G - Good subsequence

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本文介绍了一个算法问题，即在给定序列中找到最大长度的好子序列，其中好子序列定义为连续子序列且其最大值与最小值之差不大于给定阈值k。文章通过示例解释了问题，并提供了一段C++代码实现解决方案。

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G - Good subsequence

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%lld & %llu

Submit Status Practice CSU 1553

Description

Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a continuous subsequence of the given sequence and its maximum value - minimum value<=k. For example n=5, k=2, the sequence ={5, 4, 2, 3, 1}. The answer is 3, the good subsequence are {4, 2, 3} or {2, 3, 1}.

Input

There are several test cases.
Each test case contains two line. the first line are two numbers indicates n and k (1<=n<=10,000, 1<=k<=1,000,000,000). The second line give the sequence of n numbers a[i] (1<=i<=n, 1<=a[i]<=1,000,000,000).
The input will finish with the end of file.

Output

For each the case, output one integer indicates the answer.

Sample Input

Sample Output

3
1

//HDU1171：Big Event in HDU

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;

int a[10005];

int main()
{
int n,k,i,j;
while(scanf("%d%d",&n,&k)!=EOF)
{
for (i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}

/// sort(a+1,a+1+n); 不能手残排序
int max;
int min;
int count;
int ans=0;
for (i=1;i<=n;i++)
{
count=1;
max=min=a[i];
for (j=i+1;j<=n;j++)
{
if (a[j]>max)
max=a[j];
if (a[j]<min)
min=a[j];

if (max-min>k) break; //很重要的一个剪枝
else
count++;
}
if (ans<count)
ans=count;

}

printf("%d\n",ans);

}
return 0;

}