PAT 1081 Rational Sum[分子求和][比较]

1081 Rational Sum （20 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

（20 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

题目大意：给出几个分数求和。

//又是分数求和的题目，细节会很多啊。

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

int n;
int nums[];

int getSum(int s,int f){
    if(f>=n){//如果已经到了尽头，那么就对s约分并且输出
        //还是需要先根据分母找出所有的最小公倍数呢？
        //一点也不会。写不下去。
    }
}

int main() {

    cin>>n;
    for(int i=;i<n;i++){
        cin>>nums[i];
    }
    cout<<getSum(,);

    return ;
}

代码转自：https://www.liuchuo.net/archives/2108

#include <iostream>
#include <cstdlib>
using namespace std;
long long gcd(long long a, long long b) {return b ==  ? abs(a) : gcd(b, a % b);}
int main() {
    long long n, a, b, suma = , sumb = , gcdvalue;//注意数据类型的定义。
    scanf("%lld", &n);
    for(int i = ; i < n; i++) {
        scanf("%lld/%lld", &a, &b);
        gcdvalue = gcd(a, b);//找到最大公约数进行约分。
        a = a / gcdvalue;
        b = b / gcdvalue;
        suma = a * sumb + suma * b;//分子
        sumb = b * sumb;//计算分母
        gcdvalue = gcd(suma, sumb);
        sumb = sumb / gcdvalue;//再约分。
        suma = suma / gcdvalue;
    }
    long long integer = suma / sumb;
    suma = suma - (sumb * integer);//计算余下的分子。
    if(integer != ) {
        printf("%lld", integer);
        if(suma != ) printf(" ");//如果余下的分子不为0.
    }
    if(suma != )
        printf("%lld/%lld", suma, sumb);
    if(integer ==  && suma == )
        printf("");
    return ;
}

//学习了，需要经常复习吧，要不还是不会。