PAT A1081. Rational Sum (20)

最新推荐文章于 2020-07-17 09:57:49 发布

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本文介绍了一种算法，用于计算多个有理数的和，并以最简形式输出结果。该算法适用于给定一系列分数（分子/分母形式），计算它们的总和并输出最简整数部分及分数部分。

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

思路：跟PAT A1088 Rational Arithmetic类似，也就是多了一个连续相加的过程

#include<cstdio>
#include<iostream>
using namespace std;

void calculate(long int &A,long int &B,long int a2,long int b2){
    long int a,b;   a=A;b=B;
    A=a*b2+a2*b;    B=b*b2;
}//两个分数加和，非最简 

long int common(long int x,long int y){
    if(x%y==0)
        return y;
    else
        common(y,x%y);
}//选取最大公约数 

void simplest(long int &a,long int &b){
    long int t=common(a,b);
    long int num;
    a/=t;b/=t;
    if(a<b){
        cout<<a<<'/'<<b<<endl;
    }
    else if(b==1){
        cout<<a<<endl;
    }//分母为1的情形 
    else if(a>b){
        num=a/b;
        a-=b*num;
        cout<<num<<' '<<a<<'/'<<b<<endl;
    }

}

void output(long int &a,long int &b){
    if(a==0){
        cout<<'0'<<endl;
        return;
    }
    if(a<0){
        cout<<'-';
        a=-a;
    }
    simplest(a,b);
}

int main(){
//  freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
    int N;
    long int A,B,A2,B2;
    while(cin>>N){
        cin.get();
        cin>>A;cin.get();cin>>B;
        for(int i=1;i<N;i++){
            cin>>A2;cin.get();cin>>B2;
            calculate(A,B,A2,B2);//每重新输入一个分数，都加到第一个分数上 
        }
        output(A,B);//将第一个分数输出 
    }
    return 0;
}