题目:
Wavio is a sequence of integers. It has some interesting properties.
Wavio is of odd length i.e. L = 2 n + 1.
The rst (n + 1) integers of Wavio sequence makes a strictly increasing sequence.
The last (n + 1) integers of Wavio sequence makes a strictly decreasing sequence.
No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will
be given a sequence of integers. You have to nd out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider,
the given sequence as :
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.
Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be `9'.
Input
The input le contains less than 75 test cases. The description of each test case is given below. Input is terminated by end of le.
Each set starts with a postive integer, N (1 N 10000). In next few lines there will be N integers.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input
10
1 2 3 4 5 4 3 2 1 10
19
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1
5
1 2 3 4 5
Sample Output
9
9
1
题目大意:
给一个字符串,求左边连续增等于右边连续减的的最大长度
题目思路:
1、方法一:从i点向两边扩展.时间复杂度为n*n
2、方法二:分别从左边向右搜,从右向左搜.当前最大连续.再从其中中选出左右最小中最大的值.
程序:
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cctype>
#include <fstream>
#include <limits>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cassert>
using namespace std;
#define maxn 11000
const int finf=-1<<31;
int a[maxn],b[maxn],up[maxn],down[maxn];
int main()
{
int n,len;
while(~scanf("%d",&n))
{
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
len=1,b[1]=a[1],up[1]=1;
for(int i=2; i<=n; i++)//正向标记
{
if(a[i]>b[len])
{
len++;
b[len]=a[i];
up[i]=len;
}
else
{
for(int j=1; j<=len; j++)
if(a[i]<=b[j])
{
b[j]=a[i];
up[i]=j;
break;
}
}
}
len=1,b[1]=a[n],down[n]=1;
for(int i=n-1; i>0; i--)//反向标记
{
if(a[i]>b[len])
{
len++;
b[len]=a[i];
down[i]=len;
}
else
{
for(int j=1; j<=len; j++)
if(a[i]<=b[j])
{
b[j]=a[i];
down[i]=j;
break;
}
}
}
int ans=-1;
for(int i=1; i<=n; i++)//求两个标记的最大值
{
ans=max(ans,min(up[i],down[i])*2-1);
//cout<<up[i]<<down[i]<<endl;
}
printf("%d\n",ans);
}
return 0;
}