简单枚举题
先枚举左端点，再枚举右端点，不断更新最大值跟最小值即为递增排序后的左端点和右端点，
当递增排序后能得到一个长度为 R−L+1 的“连续”数列，则区间数+1
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e4 + 10;
int n,res;
int a[N];

int main(){
    scanf("%d",&n);
    for(int i = 0;i < n;++i){
        scanf("%d",&a[i]);
    }
    int INF = 1e8;
    for(int i = 0;i < n;++i){
        int maxn = -INF,minn = INF;
        for(int j = i;j < n;++j){
            minn = min(minn,a[j]);
            maxn = max(maxn,a[j]);
            if(maxn - minn == j - i)    res++;
        }
    }
    printf("%d\n",res);
    return 0;
}