1034 有理数四则运算(20)(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
解析
这是我读算法笔记时做到笔记:分数类型题目解法.这道题我一开始做到时候不仅没有AC,而且代码写的混乱。但是看了这本书的总结,不仅AC,而且代码清爽干净d=====( ̄▽ ̄*)b
代码:
#include<stdio.h>
typedef struct{
long long up,down;
}Fraction;
long long Abs(long long x){
return x>0?x:-x;
}
long long gcd(long a,long b){
if(b==0)
return a;
else
return gcd(b,a%b);
}
Fraction reduction(Fraction a){
if(a.down < 0){
a.up = -a.up;
a.down = -a.down;
}
if(a.up ==0)
a.down = 1;
else{
int d = gcd(Abs(a.up),Abs(a.down));
a.up/=d;
a.down/=d;
}
return a;
}
Fraction add(Fraction a,Fraction b){
Fraction c;
c.up = a.up*b.down+b.up*a.down;
c.down = a.down*b.down;
return reduction(c);
}
Fraction sub(Fraction a,Fraction b){
Fraction c;
c.up = a.up*b.down-a.down*b.up;
c.down = a.down*b.down;
return reduction(c);
}
Fraction muiti(Fraction a,Fraction b){
Fraction c;
c.up =a.up*b.up;
c.down = a.down*b.down;
return reduction(c);
}
Fraction divide(Fraction a,Fraction b){
Fraction c;
c.up = a.up*b.down;
c.down = b.up*a.down;
return reduction(c);
}
void show(Fraction a){
if(a.up<0)
printf("(");
if(a.down == 1)
printf("%lld",a.up);
else if(Abs(a.up) > a.down)
printf("%lld %lld/%lld",a.up/a.down,Abs(a.up)%a.down,a.down);
else
printf("%lld/%lld",a.up,a.down);
if(a.up<0)
printf(")");
}
int main()
{
Fraction f1,f2;
scanf("%lld/%lld %lld/%lld", &(f1.up), &(f1.down), &(f2.up), &(f2.down));
f1 = reduction(f1);
f2 = reduction(f2);
Fraction addresult = add(f1,f2);
Fraction subresult = sub(f1,f2);
Fraction multiresult = muiti(f1,f2);
Fraction divideresult = divide(f1,f2);
show(f1);
printf(" + ");
show(f2);
printf(" = ");
show(addresult);
printf("\n");
show(f1);
printf(" - ");
show(f2);
printf(" = ");
show(subresult);
printf("\n");
show(f1);
printf(" * ");
show(f2);
printf(" = ");
show(multiresult);
printf("\n");
show(f1);
printf(" / ");
show(f2);
printf(" = ");
if(f2.up == 0)
printf("Inf");
else
show(divideresult);
return 0;
}