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;
}
来源:https://blog.csdn.net/chen_zan_yu_/article/details/102726536