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

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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==0)
    {
        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=1; 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==0) //可以整除
        {
            printf("%lld\n",fz/fm);
        }
        else
        {
            if(fz/fm>1)
            {
                printf("%lld ",fz/fm);
            }
            printf("%lld/%lld\n",fz-fz/fm*fm,fm);
        }
    }
    return 0;
}