A1081. 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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 typedef struct{
     long long up, down;
 }fra;
 long long gcd(long long a, long long b){
     a = abs(a);
     b = abs(b);
     if(b == )
         return a;
     else return gcd(b, a % b);
 }
 fra cacul(fra a, fra b){
     fra temp;
     temp.down = a.down * b.down;
     temp.up = a.down * b.up + b.down * a.up;
     long long fact = gcd(temp.up, temp.down);
     temp.down = temp.down / fact;
     temp.up = temp.up / fact;
     return temp;
 }
 int main(){
     int N;
     fra re = {,}, temp = {, };
     scanf("%d", &N);
     for(int i = ; i < N; i++){
         scanf("%lld/%lld", &temp.up, &temp.down);
         re = cacul(re, temp);
     }
     if(re.up == ){
         printf("");
     }else if(abs(re.up) == abs(re.down)){
         printf("%lld", re.up / re.down);
     }else if(abs(re.up) > abs(re.down)){
         if(re.up % re.down == )
             printf("%d", re.up / re.down);
         else
             printf("%lld %lld/%lld", re.up / re.down, re.up % re.down, re.down);
     }else{
         printf("%lld/%lld", re.up, re.down);
     }
     cin >> N;
     return ;
 }

总结：

1、分数运算化简：化简时分子分母同除最大公因数。 输出时考虑：分子为0时直接输出0；分子>=分母时（用绝对值比较，是>=而非>），可以整除则输出整数，否则输出代分数（4/1直接输出4）；

2、gcd函数：

int gcd(int a, int b){ 
    if(b == )
        return a;
    else return gcd(b, a % b);
}