PAT1081. 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
思路：一定要注意一些特殊的情况,比如说0输出的时候要注意。

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 #define MAX 110
 5 struct element
 6 {
 7     long long up;
 8     long long down;
 9 }ele[MAX];
10 long long GCD(long long a,long long b)
11 {
12     return !b?a:GCD(b,a%b);
13 }
14
15 element Add(element A,element B)
16 {
17     struct element temp;
18     long long  down=A.down*B.down;
19     long long  up=A.up*B.down+B.up*A.down;
20     long long int d=GCD(up,down);
21     up/=d;
22     down/=d;
23     temp.up=up;
24     temp.down=down;
25     return temp;
26 }
27 void Print(element A)
28 {
29     if(A.up==0)
30     {
31         printf("0\n");
32     }
33     else if(A.up<A.down)
34     {
35         printf("%lld/%lld\n",A.up,A.down);
36     }
37     else
38     {
39         long long Jia=A.up/A.down;
40         A.up=A.up%A.down;
41         if(A.up!=0)
42           printf("%lld %lld/%lld\n",Jia,A.up,A.down);
43         else
44           printf("%lld\n",Jia);
45     }
46 }
47 int main(int argc, char *argv[])
48 {
49     int N;
50     scanf("%d",&N);
51     element sum;
52     sum.up=0;
53     sum.down=0;
54     for(int i=0;i<N;i++)
55     {
56         if(i==0)
57         {
58            scanf("%lld/%lld",&ele[i].up,&ele[i].down);
59            sum.up=ele[i].up;
60            sum.down=ele[i].down;
61            continue;
62         }
63         scanf("%lld/%lld",&ele[i].up,&ele[i].down);
64         sum=Add(sum,ele[i]);
65     }
66     Print(sum);
67     return 0;
68 }