1,因为b<sqrt(2*m).
代码如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
int main(){
    int n, m;
    int a, b,c;
    int num;
    int beg[1000], dis[1000];
    char ch1 = '[';
    char ch2 = ']';
    char ch3 = ',';
    while ((cin >> n >> m), n, m){
        num = 0;
        memset(beg, 0, sizeof(beg));
        memset(dis, 0, sizeof(dis));
        
            for (int i = 0; i < sqrt(2*m); i++){
                b = i;
                if ((2 * m) % (b + 1) == 0)  c = (2 * m) / (b + 1);
                else continue;
                if ((c - b) % 2 == 0) {
                    a = (c - b) / 2;
                    if (a <= 0) continue;
                }
                else continue;
                num++;
                beg[num] = a;
                dis[num] = b;
            }
            for (int j = num; j >= 1; j--)
                cout << ch1 << beg[j] << ch3 << dis[j] + beg[j] << ch2 << endl;
            cout << endl;
        
    }
    return 0;
}