PAT A1081. Rational Sum (20)

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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -1/6 1/8

Sample Output 3:

7/24

#include <cstdio>#include <algorithm>#include <cstring>#include <string>#define Max 111using namespace std;struct Sum{long long numerator,denominator;}S[Max];long long gcd(Sum a){long long c,d,e;d=a.numerator;e=a.denominator;if(d<0) d=-d;if(e<0) e=-e;if(e>d) swap(d,e);c=e;while(d%e!=0){c=d%e;d=e;e=c;}return c;}Sum MikEase( Sum a){if(a.denominator<0){a.denominator=-a.denominator;a.numerator=-a.numerator;}if(a.numerator==0) a.denominator=1;else {long long k=gcd(a);a.numerator=a.numerator/k;a.denominator=a.denominator/k;}return a;}Sum Get(Sum a ,Sum b){a.numerator=a.numerator*b.denominator+b.numerator*a.denominator;a.denominator=a.denominator*b.denominator;return MikEase(a);}int main(){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lld/%lld",&S[i].numerator,&S[i].denominator);if(i>0)S[i]=Get(S[i],S[i-1]);}long long a,b,c=0;a=S[n-1].numerator;b=S[n-1].denominator;if(b==1) printf("%lld",a);else {if(abs(a)>abs(b)){printf("%lld %lld/%lld",a/b,abs(a)%abs(b),b);}else printf("%lld/%lld",a,b);}system("pause");return 0;}