pat 乙级 1027 打印沙漏

ac 但写得就像坨答辩 过两天我自己都忘了这些变量用来干嘛的了

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
    int num[1000];
    int i = 0;
    for (int i = 0, j = 1, k = 1; i < 1000; i++)
    {
        num[i] = j;

        j = j + (k * 2 + 1) * 2;
        k++;
    } // 此数组存储着能组成沙漏的符号个数
    int p;
    char ch;
    scanf("%d %c", &p, &ch);
    i = 0;
    if(p<7){
        printf("%c\n",ch);
        printf("%d",p-1);
        return 0;
    }
    while (p >= num[i])
    {
        i++;
    } // 按num[i-1]得到沙漏符号数
    int q = i - 1;
    int m = q * 2 + 1;
    int r = 0;
    int rs = 0;
    for (int t = 0; t < (num[i - 1] - 1) / 2; t++)
    {
        printf("%c", ch);

        if (t + 1 == m)
        {
            printf("\n");
            r++;
            rs = r;
            if (r <= i - 2)
            {
                for (; rs > 0; rs--)
                {
                    printf(" ");
                }
            }

            q--;
            m = m + q * 2 + 1;
        }
    }
    q = i - 2;
    for (int t = 0; t < i - 1; t++)
    {
        printf(" ");
    }
    printf("%c\n", ch);
    m = 3;
    int ms = 3;
    r = 0;
    rs = 0;
    int j=i-1;
    for (int t = 0; t < (num[i - 1] - 1) / 2; t++)
    {
        if (rs == r)
        {

            for (int l = j - 1; l > 0; l--)
            {
                printf(" ");
            }
            j--;
            rs++;
        }

        printf("%c", ch);
        if (t + 1 == m && m != (num[i - 1] - 1) / 2)
        {
            rs = r;
            printf("\n");
            m = m + ms * 2 - 1;
            ms++;
        }
    }
    printf("\n%d", p - num[i - 1]);
    return 0;
}