1553: Good subsequence
Summary Time Limit: 2 Sec Memory Limit: 256 Mb Submitted: 895 Solved: 335
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
5 2
5 4 2 3 1
1 1
1
Sample Output
3
1
Hint
Source
给你一个序列,长度为n
然后给一个数字k
问你符合要求的子段的最大长度是多少?
符合要求的子段:子段的最大值和最小值的差小于等于k分析:
利用set模拟该过程,或者直接暴力
不知道怎么描述这个过程
不过按照代码模拟一下样例1应该就明白了
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define max_v 10005
using namespace std;
int a[max_v];
set<int> s;
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
s.clear();
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
}
int j=;
int ans=;
for(int i=;i<=n;i++)
{
s.insert(a[i]);
if(*s.rbegin()-*s.begin()>k)
{
s.erase(s.find(a[j]));
j++;
}
ans=max(ans,i-j+);
}
printf("%d\n",ans);
}
return ;
}
/*
给你一个序列,长度为n
然后给一个数字k
问你符合要求的子段的最大长度是多少?
符合要求的子段:子段的最大值和最小值的差小于等于k分析:
利用set模拟该过程,或者直接暴力
不知道怎么描述这个过程
不过按照代码模拟一下样例1应该就明白了
*/