能不能修改一下这个源代码？血型遗传的。#include<stdio.h>
int receive()   
{
    int in;
    int count = 4;
    while(count>0)
    {
        scanf("%d",&in);
        if(in>0 && in<5) break;     
        else     
        {
            count--;
            printf("\t\t输入错误！！！您还有%d次输入机会！\n\t\t请输入1、2、3或4：",count);
        }
    }
    return in;
}
void func1(int bloodType[][6])   
{
    int in, i, j;
    int subTable[4] = {0,0,0,0};
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t+ 父亲或母亲血型如下：\n");
    printf("\t\t+ 1、A型血\n");
    printf("\t\t+ 2、B型血\n");
    printf("\t\t+ 3、AB型血\n");
    printf("\t\t+ 4、O型血\n");
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t输入：");
    in = receive();
    for(i=0; i<16; i++)
    {
        if(bloodType[i][0] == in)
        {
            for(j=2; j<6; j++)
            {
                if(bloodType[i][j] == 1)
                    subTable[j-2] = 1;
            }
        }
    }
    printf("\t\t子女可能出现的血型为：");
    if(subTable[0] == 1) printf("\tA型 ");
    if(subTable[1] == 1) printf("\tB型 ");
    if(subTable[2] == 1) printf("\tAB型 ");
    if(subTable[3] == 1) printf("\tO型 ");
    printf("\n");
}
void func2(int bloodType[][6])
{
    int in, i, j;
    char subTable[3] = {'A','B','O'};

    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t+ 子女血型如下：\n");
    printf("\t\t+ 1、A型血\n");
    printf("\t\t+ 2、B型血\n");
    printf("\t\t+ 3、AB型血\n");
    printf("\t\t+ 4、O型血\n");
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t输入：");
    in = receive();
   
    printf("\t\t父母可能的血型为：\n");
    for(i=0; i<10; i++)
    {
        if(bloodType[i][in+1] == 1)
        {
            if(bloodType[i][0] == 1) printf("\t\t\tA型 ");
            else if(bloodType[i][0] == 2) printf("\t\t\tB型 ");
            else if(bloodType[i][0] == 3) printf("\t\t\tAB型 ");
            else printf("\t\t\tO型");
            if(bloodType[i][1] == 1) printf("+ A型 \n");
            else if(bloodType[i][1] == 2) printf("+ B型 \n");
            else if(bloodType[i][1] == 3) printf("+ AB型 \n");
            else printf("+ O型 \n");
        }
    }
    printf("\n");
}
void func3(int bloodType[][6])
{
    int in1, in2;
    int i, j;

    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t+ 血型如下：\n");
    printf("\t\t+ 1、A型血\n");
    printf("\t\t+ 2、B型血\n");
    printf("\t\t+ 3、AB型血\n");
    printf("\t\t+ 4、O型血\n");
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t请输入父亲血型的选项：");
    in1 = receive();
    printf("\t\t请输入母亲血型的选项：");
    in2 = receive();

    for(i=0; i<16; i++)
    {
        if(bloodType[i][0] == in1 && bloodType[i][1] == in2)
        {
            printf("\t\t子女可能的血型为：");
            for(j=2; j<6; j++)
            {
                if(bloodType[i][j] == 1)
                {
                    if(j == 2) printf("A型 ");
                    if(j == 3) printf("B型 ");
                    if(j == 4) printf("AB型 ");
                    if(j == 5)  printf("O型 ");
                }
            }
            printf("\n");
            break;
        }
    }
}
void func4(int bloodType[][6])
{
    int in1, in2, in3;
    int i, j, flag = 0;
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t+ 血型选择如下：\n");
    printf("\t\t+ 1、A型血\n");
    printf("\t\t+ 2、B型血\n");
    printf("\t\t+ 3、AB型血\n");
    printf("\t\t+ 4、O型血\n");
    printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
    printf("\t\t请输入父亲血型的选项：");
    in1 = receive();
    printf("\t\t请输入母亲血型的选项：");
    in2 = receive();
    printf("\t\t请输入子女血型的选项：");
    in3 = receive();
   
    for(i=0; i<16; i++)
    {
        if(bloodType[i][0] == in1 && bloodType[i][1] == in2)
        {
            if(bloodType[i][in3+1] == 1)
            {
                printf("\t\t父母和子女有血缘关系\n\n");
                flag = 1;
            }
            break;
        }
    }
    if(flag == 0) printf("\t\t父母和子女没有血缘关系\n\n");
}
void console(int bloodTable[][6])
{
    int in, count, isEnd = 0;
    while(isEnd != 2)
    {
        printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
        printf("\t\t+ 1、输入父亲或母亲一个人的血型\n");
        printf("\t\t+ 2、输入子女的血型\n");
        printf("\t\t+ 3、同时输入父亲和母亲的血型\n");
        printf("\t\t+ 4、输入父母和子女血型，确认血缘关系\n");
        printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
        printf("\t\t输入：");
        in = receive();
        switch(in)
        {
        case 1 : func1(bloodTable); break;   
        case 2 : func2(bloodTable); break;
        case 3 : func3(bloodTable); break;   
        case 4 : func4(bloodTable); break;   
        default : break;
        }
        printf("\t\t是否要继续查询？\n");
        printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
        printf("\t\t+ 1、继续\n");
        printf("\t\t+ 2、结束\n");
        printf("\t\t++++++++++++++++++++++++++++++++++++++\n");
        printf("\t\t输入：");
        count = 4;
        while(count>0)     
        {
            count--;
            scanf("%d",&isEnd);
            if(isEnd == 1 || isEnd == 2) break;
            else printf("\t\t输入错误！！！您还有%d次输入机会！\n\t\t请输入1或2：",count);
        }
    }
    printf("\t\t查询结束！！\n");
}

int main()
{
    int table[16][6] = {{1,1,1,0,0,1}, {1,2,1,1,1,1}, {1,3,1,1,1,0}, {1,4,1,0,0,1}, {2,2,0,1,0,1}, {2,3,1,1,1,0}, {2,4,0,1,0,1}, {3,3,1,1,1,0},
    {3,4,1,1,0,0},{4,4,1,1,1,0},{2,1,1,1,1,1}, {3,1,1,1,1,0}, {4,1,1,0,0,1}, {3,2,1,1,1,0}, {4,2,0,1,0,1}, {4,3,1,1,0,0}};
    console(table);
    return 0;
}

这源代码是对的，我就想能不能有其他的写法。
根据血型遗传关系，编程实现：○1.输入父亲或母亲一个人的血型时，输出子女可能的血型，○2.输入子女的血型时，输出其父母可能的血型，○3.同时输入父亲和母亲的血型时，输出子女可能的血型， ○4.同时输入父母和子女血型时能判断是否具有血型血缘关系。
○1.程序运行时有友好的操作提示界面；
○2.可进行多人次输入，将血型关系总表输入文件保持，在主菜单中能选择查看血型关系总表；
    ○3.有输入错误检查，并限制出错次数，如：应输入字母时输入其它内容，能提示错误并要求重新输入；
○4.程序中有详细的注释说明；

能不能改成指针类型啊

哪个大佬帮我写一下呀，呜呜呜