优化过程的一道题

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
int a[N],b[N];
double jidian(int x) { 
    if(x >= 90 && x <= 100) return 4.0;
    else if(x >= 85 && x <= 89) return 3.7;
    else if(x >= 82 && x <= 84) return 3.3;
    else if(x >= 78 && x <= 81) return 3.0;
    else if(x >= 75 && x <= 77) return 2.7;
    else if(x >= 72 && x <= 74) return 2.3;
    else if(x >= 68 && x <= 71) return 2.0;
    else if(x >= 64 && x <= 67) return 1.5;
    else if(x >= 60 && x <= 63) return 1.0;
    else return 0;
}
int main()
{
    int n;
    double sum=0.0,cnt=0;
    scanf("%d\n", &n);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &a[i]);
        cnt+=a[i];
    }
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &b[i]);
    }
    for (int i = 0; i < n; i ++ )
    {
        sum+=jidian(b[i])*a[i];
    }
    printf("%.2f",sum/cnt);
}

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 20;
int key[10] = {0, 60, 64, 68, 72, 75, 78, 82, 85, 90};
double value[10] = {0, 1.0, 1.5, 2.0, 2.3, 2.7, 3.0, 3.3, 3.7, 4.0};
int a[N], b[N];
double jidian(int x)
{
    for (int i = 9; i >= 0; i -- )
        if (x >= key[i])
            return value[i];
    return 0;
}
int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n; i ++ ) cin >> a[i];
    for (int i = 0; i < n; i ++ ) cin >> b[i];
    double sum = 0, cnt = 0;
    for (int i = 0; i < n; i ++ )
    {
        sum += jidian(b[i]) * a[i];
        cnt += a[i];
    }
    printf("%.2lf\n", sum / cnt);
    return 0;
}