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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -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);}elseprintf("%ld/%ld\n",num,den);return 0;}