PTA甲级——1081

1081 Rational Sum

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:

1
2
5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

1
3 1/3

Sample Input 2:

1
2
2
4/3 2/3

Sample Output 2:

1
2

Sample Input 3:

1
2
3
1/3 -1/6 1/8

Sample Output 3:

1
7/24

思路

​ 基本思路：就是你小学如何计算多个个分数的和并且化简的过程，无他，直接上代码。

​ 先根据分数加法的公式累加，后分离出整数部分和分数部分

​ 分子和分母都在长整型内，所以不能用int存储，否则有一个测试点不通过

​ 一开始一直是浮点错误，按理来说应该是出现了/0或者%0的情况，找了半天也不知道错在哪里，后来注意到应该在累加的时候考虑是否会超出long long的范围，所以在累加每一步之前进行分子分母的约分处理，然后就AC了～ 应该还要考虑整数和小数部分都为0时候输出0的情况，但是测试用例中不涉及，所以如果没有最后两句也是可以AC的（PS：据说更新后的系统已经需要考虑全零的情况了～）

代码（大模拟）

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#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
long long gcd(long long a,long long b){
    return b==0?abs(a):gcd(b,a%b);
}
int main()
{
    int n;
    cin>>n;
    long long a,b,suma=0,sumb=1,gcdvalue;
    for(int i=0;i<n;i++){
        scanf("%lld/%lld",&a,&b);
        gcdvalue=gcd(a,b);
        a/=gcdvalue;
        b/=gcdvalue;
        suma=a*sumb+b*suma;
        sumb=b*sumb;
        gcdvalue=gcd(suma,sumb);
        suma/=gcdvalue;
        sumb/=gcdvalue;
    }
    long long integer=suma/sumb;
    suma=suma-(sumb*integer);
    if(integer){
        cout<<integer;
        if(suma) cout<<" ";
    }
    if(suma) cout<<suma<<"/"<<sumb;
    if(integer==0&&suma==0) cout<<0;
    return 0;
}