1034. 有理数四则运算(20)
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 <cstdio>
#include <cmath>
typedef long long ll;
using namespace std;
struct Fraction{
ll up, down;
}f1, f2;
ll gcd(ll a, ll b){
return !b ? a : gcd(b, a%b);
}
Fraction reduction(Fraction result){
if(result.down < 0){
result.up = -result.up;
result.down = -result.down;
}
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;
}
Fraction add(Fraction f1, Fraction f2){
Fraction result;
result.up = f1.up * f2.down + f2.up * f1.down;
result.down = f1.down * f2.down;
return reduction(result);
}
Fraction minu(Fraction f1, Fraction f2){
Fraction result;
result.up = f1.up * f2.down - f2.up * f1.down;
result.down = f1.down * f2.down;
return reduction(result);
}
Fraction multi(Fraction f1, Fraction f2){
Fraction result;
result.up = f1.up * f2.up;
result.down = f1.down * f2.down;
return reduction(result);
}
Fraction divide(Fraction f1, Fraction f2){
Fraction result;
result.up = f1.up * f2.down;;
result.down = f1.down * f2.up;
return reduction(result);
}
void showResult(Fraction 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, (ll)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", &f1.up, &f1.down, &f2.up, &f2.down);
showResult(f1);printf(" + ");showResult(f2);printf(" = ");showResult(add(f1, f2));printf("\n");
showResult(f1);printf(" - ");showResult(f2);printf(" = ");showResult(minu(f1, f2));printf("\n");
showResult(f1);printf(" * ");showResult(f2);printf(" = ");showResult(multi(f1, f2));printf("\n");
showResult(f1);printf(" / ");showResult(f2);printf(" = ");
if(f2.up == 0) printf("Inf");
else showResult(divide(f1, f2));
return 0;
}