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 (≤), 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:
7/241 #include <iostream> 2 #include <vector> 3 #include <math.h> 4 using namespace std; 5 long int gcd(long int a, long int b) 6 { 7 if(b==0) return a; 8 else return gcd(b,a%b); 9 } 10 int main() 11 { 12 int N; 13 cin >> N; 14 long long Inter = 0, resa = 0, resb = 1, a, b, Div, Mul; 15 for (int i = 0; i < N; ++i) 16 { 17 char c; 18 cin >> a >> c >> b; 19 resa = resa * b + a * resb;//同分相加的分子 20 resb = resb * b; 21 Inter += resa / resb;//简化 22 resa = resa - resb * (resa / resb); 23 Div = gcd(resb, resa); 24 resa /= Div; 25 resb /= Div; 26 } 27 if (Inter == 0 && resa == 0) 28 cout << 0 << endl; 29 else if (Inter != 0 && resa == 0) 30 cout << Inter << endl; 31 else if (Inter == 0 && resa != 0) 32 cout << resa << "/" << resb << endl; 33 else 34 cout << Inter << " " << resa << "/" << resb << endl; 35 return 0; 36 }