C :

#include <stdio.h> 
int Count(int arr[], int n)  
{  
    int L = 0, R = 0, max = 0, min = 0, num = 0;  
      
    for (L = 0; L < n; ++L)    //先从[1-n],然后[2-n],[3-n]，如此搜索...   
    {   
        max = -1, min = 50001;            
        for (R = L; R < n; ++R)      
        {  
            if (max < arr[R])  //找出区间[L,R]之间的最大和最小值   
                max = arr[R];  
            if (min > arr[R])  
                min = arr[R];  
            if (max - min + 1 == R - L + 1)  //判断是否为连号区间数   
            {  
            //  printf("[%d,%d]\n",L,R);     //打印每种情况   
                ++num;  
            }  
        }  
    }   
    return num;   
}
int main()  
{  
    int arr[50000] = {0}, n = 0, i = 0;  
    scanf("%d",&n);  
    for (i = 0; i < n; ++i)  
        scanf("%d",&arr[i]);  
    printf("%d",Count(arr,n));   //打印连号区间数   
    return 0;  
}   

C++ :

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

const int N = 50005;

int a[N];

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