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  • PAT 1034 有理数四则运算

    https://pintia.cn/problem-sets/994805260223102976/problems/994805287624491008

    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

    代码:
    #include <bits/stdc++.h>
    using namespace std;
    
    long long int gcd(long long int x,long long int y) {
        long long int z = x % y;
        while(z)
        {
            x = y;
            y = z;
            z = x % y;
        }
        return y;
    }
    
    void out(long long int x,long long int y) {
        if(y < 0) {
            y = y * (-1);
            x = x * (-1);
        }
        if(x > 0) {
            if(y == 1) printf("%lld", x);
            else {
                if(x % y == 0) printf("%lld", x / y);
                else {
                    if(x < y) printf("%lld/%lld", x / gcd(x, y), y / gcd(x, y));
                    else printf("%lld %lld/%lld", x / y, (x % y) / gcd(x % y, y), y / gcd(x % y, y));
                }
            }
        }
        else if(x == 0) printf("0");
        else {
            if(y == 1) printf("(%lld)", x);
            else {
                if(x % y == 0) printf("(%lld)", x / y);
                else {
                    if(abs(x) < y) printf("(%lld/%lld)", x / abs(gcd(x, y)), y / abs(gcd(x, y)));
                    else printf("(%lld %lld/%lld)", x / y, abs(x) % y/gcd(abs(x) % y, y), y / gcd(abs(x) % y, y));
                }
            }
        }
    }
    
    void add(long long int a1, long long int b1, long long int a2, long long int b2) {
        out(a1, b1);
        printf(" + ");
        out(a2, b2);
        printf(" = ");
        long long int a3 = a1 * b2 + a2 * b1;
        long long int b3 = b1 * b2;
        out(a3, b3);
        printf("
    ");
    }
    
    void subtract(long long int a1, long long int b1, long long int a2, long long int b2) {
        out(a1, b1);
        printf(" - ");
        out(a2, b2);
        printf(" = ");
        long long int a3 = a1 * b2 - a2 * b1;
        long long int b3 = b1 * b2;
        out(a3, b3);
        printf("
    ");
    }
    
    void multiplication(long long int a1, long long int b1, long long int a2, long long int b2) {
        out(a1, b1);
        printf(" * ");
        out(a2, b2);
        printf(" = ");
        long long int a3 = a1 * a2;
        long long int b3 = b1 * b2;
        out(a3, b3);
        printf("
    ");
    }
    
    void division(long long int a1, long long int b1, long long int a2, long long int b2) {
        out(a1, b1);
        printf(" / ");
        out(a2, b2);
        printf(" = ");
        if(a2 * b2 == 0) {
            printf("Inf");
            return ;
        }
        long long int a3 = a1 * b2;
        long long int b3 = a2 * b1;
        out(a3, b3);
        printf("
    ");
    }
    
    int main() {
        long long int a, b, c, d;
        scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
        add(a, b, c, d);
        subtract(a, b, c, d);
        multiplication(a, b, c, d);
        division(a, b, c, d);
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9388627.html
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