//罕见的C程序题解
#include<stdio.h>
int main()
{
    int n;
    int a[100];
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    double sum1 = 0,sum2 = 0,sum3=0,sum4=0;
    for (int i = 0; i < n; i++)
    {
        if (a[i] >= 0 && a[i] <= 18)
            sum1++;
        else if (a[i] >= 19 && a[i] <= 35)
            sum2++;
        else if (a[i] >= 36 && a[i] <= 60)
            sum3++;
        else if (a[i] >= 61 )
            sum4++;
    }
    printf("%.2f%%\n%.2f%%\n%.2f%%\n%.2f%%\n", 100*sum1 / n, 100*sum2 / n, 100*sum3 / n, 100*sum4 / n);

    return 0;
}

嗨嗨嗨

#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,m,a,s1=0,s2=0,s3=0,s4=0;
cin>>n;
m=n;
while(n--)
{
cin>>a;
if(a>=61)
s1++;
else if(a>=36)
s2++;
else if(a>=19)
s3++;
else
s4++;
}
printf("%.2lf%\n%.2lf%\n%.2lf%\n%.2lf%\n",100.0*s4/m,100.0*s3/m,100.0*s2/m,100.0*s1/m);

return 0;
}

很好，很有精神！