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("\n");
}

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("\n");
}

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("\n");
}

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("\n");
}

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;
}

  

posted @ 2018-07-30 07:33  丧心病狂工科女  阅读(282)  评论(0编辑  收藏  举报