hdu 2058 The sum problem

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9078    Accepted Submission(s): 2776

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]

 

 

Author

8600

 

 

Source

校庆杯Warm Up

 

 

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linle

//枚举序列长度

//等差数列呀，开始用队列超时了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 20000
using namespace std;
int main()
{
    int n,m;
    int len,a;
    while(scanf("%d%d",&n,&m),n||m)
    {
        len=sqrt(double(2*m))+1;
        while(--len)
        {
           a=m/len-(len-1)/2;
           if((((a<<1)+len-1)*len>>1)==m)
               a<=n?printf("[%d,%d]\n",a,a+len-1):0;
        }
      printf("\n");
    }
    return 0;
}

//下面错误的代码也过了、、呵呵

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 20000
using namespace std;
int main()
{
    int n,m;
    int len,a;
    while(scanf("%d%d",&n,&m),n||m)
    {
        len=sqrt(double(2*m))+1;
        while(--len)
        {
           a=m/len-(len-1)/2;
           if((((a<<1)+len-1)*len>>1)==m)
               printf("[%d,%d]\n",a,a+len-1);//这里没判断a要不大于n噢
        }
      printf("\n");
    }
    return 0;
}