pat1081. Rational Sum (20)

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

---

提交代码

测试数据比较弱。如果要算：(2^63)/(3)+(1)/(5)    怎么办？？

 #include<cstdio>
 #include<stack>
 #include<algorithm>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 using namespace std;
 long long gcd(long long a,long long b)
 {
     if(b==)
     {
         return a;
     }
     return gcd(b,a%b);
 }
 int main()
 {
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,i;
     long long fz,ffz,fm,ffm,com;
     while(scanf("%d",&n)!=EOF)
     {
         scanf("%lld/%lld",&fz,&fm);
         com=gcd(fz,fm);
         fz/=com;
         fm/=com;
         for(i=; i<n; i++)
         {
             //cout<<"i: "<<i<<endl;
             scanf("%lld/%lld",&ffz,&ffm);
             com=gcd(fm,ffm);
             //cout<<"com:  "<<com<<endl;
             ffz=ffz*(fm/com);
             //cout<<"ffz:  "<<ffz<<endl;
             fm=fm*(ffm/com);
             //cout<<"fm:  "<<fm<<endl;
             fz=fz*(ffm/com);
             //cout<<"fz:  "<<ffm<<endl;
             fz+=ffz;
             //cout<<"fz:  "<<fz<<endl;
             com=gcd(fz,fm);
             //cout<<"com:  "<<com<<endl;
             fz/=com;
             //cout<<"fz:  "<<fz<<endl;
             fm/=com;
             //cout<<"fm:  "<<fm<<endl;
         }

         //cout<<fz<<" "<<fm<<endl;

         if(fz%fm==) //可以整除
         {
             printf("%lld\n",fz/fm);
         }
         else
         {
             if(fz/fm>)
             {
                 printf("%lld ",fz/fm);
             }
             printf("%lld/%lld\n",fz-fz/fm*fm,fm);
         }
     }
     return ;
 }