★ experimental task

In the final exam of the first grade of primary school, the teacher gave a question: give a row of numbers, select several consecutive numbers, and calculate their sum. The sum of these numbers is required to meet the conditions of a and b. Quickly count out the number of subsequences of students, you can get full marks! 
After Xiaoming went home, he asked his cousin, mcginn, who was studying as a graduate student. mcginn wrote a piece of code and solved the problem quickly and efficiently. Since then, Xiaoming is determined to become a code farmer! 

★ data input

Enter a n n for the number of numbers, and a and B (|a|, |b| < = 100000) enter an array of nums [], (|nums [i] | < = 100000) for the number

★ data output

There are several kinds of continuous subsequences that output a number to indicate that the condition is satisfied 

Note: statistical results do not include empty sequences 

★ data range

80% of points, n < = 1000, the remaining 20%, n < = 100000 

Code 1:

#include <iostream>  
#include <stdio.h>  
#include <string.h>  
#include <algorithm>  
#include <math.h>  
#include <stdlib.h>  
#include <vector>  
#include <queue>  
  
using namespace std;  
  
class Search {  
public:  
    int mergeSort(vector<long>& sum, int left, int right, int lo, int hi)  
    {  
        if(hi-lo <= 1) return 0;  
        int mid = (lo+hi)/2, m = mid, n = mid, count =0;  
        count = mergeSort(sum,left,right,lo,mid) + mergeSort(sum,left,right,mid,hi);  
        for(int i =lo; i< mid; i++)  
        {  
            while(m < hi && sum[m] - sum[i] < left) m++;  
            while(n < hi && sum[n] - sum[i] <= right) n++;  
            count += n - m;  
        }  
        inplace_merge(sum.begin()+lo, sum.begin()+mid, sum.begin()+hi);  
        return count;  
    }  
   
    int countRangeSum(vector<int>& nums, int left, int right) {  
        int len = nums.size();  
        vector<long> sum(len + 1, 0);  
        for(int i =0; i< len; i++) sum[i+1] = sum[i]+nums[i];  
        return mergeSort(sum, left, right, 0, len+1);  
    }  
}s;  
  
  
  
int main(){  
    int n,a,b,i;  
    scanf("%d %d %d",&n,&a,&b);  
    vector<int> nums(n,0);  
    for(i=0;i<n;i++){  
        scanf("%d",&nums[i]);  
    }   
    int count=s.countRangeSum(nums,a,b);  
    printf("%d",count);  
    return 0;  
}