标签:有理数 运算 gcd LL cout printf minus lld
https://www.acwing.com/problem/content/description/1580/
思路:
这题思路并不难,但如果你傻乎乎的一种一种情况的输出,那会非常的繁琐,巧妙的利用一个函数来统一起来实现。
#include <iostream>
using namespace std;
typedef long long LL;
LL gcd(LL a, LL b)
{
return b ? gcd(b, a % b) : a;
}
void print(LL a, LL b)
{
LL d = gcd(a, b);
a /= d, b /= d;
if (b < 0) a *= -1, b *= -1;
bool is_minus = a < 0;
if (is_minus) cout << "(";
if (b == 1) cout << a;
else
{
if (abs(a) >= b) printf("%lld ", a / b), a = abs(a) % b;
printf("%lld/%lld", a, b);
}
if (is_minus) cout << ")";
}
void add(LL a, LL b, LL c, LL d)
{
print(a, b), cout << " + ", print(c, d), cout << " = ";
a = a * d + b * c;
b = b * d;
print(a, b), cout << endl;
}
void sub(LL a, LL b, LL c, LL d)
{
print(a, b), cout << " - ", print(c, d), cout << " = ";
a = a * d - b * c;
b = b * d;
print(a, b), cout << endl;
}
void mul(LL a, LL b, LL c, LL d)
{
print(a, b), cout << " * ", print(c, d), cout << " = ";
a = a * c;
b = b * d;
print(a, b), cout << endl;
}
void div(LL a, LL b, LL c, LL d)
{
print(a, b), cout << " / ", print(c, d), cout << " = ";
if (!c) puts("Inf");
else
{
a = a * d;
b = b * c;
print(a, b), cout << endl;
}
}
int main()
{
LL a, b, c, d;
scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
add(a, b, c, d);
sub(a, b, c, d);
mul(a, b, c, d);
div(a, b, c, d);
return 0;
}
标签:有理数,运算,gcd,LL,cout,printf,minus,lld 来源: https://www.cnblogs.com/xjtfate/p/16612432.html
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