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;}