血型遗传关系c语言编程,血型遗传

最新推荐文章于 2021-05-24 05:18:54 发布

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最新推荐文章于 2021-05-24 05:18:54 发布

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文章标签：血型遗传关系c语言编程

这是一个C语言实现的血型遗传计算程序，用户可以输入父亲、母亲或子女的血型，程序将根据血型遗传规则推断可能的血型组合。程序提供了四个功能：输入一方血型推断子女可能血型、输入子女血型推断父母可能血型、同时输入父母血型推断子女可能血型以及验证父母和子女的血缘关系。

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血型遗传

能不能修改一下这个源代码？血型遗传的。#include

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;

}