PAT 1081. Rational Sum (20)

1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

这道题先用辗转相除法求GCD，然后就可以求出LCM，然后就是要小心一些细节上的判断就好了。代码如下：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
long int gcd(long int x,long int y)
{
while(y)
{
long int rem=x%y;
x=y;
y=rem;
}
return x;
}
long int lcm(long int a,long int b)
{
return a*b/gcd(a,b);
}
int main(void)
{
int N;
cin>>N;
long int num[100],den[100];
for(int i=0;i<N;i++)
scanf("%ld/%ld",&num[i],&den[i]);
long int resultNum=0,resultDen=1;
for(int i=0;i<N;i++)
resultDen=lcm(resultDen,den[i]);
for(int i=0;i<N;i++)
resultNum+=resultDen/den[i]*num[i];
long int first=0;
bool flag=true;
if(resultNum<0)
{
flag=false;
resultNum=-resultNum;
}
if(resultNum>resultDen)
{
first=resultNum/resultDen;
resultNum%=resultDen;
}
if(resultNum==0)
{
if(first>0)
cout<<first;
else
cout<<0;
return 0;
}
long int g=gcd(resultNum,resultDen);
resultNum/=g;
resultDen/=g;
if(flag==false)
cout<<"-";
if(first!=0)
cout<<first<<" ";
cout<<resultNum<<"/"<<resultDen;
}