1034 有理数四则运算 (20 分)

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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);

}
struct st
{
    ll up;
    ll down;
} a,b;
st reduction(st result)
{
    if(result.down<0)
    {
        result.down=-result.down;
        result.up=-result.up;
    }
    if(result.up==0)
    {
        result.down=1;
    }
    else
    {
        int d=gcd(abs(result.up),abs(result.down));
        result.up/=d;
        result.down/=d;
    }
    return result;
}
st add(st f1,st f2)
{
    st result;
    result.up=f1.up*f2.down+f2.up*f1.down;
    result.down=f1.down*f2.down;
    return reduction(result);
}
st minu(st f1,st f2)
{
    st result;
    result.up=f1.up*f2.down-f2.up*f1.down;
    result.down=f1.down*f2.down;
    return reduction(result);
}
st multi(st f1,st f2)
{
    st result;
    result.up=f1.up*f2.up;
    result.down=f1.down*f2.down;
    return reduction(result);
}
st divide(st f1,st f2)
{
    st result;
    result.up=f1.up*f2.down;
    result.down=f1.down*f2.up;
    return reduction(result);
}
void show(st 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);
    show(a);
    printf(" + ");
    show(b);
    printf(" = ");
    show(add(a,b));
    printf("\n");

    show(a);
    printf(" - ");
    show(b);
    printf(" = ");
    show(minu(a,b));
    printf("\n");

    show(a);
    printf(" * ");
    show(b);
    printf(" = ");
    show(multi(a,b));
    printf("\n");

    show(a);
    printf(" / ");
    show(b);
    printf(" = ");
    if(b.up==0)printf("Inf");
    else
        show(divide(a,b));
    printf("\n");
    //printf("%d",b.up);
    return 0;
}

 

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