Minimum ticket price (day 80)

1： subject

 In a country where train travel is very popular , You planned some train trips a year in advance .

 In the next year , The day you are going to travel will be called  days  The array of gives .

 Each item is from  1  To  365  The integer of .

 There are three different ways of selling train tickets ：

 A one-day pass costs  costs[0]  dollar ;
 A seven day pass costs  costs[1]  dollar ;
 A 30 day pass costs  costs[2]  dollar .
 The pass allows unlimited travel for days .

 for example , If we were in  2  Days to get a duration of  7  Day's pass , Then we can travel together  7  God ： The first  2  God 、 The first  3  God 、 The first  4  God 、 The first  5  God 、 The first  6  God 、 The first  7  Day and day  8  God .

 Return what you want to complete in the given list  days  The minimum cost of travel for each day listed in .

 Input format 
 The first line contains an integer  n, Express  days  Length of array .

 The second line contains  n  It's an integer , Express  days[i].

 The third line contains  3  It's an integer , Express  costs[i].

 Output format 
 An integer , Indicates the minimum consumption required for travel .

 Data range 
1≤n≤365,
1≤days[i]≤365,
days  Strictly increasing in order ,
1≤costs[i]≤1000.

 sample input ：
6
1 4 6 7 8 20
2 7 15
 sample output ：
11

2： Topic realization

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 400;

int n;
int days[N], f[N];
int c1, c7, c30;

int main()
{
    
    cin >> n;
    for (int i = 1; i <= n; i ++ ) cin >> days[i];
    cin >> c1 >> c7 >> c30;

    for (int i = 1, j7 = 0, j30 = 0; i <= n; i ++ )
    {
    
        while (days[i] - days[j7 + 1] >= 7) j7 ++ ;
        while (days[i] - days[j30 + 1] >= 30) j30 ++ ;
        f[i] = min({
    f[i - 1] + c1, f[j7] + c7, f[j30] + c30});
    }

    cout << f[n] << endl;

    return 0;
}