A1081 Rational Sum [分数计算]
题意:分数的计算
注意
- 每步都要化简(求最大公约数),防止溢出
- 真分数假分数要判断输出
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
struct fraction
{
ll up, down;
};
fraction simplify(fraction result)
{
int d = gcd(abs(result.up),abs(result.down)); //记得绝对值
result.up /= d;
result.down /= d;
return result;
}
fraction add(fraction a, fraction b)
{
fraction result;
result.up = a.up * b.down + b.up * a.down;
result.down = a.down * b.down;
return simplify(result);
}
void showresult(fraction r)
{
if (r.down == 1)
printf("%lld\n", r.up);
else if (abs(r.up) > abs(r.down))
{
printf("%lld %lld/%lld\n", r.up / r.down, abs(r.up) % r.down, r.down);
}
else
{
printf("%lld/%lld\n", r.up, r.down);
}
}
int main()
{
int n;
cin >> n;
fraction sum, temp;
sum.up = 0, sum.down = 1;
for (int i = 0; i < n; i++)
{
scanf("%lld/%lld", &temp.up, &temp.down);
sum = add(sum, temp);
}
showresult(sum);
return 0;
}