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。
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#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 testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testint 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 testclockid_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 //testreturn 0;
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