PAT (Advanced) 1088. Rational Arithmetic (20)

本文链接：https://blog.youkuaiyun.com/cx86918626/article/details/79915992

本文提供了一种解决1088.RationalArithmetic问题的方法，通过计算分数的加减乘除并简化结果来实现。使用了long long类型避免溢出，并通过求最大公约数的方法简化分数。

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解题思路：

按题意计算即可。用公约数一直化成最简，注意数的输出格式，用long long 防溢出。

代码如下：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

long long gcd(long long  a, long long b)
{
    if(b == 0) return a;
    else return gcd(b, a%b);
}

void printNum(long long a, long long b)
{
    if((a > 0 && b < 0) || (a < 0 && b < 0))
    {
        a = -a;
        b = -b;
    }
    if(a < 0) printf("(");
    if(b == 1) printf("%lld", a);
    else
    {
        if(abs(a) > abs(b))
            printf("%lld %lld/%lld", a/b, abs(a)%abs(b), abs(b));
        else
            printf("%lld/%lld", a, b);
    }
    if(a < 0) printf(")");
}


int main()
{
    long long a, b, c, d;
    while(scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d) == 4)
    {
        long long g1 = gcd(a, b);
        long long g2 = gcd(c, d);
        if(g1 > 0)
        {
            a /= g1;
            b /= g1;
        }
        if(g2 > 0)
        {
            c /= g2;
            d /= g2;
        }
        long long tmp1, tmp2;
        printNum(a, b); printf(" + "); printNum(c, d); printf(" = ");
        long long g = gcd(b, d);
        tmp1 = d / g * a + b / g * c;
        tmp2 = b / g * d;
        printNum(tmp1/gcd(abs(tmp1), abs(tmp2)), tmp2/gcd(abs(tmp1), abs(tmp2)));
        printf("\n");
        printNum(a, b); printf(" - "); printNum(c, d); printf(" = ");
        g = gcd(b, d);
        tmp1 = d / g * a - b / g * c;
        tmp2 = b / g * d;
        printNum(tmp1/gcd(abs(tmp1), abs(tmp2)), tmp2/gcd(abs(tmp1), abs(tmp2)));
        printf("\n");
        printNum(a, b); printf(" * "); printNum(c, d); printf(" = ");
        tmp1 = a*c;
        tmp2 = b*d;
        printNum(tmp1/gcd(abs(tmp1), abs(tmp2)), tmp2/gcd(abs(tmp1), abs(tmp2)));
        printf("\n");
        printNum(a, b); printf(" / "); printNum(c, d); printf(" = ");
        if(c == 0) printf("Inf\n");
        else
        {
            tmp1 = a*d;
            tmp2 = b*c;
            printNum(tmp1/gcd(abs(tmp1), abs(tmp2)), tmp2/gcd(abs(tmp1), abs(tmp2)));
            printf("\n");
        }
    }
    return 0;
}