PAT (Advanced Level) Practice 1081 Rational Sum
1081 Rational Sum
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
Sample Output 3:
7/24
AC代码:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
struct Fraction{
LL up,down;
};
LL gcd( LL a , LL b )
{
if( b == 0 ) return a;
else return gcd( b, a%b );
}
Fraction reduction(Fraction res){
if(res.down < 0) {
res.up = -res.up;
res.down = -res.down;
}
if(res.up == 0){
res.down = 1;
}else{
int d = gcd(abs(res.up),abs(res.down));
res.up /= d;
res.down /= d;
}
return res;
}
Fraction add( Fraction f1,Fraction f2){
Fraction res;
res.up = f1.up * f2.down+f2.up * f1.down;//分数和的分子
res.down = f1.down * f2.down; //分数和的分母
return reduction(res);
}
void showResult(Fraction r){
r = reduction(r);
if(r.down == 1) printf("%lld",r.up);//整数
else if(abs(r.up) > r.down) printf("%lld %lld/%lld",r.up/r.down,abs(r.up)%r.down,r.down);//假分数
else printf("%lld/%lld",r.up,r.down); //真分数
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
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;
}