1034 有理数四则运算 (20 分)

1034 有理数四则运算 (20 分)

重点在对分数的处理

 

#include <iostream>
#include <cmath>

using namespace std;

//辗转相除法 求最大公约数 
int gcd(long long a, long long b){
    return b == 0 ? a : gcd(b, a % b);
}

void print(long long a, long long b){
    long long c = 0; //带分数前面的整数部分,默认是0
    if(a > 0){ //正数
        if(b == 1){ //形如3/1
            printf("%lld", a);
        }
        else if(a > b){ //形如5/3
            c = a / b;
            a -= b * c;
            printf("%lld %lld/%lld", c, a, b);
        }
        else{ //真分数 形如3/5
            printf("%lld/%lld", a, b);
        }
    }
    else if(a == 0){ //形如0/3
        printf("%c", '0');
    }
    else{ //负数
        if(b == 1){ //形如-3/1
            printf("(%lld)", a);
        }
        else if(-1 * a > b){ //形如-5/3
            c = a / b;
            a = (-1 * a) % b;
            printf("(%lld %lld/%lld)", c, a, b);
        }
        else{ //真分数
            printf("(%lld/%lld)", a, b);
        }
    }
}

void add(long long a1, long long b1, long long a2, long long b2){
    print(a1, b1);
    printf(" + ");
    print(a2, b2);
    printf(" = ");
    long long a3 = a1 * b2 + a2 * b1;
    long long b3 = b1 * b2;
    //化简到最简形式,非带分数形式
    long long gcd3 = abs(gcd(a3, b3));
    a3 /= gcd3;
    b3 /= gcd3;
    print(a3, b3);
    printf("
");
}

void subtract(long long a1, long long b1, long long a2, long long b2){
    print(a1, b1);
    printf(" - ");
    print(a2, b2);
    printf(" = ");
    long long a3 = a1 * b2 - a2 * b1;
    long long b3 = b1 * b2;
    //化简到最简形式,非带分数形式
    long long gcd3 = abs(gcd(a3, b3));
    a3 /= gcd3;
    b3 /= gcd3;
    print(a3, b3);
    printf("
");
}

void multiply(long long a1, long long b1, long long a2, long long b2){
    print(a1, b1);
    printf(" * ");
    print(a2, b2);
    printf(" = ");
    long long a3 = a1 * a2;
    long long b3 = b1 * b2;
    //化简到最简形式,非带分数形式
    long long gcd3 = abs(gcd(a3, b3));
    a3 /= gcd3;
    b3 /= gcd3;
    print(a3, b3);
    printf("
");
}

void divide(long long a1, long long b1, long long a2, long long b2){
    print(a1, b1);
    printf(" / ");
    print(a2, b2);
    printf(" = ");
    if(a2 == 0){
        printf("Inf");
    }
    else if(a2 < 0){
        long long a3 = -1 * a1 * b2;
        long long b3 = -1 * b1 * a2;
        //化简到最简形式,非带分数形式
        long long gcd3 = abs(gcd(a3, b3));
        a3 /= gcd3;
        b3 /= gcd3;
        print(a3, b3);
    }
    else{
        long long a3 = a1 * b2;
        long long b3 = b1 * a2;
        //化简到最简形式,非带分数形式
        long long gcd3 = abs(gcd(a3, b3));
        a3 /= gcd3;
        b3 /= gcd3;
        print(a3, b3);
    }
    printf("
");
}

int main(){
    long long a1, b1, a2, b2;
    long long c1 = 0, c2 = 0;
    scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);
    //先化简到最简形式,非带分数形式
    long long gcd1 = abs(gcd(a1, b1));//abs绝对值
    a1 /= gcd1;
    b1 /= gcd1;
    long long gcd2 = abs(gcd(a2, b2));
    a2 /= gcd2;
    b2 /= gcd2;
    //统一用最简形式参与运算
    add(a1, b1, a2, b2);
    subtract(a1, b1, a2, b2);
    multiply(a1, b1, a2, b2);
    divide(a1, b1, a2, b2);
    return 0;
}

 

原文地址:https://www.cnblogs.com/siro/p/11199942.html