1081. Rational Sum (20)-PAT甲级真题

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:

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

分析：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～
以及：abs()在stdlib.h头文件里面

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～）

#include <iostream>
#include <cstdlib>
using namespace std;
long long gcd(long long a, long long b) {return b == 0 ? abs(a) : gcd(b, a % b);}
int main() {long long n, a, b, suma = 0, sumb = 1, gcdvalue;scanf("%lld", &n);for(int i = 0; i < n; i++) {scanf("%lld/%lld", &a, &b);gcdvalue = gcd(a, b);a = a / gcdvalue;b = b / gcdvalue;suma = a * sumb + suma * b;sumb = b * sumb;gcdvalue = gcd(suma, sumb);sumb = sumb / gcdvalue;suma = suma / gcdvalue;}long long integer = suma / sumb;suma = suma - (sumb * integer);if(integer != 0) {printf("%lld", integer);if(suma != 0) printf(" ");}if(suma != 0)printf("%lld/%lld", suma, sumb);if(integer == 0 && suma == 0)printf("0");return 0;
}

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