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  • 1034 有理数四则运算 (20 分)

    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
    #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 == 0?a:gcd(b,a%b);
    }
    
    Fraction reduction(Fraction result){
        if(result.down < 0){
            result.down = -result.down;
            result.up = -result.up;
        }
        if(result.up == 0) result.down = 1;
        int 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 result;
    }
    
    Fraction minu(Fraction f1,Fraction f2){
        Fraction result;
        result.up = f1.up*f2.down - f1.down * f2.up;
        result.down = f1.down*f2.down;
        return result;
    }
    
    Fraction multi(Fraction f1,Fraction f2){
        Fraction result;
        result.up = f1.up*f2.up;
        result.down = f1.down*f2.down;
        return result;
    }
    
    Fraction divide(Fraction f1,Fraction f2){
        Fraction result;
        result.up = f1.up*f2.down;
        result.down = f1.down*f2.up;
        return 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("
    ");
        
        showResult(a);
        printf(" - ");
        showResult(b);
        printf(" = ");
        showResult(minu(a,b));
        printf("
    ");
        
        showResult(a);
        printf(" * ");
        showResult(b);
        printf(" = ");
        showResult(multi(a,b));
        printf("
    ");
        
        showResult(a);
        printf(" / ");
        showResult(b);
        printf(" = ");
        if(b.up == 0) printf("Inf");
        else showResult(divide(a,b));
        printf("
    ");
        
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/10292915.html
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