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1081. Rational Sum (20)

1081. Rational Sum (20)

作者: cheerss | 来源:发表于2017-12-14 23:33 被阅读0次

PAT-A 1081,题目地址:https://www.patest.cn/contests/pat-a-practise/1081
这道题目不难,细心点别写错就好。
思路:

  1. 每输入一个分数,就往最终结果里加一个,不过此时还不急着化为假分数
  2. 等加完所有的数,在统一约分,化为假分数等等

注意:分数相加一般都要使得分母相乘,为了防止100个分数相加使得分母过大超出数据范围,建议每次加完都约分一下,但是结果不化为假分数。代码直接AC了,也不知道不进行约分会不会有case出错

代码如下:

#include <iostream>

using namespace std;
 
//计算出两个数a,b的最大公约数
int max_factor(int a, int b){
    a = a < 0 ? -a : a;
    b = b < 0 ? -b : b;
    if(a > b){
        int temp = a;
        a = b;
        b = temp;
    }
    for(int i = a; i >= 2; i--){
        if(i > 0 && a % i == 0 && b % i == 0){
            return i;
        }
    }
    return 1;
}

class Number{
    public:
        int integer; //整数部分
        int numrator; //分子
        int deno; //分母
        Number(): integer(0), numrator(0), deno(1){}
        Number(int a, int b, int c): integer(a), numrator(b), deno(c){}

        //化为假分数并约分
        void change(){
            integer = numrator / deno;
            numrator %= deno;
            int factor = max_factor(numrator, deno);
            numrator /= factor;
            deno /= factor;
        }
        //仅仅约分
        void simple(){
            int factor = max_factor(numrator, deno);
            numrator /= factor;
            deno /= factor;
        }
};

//两个分数相加
Number add(const Number& a, const Number& b){
    int deno = a.deno * b.deno;
    int numrator = a.deno * b.numrator + a.numrator * b.deno;
    if(deno < 0)
        numrator = -numrator;
    return Number(0, numrator, deno);
}

int main(){
    int n;
    scanf("%d", &n);
    Number each;
    Number all;
    for(int i = 0; i < n; i++){
        scanf("%d/%d", &each.numrator, &each.deno);
        all = add(all, each); //输入一个,加一个
        all.simple();
    }
    all.change();
    if(all.integer == 0 && all.numrator == 0){
        cout << "0" << endl;
    }
    else if(all.integer != 0 && all.numrator == 0){
        cout << all.integer << endl;
    }
    else if(all.integer == 0 && all.numrator != 0){
        cout << all.numrator << "/" << all.deno << endl;
    }
    else{
        cout << all.integer << " " << all.numrator << "/" << all.deno << endl;
    }
    return 0;
}

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