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

代码如下：

#include<iostream>
#include<cstdlib>
using namespace std;

long long gcd(long long  a,long long b){
	return (b==0)?abs(a):gcd(b,a%b);
}

int main(){
	long long suma=0,sumb=1,a,b,n;
	scanf("%lld",&n);
	for(int i=0;i<n;++i){
		scanf("%lld/%lld",&a,&b);
		long long val=gcd(a,b);
		a/=val;b/=val;
		suma=suma*b+a*sumb;sumb=sumb*b;
		val=gcd(suma,sumb);
		suma/=val;sumb/=val;
	}
	long long c=suma/sumb;
	suma=suma-(sumb*c);
	if(c!=0){
		printf("%lld",c);
		if(suma!=0){
			printf(" ");
		}
	}
	if(suma!=0){
		printf("%lld/%lld",suma,sumb);
	}
	if(suma==0&&c==0){
		printf("0");
	}
	return 0;
}