1034. 有理数四则运算(20) PAT乙级真题

1034. 有理数四则运算(20)

本题要求编写程序，计算2个有理数的和、差、积、商。

输入格式：

输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为0。

输出格式：

分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”，其中k是整数部分，a/b是最简分数部分；若为负数，则须加括号；若除法分母为0，则输出“Inf”。题目保证正确的输出中没有超过整型范围的整数。

输入样例1：

2/3 -4/2

输出样例1：

2/3 + (-2) = (-1 1/3)2/3 - (-2) = 2 2/32/3 * (-2) = (-1 1/3)2/3 / (-2) = (-1/3)

输入样例2：

5/3 0/6

输出样例2：

1 2/3 + 0 = 1 2/31 2/3 - 0 = 1 2/31 2/3 * 0 = 01 2/3 / 0 = Inf

这道题真的是逼着新手去练习函数了。当然，不觉得主函数函数体太长看着难受还可以继续反抗。把加、减、乘、除四个模块化为函数能使程序可读性大大加强。我是分是否为0写的函数，还有几个负责求最大公因数和输出的函数，这次的代码就有点长了：

#include<stdio.h>#include<math.h>void case1();   //a1=0,a2=0void case2(long a2,long b2);   //a1=0,a2!=0void case3(long a1,long b1);   //a1!=0,a2=0void case4(long a1,long b1,long a2,long b2);   //a1!=0,a2!=0void pri(long a,long b,long c);int zuida(long a,long b);int main(void){    long a1,a2,b1,b2;    scanf("%ld%*c%ld %ld%*c%ld",&a1,&b1,&a2,&b2);    if (a1==0 && a2 ==0) case1();    else if (a1==0 && a2!=0) case2(a2,b2);    else if (a1!=0 && a2==0) case3(a1,b1);    else case4(a1,b1,a2,b2);    return 0;}void case1(){    printf("0 + 0 = 0\n");    printf("0 - 0 = 0\n");    printf("0 * 0 = 0\n");    printf("0 / 0 = Inf");}void case2(long a2,long b2)       //a1=0,a2!=0{    printf("0 + ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));    printf("\n0 - ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(-a2,b2,zuida(fabs(a2),fabs(b2)));    printf("\n0 * ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = 0");    printf("\n0 / ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = 0");}void case3(long a1,long b1)       //a1!=0,a2=0{    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" + 0 = ");pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" - 0 = ");pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" * 0 = 0");printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" / 0 = Inf");}void case4(long a1,long b1,long a2,long b2)     //a1!=0,a2!=0{    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" + ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(a1*b2+a2*b1,b1*b2,zuida(fabs(a1*b2+a2*b1),fabs(b1*b2)));printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" - ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(a1*b2-a2*b1,b1*b2,zuida(fabs(a1*b2-a2*b1),fabs(b1*b2)));printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" * ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(a1*a2,b1*b2,zuida(fabs(a1*a2),fabs(b1*b2)));printf("\n");    pri(a1,b1,zuida(fabs(a1),fabs(b1)));printf(" / ");pri(a2,b2,zuida(fabs(a2),fabs(b2)));printf(" = ");pri(a1*b2,b1*a2,zuida(fabs(a1*b2),fabs(b1*a2)));}void pri(long a,long b,long c){    a=a/c;b=b/c;    if (b<0)    {        a=-a;b=-b;    }    if (a%b==0)    {        if (a<0)        {            printf("(%d)",a/b);        }        else        {            printf("%d",a/b);        }    }    else    {        if (a<0)        {            if (-a>b)            {                printf("(%d %d/%d)",a/b,-a%b,b);            }            else            {                printf("(%d/%d)",a,b);            }        }        else        {            if (a>b)            {                printf("%d %d/%d",a/b,a%b,b);            }            else            {                printf("%d/%d",a,b);            }        }    }}int zuida(long a,long b){    long c;    if (a<b) {c=a;a=b;b=c;}    while(b!=0)    {        c=a%b;        a=b;        b=c;    }    return a;}