1034 有理数四则运算

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:

5/3 0/6

 

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

 

思路:

测试点2、3运行时错误:a1,b1,a2,b2的类型应该定义为long long,此时s1、s2数组的长度也应相应增大

运行时错误应该首先检查数组是否越界

 

代码:

 

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
using namespace std;
struct Fraction{
    long long up,down;
    Fraction(long long up, long long down):up(up),down(down){};
    Fraction(){};
};
long long gcd(long long a, long long b){
    if(b==0){
        return a;
    }
    gcd(b,a%b);
}
Fraction reduction(Fraction fraction){//化简
    if(fraction.down < 0){
        fraction.down = - fraction.down;
        fraction.up = - fraction.up;
    }
    if(fraction.up == 0){
        fraction.down = 1;
    }else{
        long long d = gcd(abs(fraction.up), abs(fraction.down));
        fraction.up /= d;
        fraction.down /= d;
    }
    return fraction;
}
void show(Fraction fraction){
    fraction = reduction(fraction);
    if(fraction.up < 0){
        cout<<"(";
    }
   
    if(fraction.down == 1){//整数
        cout<<fraction.up;
    }
    else if(abs(fraction.up) > abs(fraction.down)){ //假分数
        cout<<fraction.up/fraction.down<<" "<<abs(fraction.up) % abs(fraction.down)<<"/"<<fraction.down; 
    }else{ //真分数
        cout<<fraction.up<<"/"<<fraction.down;
    }
    
    if(fraction.up < 0){
        cout<<")";
    }
}
void add(Fraction a, Fraction b){
    Fraction result;
    show(a);
    cout<<" + ";
    show(b);
    cout<<" = ";
    result.up = a.up * b.down + a.down * b.up;
    result.down = a.down * b.down;
    show(reduction(result));
}
void substract(Fraction a, Fraction b){
    Fraction result;
    show(a);
    cout<<" - ";
    show(b);
    cout<<" = ";
    result.up = a.up * b.down - b.up * a.down;
    result.down = a.down * b.down;
    show(reduction(result));
}
void multiply(Fraction a, Fraction b){
    Fraction result;
    show(a);
    cout<<" * ";
    show(b);
    cout<<" = ";
    result.up = a.up * b.up;
    result.down = a.down * b.down;
    show(reduction(result));
}
void divide(Fraction a, Fraction b){
    Fraction result;
    show(a);
    cout<<" / ";
    show(b);
    cout<<" = ";
    result.up = a.up * b.down;
    result.down = a.down * b.up;
    if(result.down == 0){
        cout<<"Inf";
        return;
    }
    show(reduction(result));
}
int main(){
    char s1[30],s2[30];
    long long a1,b1,a2,b2;
    cin>>s1;
    cin>>s2;
    sscanf(s1, "%lld/%lld", &a1, &b1);
    sscanf(s2, "%lld/%lld", &a2, &b2);
    Fraction f1 = Fraction(a1,b1);
    Fraction f2 = Fraction(a2,b2);
    add(f1, f2);
    cout<<endl;
    substract(f1, f2);
    cout<<endl;
    multiply(f1, f2);
    cout<<endl;
    divide(f1, f2);
    return 0;
}

 

posted @ 2022-11-02 09:21  Yohoc  阅读(34)  评论(0编辑  收藏  举报