PAT-B1012代码优化

题目：https://pintia.cn/problem-sets/994805260223102976/exam/problems/994805311146147840
经过千辛万苦终于拿满了，但是代码行数太多了233，望各位能给个优化的意见。

#include 
int main()
{
    int N,i,A[6]={0},flag=1,count=0,temp=0;
    scanf("%d",&N);
    int num[N];
    for(i=0;iscanf("%d",&num[i]);
    }
    for(i=0;iswitch(num[i] % 5)
        {
            case 0:
                {
                    if(num[i] % 2 == 0)
                        A[1] += num[i];
                    break;
                }
            case 1:
                {
                    A[2] += flag*num[i];
                    flag=-flag;
                    temp=1;
                    break;
                }
            case 2:
                {
                    A[3]++;
                    break;
                }
            case 3:
                {
                    A[4] += num[i];
                    count++;
                    break;     
                }
            case 4:
                {
                    if(num[i]>A[5])
                       A[5]=num[i];
                    break;
                }
        }
    }  
    for(i=1;i<6;i++)
    {
        if(i != 1)
            printf(" ");
        switch(i)
        {
            case 1:
            case 3:
            case 4:
            case 5:
                {
                    if(A[i] != 0)
                    {
                        if(i==4)
                        {
                            printf("%.1f",(double)A[4]/count);
                            break;
                        }
                        printf("%d",A[i]);
                    }
                    else
                        printf("N");
                    break;
                }
            case 2:
                {
                    if(temp == 1)
                        printf("%d",A[2]);
                    else
                        printf("N");
                    break;
                }
        }
    }
    return 0;
}

减少一次循环，其他没什么好优化的，试试：

#include <stdio.h>
#define N 1000
int main()
{
    int n, i, A[6] = { 0 }, num[N] = { 0 }, flag = 1, count = 0, temp = 0;
    scanf("%d", &n);
    for (i = 0; i < n; i++)
    {
        scanf("%d", &num[i]);
        switch (num[i] % 5) {
        case 0: {
            if (num[i] % 2 == 0)
                A[1] += num[i];
            break;
        }
        case 1: {
            A[2] += flag * num[i];
            flag = -flag;
            temp = 1;
            break;
        }
        case 2: {
            A[3]++;
            break;
        }
        case 3: {
            A[4] += num[i];
            count++;
            break;
        }
        case 4: {
            if (num[i] > A[5])
                A[5] = num[i];
            break;
        }
        }
    }
    for (i = 1; i < 6; i++)
    {
        if (i != 1)
            printf(" ");
        switch (i)
        {
        case 1:
        case 3:
        case 4:
        case 5:
        {
            if (A[i] != 0)
            {
                if (i == 4)
                {
                    printf("%.1f", (double)A[4] / count);
                    break;
                }
                printf("%d", A[i]);
            }
            else
                printf("N");
            break;
        }
        case 2:
        {
            if (temp == 1)
                printf("%d", A[2]);
            else
                printf("N");
            break;
        }
        }
    }
    return 0;
}