Increasing Pair

Increasing Pair

Time limit: 1000 ms
Memory limit: 128 MB

You are given an array A of length N representing a permutation. Find two indices ii and jj such that:

1. i<j 
2. A_i < A_j
3. j - i is maximized

Standard input

The first line contains a single integer N.

The second line contains N integers representing the permutation.

Standard output

If there is no solution print -1−1.

Otherwise, output a single integer representing the maximum difference j-i between a valid pair of indices.

Constraints and notes

 

• 2<=N<=10^5

 

Simple Input:

4
1 3 4 2

Simple Ouput:

3

题意：给你一个有N个元素的数组，元素是1-N的一个排列，问是否存在一组i,j(i<j)使得a[i]<a[j]并且j-i的值最大

做法：首先用一个pos数组记录下每个元素的位置，然后从N-1扫描，用mx记录下当前位置最大的元素的位置(因为后面的元素都不会比现在的元素大)，遇到更大的位置，就更新mx的值，否则就更新maximum

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxN=100005;
int pos[maxN];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int a;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a);
            pos[a]=i;
        }
        int mx=0,ans=0;
        for(int i=n;i>0;i--)
        {
            if(pos[i]>mx)mx=pos[i];
            else ans=max(ans,mx-pos[i]);
        }
        if(ans==0)printf("-1\n");
        else printf("%d\n",ans);
    }
    return 0;
}