hdu 2058 The sum problem

最新推荐文章于 2025-08-08 22:01:10 发布

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最新推荐文章于 2025-08-08 22:01:10 发布

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文章标签： java

原文链接：http://www.cnblogs.com/372465774y/archive/2012/07/19/2599898.html

本文介绍了一道关于寻找序列中所有可能的子序列，使得这些子序列的和等于给定值M的问题。输入包含多个测试案例，每个案例由两个整数N和M组成，输出则是所有符合条件的子序列。文章提供了两种解决方案，一种使用了正确的边界条件判断，另一种虽然逻辑上有缺陷但仍然通过了测试。

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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;
}