PAT甲级1081 Rational Sum (20 分)

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

      题目大意：

         给定n个的分数，要求分数的和，并以带分数的形式输出。

 

思路：

         由于标明分子分母都是长整型，那么就有可能是全为最大的长整型数。那么最好还是用string存储，并进行处理。

 

参考代码：

 

#include<cstdio>
#include<cstdlib>
long long num, denum, integer, gcdv;
long long gcd(long long a, long long b){return b == 0? abs(a): gcd(b, a % b);}
int main(){
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; ++i){
		long long t1, t2;
		scanf("%lld/%lld", &t1, &t2);
		gcdv = gcd(t1, t2);
		t1 /= gcdv, t2 /= gcdv;
		if(i == 0){
			num = t1, denum = t2;
			continue;
		}
		num = t1 * denum + num * t2;
		denum = t2 * denum;
		gcdv = gcd(num, denum);
		num /= gcdv, denum /= gcdv;
	}
	integer = num / denum;
	num = num - (denum * integer);
	if(integer != 0){
		printf("%lld", integer);
		if(num != 0)	printf(" ");
	}
	if(num != 0)	printf("%lld/%lld", num, denum);
	if(integer == 0 && num ==0)	printf("0");
	return 0;
}

参考代码:https://blog.csdn.net/liuchuo/article/details/52139107