PAT乙级.1034. 有理数四则运算(20)

1034. 有理数四则运算(20)

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题目

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 2/3
2/3 * (-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 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

PAT链接

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思路

分数数据结构存储

struct Fraction{    LL up, down;}a, b;

运算

Fraction reduction(Fraction result)Fraction add(Fraction f1, Fraction f2)Fraction sub(Fraction f1, Fraction f2)Fraction mul(Fraction f1, Fraction f2)Fraction div(Fraction f1, Fraction f2)void showResult(Fraction r)

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代码

/*** @tag     PAT_B_1034* @authors R11happy ([e-mail])* @date    2016-9-8 19:29-22:04* @version 1.0* @Language C++* @Ranking  890/371* @function null*/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>    //头文件不能用<cmath>,用<algorithm>才能过using namespace std;typedef long long LL;struct Fraction{    LL up, down;}a, b;LL gcd(LL a, LL b){    if (b == 0)  return a;    else return gcd(b, a%b);}//分数化简Fraction reduction(Fraction result){    if (result.down < 0)    {        result.up = -result.up;        result.down = -result.down;    }    if (result.up == 0)    {        result.down = 1;    }    else    {        LL d = gcd(abs(result.up), abs(result.down));   //要计算绝对值的公约数        result.up /= d;        result.down /= d;    }    return  result;}Fraction add(Fraction f1, Fraction f2){    Fraction result;    result.up = f1.up*f2.down + f2.up*f1.down;    result.down = f1.down * f2.down;    return reduction(result);}Fraction sub(Fraction f1, Fraction f2){    Fraction result;    result.up = f1.up*f2.down - f2.up*f1.down;    result.down = f1.down * f2.down;    return reduction(result);}Fraction mul(Fraction f1, Fraction f2){    Fraction result;    result.up = f1.up * f2.up;    result.down = f1.down * f2.down;    return reduction(result);}Fraction div(Fraction f1, Fraction f2){    Fraction result;    result.up = f1.up * f2.down;    result.down = f1.down * f2.up;    return reduction(result);}void showResult(Fraction r){    r = reduction(r);    if (r.up < 0)   printf("(");    if (r.down == 1) printf("%lld", r.up);    else if (abs(r.up) > r.down)   printf("%lld %lld/%lld", r.up / r.down, abs(r.up) % r.down, r.down);    //是else if 不是 if 而且abs不能少    else        printf("%lld/%lld", r.up, r.down);    if (r.up < 0)    printf(")");}int main(int argc, char const *argv[]){    scanf("%lld/%lld %lld/%lld", &a.up, &a.down, &b.up, &b.down);    // 加    showResult(a);    printf(" + ");    showResult(b);    printf(" = ");    showResult(add(a, b));    printf("\n");    // 减    showResult(a);    printf(" - ");    showResult(b);    printf(" = ");    showResult(sub(a, b));    printf("\n");    //乘    showResult(a);    printf(" * ");    showResult(b);    printf(" = ");    showResult(mul(a, b));    printf("\n");    // 除    showResult(a);    printf(" / ");    showResult(b);    printf(" = ");    if(b.up == 0)   printf("Inf");    else showResult(div(a, b));    printf("\n");    return 0;}

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