1081 Rational Sum (20)

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N个有理数（以分子/分母的形式给出），计算这N个数的总和，最后总和要以（整数 分子/分母）的形式给出。如果整数部分为0，则直接输出小数部分。

分析：用分数加法进行模拟，每次先把分母变成两个分数的最小公倍数，再把分子乘以相应倍数相加。

/***********************************************
 *                   _ooOoo_                   *
 *                  o8888888o                  *
 *                  88" . "88                  *
 *                  (| -_- |)                  *
 *                  O\  =  /O                  *
 *               ____/`---'\____               *
 *             .'  \\|     |//  `.             *
 *            /  \\|||  :  |||//  \            *
 *           /  _||||| -:- |||||-  \           *
 *           |   | \\\  -   * |    |           *
 *           | \_|  ''\---/''  |   |           *
 *           \  .-\__  `-`  ___/-. /           *
 *         ___`. .'  /--.--\  `. . __          *
 *      ."" '<  `.___\_<|>_/___.'  >'"".       *
 *     | | :  `- \`.;`\ _ /`;.`/ - ` : | |     *
 *     \  \ `-.   \_ __\ /__ _/   .-` /  /     *
 *======`-.____`-.___\_____/___.-`____.-'======*
 *                   `=---='                   *
 *^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

             佛祖保佑       永无BUG

   本程序已经经过开光处理，绝无可能再产生bug
 **********************************************/
#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
using namespace std;

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

long long lcm(long long a,long long b)
{
    return a/gcd(a,b)*b;
}

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int n;scanf("%d",&n);
    long long numerator=0,denominator=1;
    for(int i=0;i<n;++i)
    {
        long long a,b;scanf("%lld/%lld",&a,&b);
        long long t1,t2,temp=lcm(denominator,b);
        t1=temp/denominator,t2=temp/b;
        denominator=temp;
        numerator=numerator*t1+a*t2;
        long long g=gcd((long long)fabs(denominator),(long long)fabs(numerator));
        denominator/=g,numerator/=g;
    }
    int f=0;
//    db2(denominator,numerator);
    if(numerator%denominator==0)printf("%lld\n",numerator/denominator);
    else
    {
        long long x=numerator/denominator;
//        db3(x,numerator,denominator);
        numerator=numerator-x*denominator;
        if(x!=0)printf("%lld ",x);
        if(denominator<0)
        {
            if(numerator<0)printf("%lld/%lld\n",-1*numerator,-1*denominator);
            else printf("-%lld/%lld\n",numerator,-1*denominator);
        }
        else printf("%lld/%lld\n",numerator,denominator);
    }

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}

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