PAT (Advanced) 1081. Rational Sum (20)

原题：1081. Rational Sum (20)

解题思路：
用long long 去模拟分数加法即可，注意最后一个样例和为0

代码如下：#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

long long gcd(long long a, long long b)
{
if(b == 0) return a;
return gcd(b, a%b);
}

int main()
{
int n;
while(scanf("%d", &n) == 1)
{
long long num, denum, a, b;
num = 0;
denum = 1;
for(int i = 0; i < n; i++)
{
scanf("%lld/%lld", &a, &b);
num = num*b + a*denum;
denum = denum * b;
long long g = gcd(num, denum);
num /= g;
denum /= g;
}
if(num == 0) printf("0\n");
else
if(num / denum && num % denum) printf("%lld %lld/%lld\n", num/denum, num%denum, denum);
else if(num / denum) printf("%lld\n", num/denum);
else printf("%lld/%lld\n", num, denum);
}
return 0;
}