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
结尾无空行
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)
结尾无空行
输入样例 2:
5/3 0/6
结尾无空行
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
结尾无空行
代码:
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
long long gcf(long long m, long long n) {
return n == 0 ? m : gcf(n, m % n);
}
long long gcm(long long m, long long n) {
long long a = gcf(m, n);
return m % n == 0 ? m < n ? n : m : m / a * n / a * a;
}
void format(long long m, long long n) {
if (m * n == 0) {
n == 0 ? cout << "Inf" : cout << 0;
return;
}
if (n < 0) m = -1 * m; n = -1 * n;
bool flag = m < 0; bool isout = false;
m = abs(m); n = abs(n);
if (m >= n) {
if (flag) cout << "(-"; isout = true;
cout << m / n;
m = m - (m / n) * n;
if (m == 0) {
if(flag) cout << ")";
return;
}
if (m != 0) cout << " ";
}
if (flag && !isout) cout << "(-";
cout << m / gcf(m, n) << "/" << n / gcf(m, n);
if (flag) cout << ")";
}
int main() {
long long a, b,c,d,m,n;
scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
format(a, b); cout << " + "; format(c, d); cout << " = "; format((gcm(b, d) / b) * a + (gcm(b, d) / d) * c, gcm(b, d)); cout << endl;
format(a, b); cout << " - "; format(c, d); cout << " = "; format((gcm(b, d) / b) * a - (gcm(b, d) / d) * c, gcm(b, d)); cout << endl;
format(a, b); cout << " * "; format(c, d); cout << " = "; format(a / gcf(a, b) * c / gcf(c, d), b / gcf(a, b) * d / gcf(c, d)); cout << endl;
format(a, b); cout << " / "; format(c, d); cout << " = "; format(a / gcf(a, b) * d / gcf(c, d), b / gcf(a, b) * c / gcf(c, d)); cout << endl;
return 0;
}