PAT (Basic Level) Practise （中文）1034. 有理数四则运算(20)

1034. 有理数四则运算(20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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 = 0
1 2/3 / 0 = Inf
#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;struct Fraction{    ll up,down;}a,b;ll gcd(ll a,ll b){    return !b?a:gcd(b,a%b);}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));        result.up/=d;        result.down/=d;    }    return result;}Fraction add(Fraction f1,Fraction f2){    Fraction result;    result.up=f1.up*f2.down+f1.down*f2.up;    result.down=f1.down*f2.down;    return reduction(result);}Fraction sub(Fraction f1,Fraction f2){    Fraction result;    result.up=f1.up*f2.down-f1.down*f2.up;    result.down=f1.down*f2.down;    return reduction(result);}Fraction multi(Fraction f1,Fraction f2){    Fraction result;    result.up=f1.up*f2.up;    result.down=f1.down*f2.down;    return reduction(result);}Fraction divide(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);    }    else{        printf("%lld/%lld",r.up,r.down);    }    if(r.up<0) printf(")");}int main(){    scanf("%lld/%lld %lld/%lld",&a.up,&a.down,&b.up,&b.down);    showResult(a);    printf(" + ");    showResult(b);    printf(" = ");    showResult(add(a,b));    printf("\n");    showResult(a);    printf(" - ");    showResult(b);    printf(" = ");    showResult(sub(a,b));    printf("\n");    showResult(a);    printf(" * ");    showResult(b);    printf(" = ");    showResult(multi(a,b));    printf("\n");    showResult(a);    printf(" / ");    showResult(b);    printf(" = ");    if(b.up==0) printf("Inf");    else showResult(divide(a,b));    return 0;}