1081. Rational Sum (20)【模拟】——PAT (Advanced Level) Practise

题目信息

1081. Rational Sum (20)

时间限制400 ms

内存限制65536 kB

代码长度限制16000 B

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

解题思路

模拟加法，注意结果中分母负号和分子为0情况

AC代码

#include <cstdio>
struct node{
    long long a, b;
    node(long long a, long long b):a(a), b(b){}
};
long long gcd(long long a, long long b){
    return b ? gcd(b, a%b) : a;
}
node add(node& a, node& b){
    node r(a.a*b.b + a.b*b.a, a.b*b.b);
    long long t = gcd(r.a, r.b);
    r.a /= t;
    r.b /= t;
    return r;
}
int main()
{
    int n;
    long long a, b;
    scanf("%d", &n);
    scanf("%lld/%lld", &a, &b);
    node p(a, b);
    for (int i = 1; i < n; ++i){
        scanf("%lld/%lld", &a, &b);
        node t(a, b);
        p = add(p, t);
    }
    if (p.b < 0){
        p.b *= -1;
        p.a *= -1;
    }
    if (p.a >= p.b){
        printf("%lld", p.a/p.b);
        if (p.a % p.b){
            printf(" %lld/%lld", p.a%p.b, p.b);
        }
    }else{
        if (p.a == 0){
            printf("0");
        }else{
            printf("%lld/%lld", p.a, p.b);
        }
    }
    printf("\n");
    return 0;
}