PAT-A 1081. Rational Sum

1081. 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

程序代码:

#include<stdio.h>
long long ans[2]={0,1};
long long factor(long long a,long long b);
void add(long long ans[],long long a1,long long a2,long long b1,long long b2);
void print(long long a,long long b);
int main()
{
	int n;
	long long a,b;
	scanf("%d",&n);
	while(n--){
		scanf("%lld/%lld",&a,&b);
		add(ans,ans[0],ans[1],a,b);
	}
	print(ans[0],ans[1]);
	return 0;
}
long long factor(long long a,long long b)
{
	if(a<0)
		a=-a;
	if(b<0)
		b=-b;
	long long max=a>b?a:b;
	long long min=a<b?a:b;
	int r=max%min;
	while(r){
		max = min;
		min = r;
		r= max%min;
	}
	return min;
}
void add(long long ans[],long long a1,long long a2,long long b1,long long b2)
{
	ans[0]= a1*b2+a2*b1;
	ans[1]= a2*b2;
	if(ans[0]==0){
		ans[1]=1;
		return;
	}
	long long tmp = factor(ans[0],ans[1]);
	ans[0]=ans[0]/tmp;
	ans[1]=ans[1]/tmp;		
	return ;
}
void print(long long a,long long b)
{
	if(b==0)
		return ;
	if(a==0){
		printf("0");
		return;
	}
	int flag = 0;
	if(a<0){
		flag = 1;
		a=-a;
	}
	long long tmp = factor(a,b),c;
	a=a/tmp;
	b=b/tmp;
	c=a/b;
	if(flag){
		if(b==1)
			printf("-%lld",c);
		else{
			if(c==0)
				printf("-%lld/%lld",a,b);
			else
				printf("-%lld %lld/%lld",c,a%b,b);
		}
	}
	else{
                if(b==1)
                        printf("%lld",c);
                else{
                        if(c==0)
                                printf("%lld/%lld",a,b);
                        else
                                printf("%lld %lld/%lld",c,a%b,b);
		}
	}
}