Pat(Advanced Level)Practice--1081(Rational Sum)

Pat1081代码

题目描述：

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

AC代码：

#include<cstdio>
#include<cstdlib>

using namespace std;

long int GCD(long int num,long int den)
{
	long int temp;
	while(den!=0)
	{
		temp=num%den;
		num=den;
		den=temp;
	}
	return num;
}

void GetSum(long int a,long int b,long int *num,long int *den)
{
	long int ret;
	long int upsum,downsum;
	upsum=a*(*den)+b*(*num);
	downsum=b*(*den);
	ret=GCD(upsum,downsum);
	(*num)=upsum/ret;
	(*den)=downsum/ret;
}

int main(int argc,char *argv[])
{
	int n;
	long int a,b;
	long int num=0,den=0;
	scanf("%d",&n);
	scanf("%ld/%ld",&num,&den);
	for(int i=1;i<n;i++)
	{
		scanf("%ld/%ld",&a,&b);
		GetSum(a,b,&num,&den);
	}
	if(num%den==0)
		printf("%ld\n",num/den);
	else if(num>den)
	{
		printf("%ld",num/den);
		printf(" %ld/%ld\n",num-num/den*den,den);
	}
	else
		printf("%ld/%ld\n",num,den);

	return 0;
}