pat-a1081. Rational Sum (20)

以前做这种题没有用struct。。两个int相乘不会超过long long。。以前还是先约分再加的。。可以直接加了再约分。

#include<cstdio>
typedef long long int LL;
struct node{
LL fenz,fenm;
node(LL a=0,LL b=1):fenz(a),fenm(b){
}
};
LL gcd(LL a,LL b){
if(b==0) return a;
return gcd(b,a%b);
}
void add(node& sum,const node& t){
sum.fenz=sum.fenz*t.fenm+sum.fenm*t.fenz;
sum.fenm=sum.fenm*t.fenm;
LL k=gcd(sum.fenz,sum.fenm);
sum.fenz/=k;
sum.fenm/=k;
}
int main(){
int n;
node sum(0,1);
LL a,b;
scanf("%d",&n);
while(n--){
scanf("%lld/%lld",&a,&b);
node temp(a,b);
add(sum,temp);
}
int t=sum.fenz/sum.fenm;
if(t*sum.fenm==sum.fenz) printf("%d\n",t);
else{
sum.fenz=sum.fenz%sum.fenm;
if(sum.fenz<0){
printf("-");
sum.fenz=-sum.fenz;
}
if(t!=0) printf("%d ",t);
if(sum.fenz) printf("%lld/%lld\n",sum.fenz,sum.fenm);
}
return 0;
}

Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range of "long int". If there is a negative number,
then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form "integer numerator/denominator" where "integer" is the integer part of the sum, "numerator" < "denominator", and the numerator and the denominator have no common factor. You must output only the fractional
part if the integer part is 0.
Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

2

Sample Input 3:

3
1/3 -1/6 1/8