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Chrysanthemum chain (winter vacation daily question 39)

2022-06-25 01:50:00 【51CTO】

Every day , As part of her walk around the farm , cow Bessie Will pass her favorite grass , One of them is Chrysanthemum chain （ Winter vacation every day 39）_hash A flower （ Colorful daisies ）, The number is , Line up .

flowers Chrysanthemum chain （ Winter vacation every day 39）_#include_03 Yes A flower petal .

As a budding photographer ,Bessie Decided to take some pictures of these flowers .

To be specific , For each pair, meet Chrysanthemum chain （ Winter vacation every day 39）_i++_05 The flowers of ,Bessie Will give you flowers To flower All the flowers between （ Include and ） Take a picture .

later Bessie When you look at these photos, you notice that there are... In some photos 「 Average 」 The flowers of —— One happens to have Chrysanthemum chain （ Winter vacation every day 39）_i++_11 A petaled flower , among Equal to the average number of petals of all flowers in the picture .

Bessie There are several photos with average flowers ？

Input format
The first line of input contains Chrysanthemum chain （ Winter vacation every day 39）_hash .

The second line contains Chrysanthemum chain （ Winter vacation every day 39）_hash Space separated integers .

Output format
Output the number of photos with average flowers .

Data range
Chrysanthemum chain （ Winter vacation every day 39）_#include_16

sample input ：

4
1 1 2 3

sample output ：

6

Sample explanation
Each photo containing only one flower will be included in the answer （ In this example, there are Chrysanthemum chain （ Winter vacation every day 39）_i++_17 Zhang ）.

in addition , In this example Chrysanthemum chain （ Winter vacation every day 39）_hash_06 by and The corresponding photos also have average flowers .

       
       #include<iostream>
       
       #include<unordered_set>
       
       using 
       
       namespace 
       
       std;
       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       const 
       
       int 
       
       N 
       
       = 
       
       110;
       
       int 
       
       n;
       
       int 
       
       q[
       
       N], 
       
       s[
       
       N];
       
       //  Seeking for  i  There are several sections that meet the requirements of the topic 
       
       LL 
       
       solve(
       
       int 
       
       u){
       
       LL 
       
       res 
       
       = 
       
       0;
       
       unordered_set
       
       <
       
       int
       
       > 
       
       hash;
       
       for(
       
       int 
       
       i 
       
       = 
       
       u; 
       
       i 
       
       <= 
       
       n; 
       
       i
       
       ++){
       
       hash.
       
       insert(
       
       q[
       
       i]);
       
       if((
       
       s[
       
       i] 
       
       - 
       
       s[
       
       u 
       
       - 
       
       1]) 
       
       % (
       
       i 
       
       - 
       
       u 
       
       + 
       
       1)) 
       
       continue;
       
       int 
       
       avg 
       
       = (
       
       s[
       
       i] 
       
       - 
       
       s[
       
       u 
       
       - 
       
       1]) 
       
       / (
       
       i 
       
       - 
       
       u 
       
       + 
       
       1);
       
       if(
       
       hash.
       
       count(
       
       avg)) 
       
       res
       
       ++;
       
    }
       
       return 
       
       res;
       
}
       
       int 
       
       main(){
       
       scanf(
       
       "%d", 
       
       &
       
       n);
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       n; 
       
       i
       
       ++){
       
       scanf(
       
       "%d", 
       
       &
       
       q[
       
       i]);
       
       s[
       
       i] 
       
       = 
       
       s[
       
       i 
       
       - 
       
       1] 
       
       + 
       
       q[
       
       i];
       
    }
       
       LL 
       
       res 
       
       = 
       
       0;
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       n; 
       
       i
       
       ++) 
       
       res 
       
       += 
       
       solve(
       
       i);  
       
       printf(
       
       "%d\n", 
       
       res);
       
       return 
       
       0;
       
}
      
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