1081 Rational Sum （20 分）
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
題目大意：給定一個自然數n，然後讀入n個分數，求n個分數的和，如果是整數的話，直接輸出；如果是假分數，則按照帶分數形式輸出，真分數也直接輸出。
題目思路：設定數據結構fenshu，兩個成員up和down，設定規則，僅讓up有可能爲負，所以當down爲負數的話，up和down都設爲相反數。而當up=0時，設置down=1；找出最大公約數然後讓up和down同除。而分數加法運算f1+f2sum.up =f1.up * f2.down + f1.down * f2.up，sum.down=f1.down*f2.down.
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
struct fenshu{
ll up,down;
};
ll gcd(ll a,ll b){
return b==0?a:gcd(b,a%b);
}
fenshu reduct(fenshu result){
//化簡
if(result.down<0){
result.up=-result.up;
result.down=-result.down;
}
if(result.up==0){
result.down=1;
}
else{
int yueshu=gcd(abs(result.up),abs(result.down));
result.up/=yueshu;
result.down/=yueshu;
}
return result;
}
fenshu add(fenshu f1,fenshu f2){
fenshu result;
result.up=f1.up*f2.down+f1.down*f2.up;
result.down=f1.down*f2.down;
return reduct(result);
}
void showresult(fenshu f){
reduct(f);
if(f.down==1){
printf("%lld",f.up);
}
else if(abs(f.up)>f.down){
printf("%lld %lld/%lld",f.up/f.down,abs(f.up)%f.down,f.down);
}
else{
printf("%lld/%lld",f.up,f.down);
}
}
int main(){
int n;
cin>>n;
fenshu 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;
}