题目描述
Given N rational numbers in the form numerator/denominator, you are supposed to calculate their sum.
输入
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.
输出
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.
思路
分数的加减法，注意约分（最大公约数）
代码
#include<iostream>
#include<cstdio>
#include<stdlib.h>
using namespace std;
typedef struct node {
int num;
int den;
}Ration;
Ration ration[1000];
void yue(int i)
{
int a = ration[i].num;
int b = ration[i].den;
int c = ration[i].num % ration[i].den;
while (c)
{
a = b;
b = c;
c = a % b;
}
ration[i].num /= b;
ration[i].den /= b;
}
int main()
{
int N;
cin >> N;
for (int i = 0; i < N; i++)
{
scanf("%d/%d", &ration[i].num, &ration[i].den);
}
ration[N].num = 0;
ration[N].den = 1;
for (int i = 0; i < N; i++)
{
ration[i + 1].num = ration[i].num * ration[i + 1].den + ration[i + 1].num * ration[i].den;
ration[i + 1].den = ration[i].den * ration[i + 1].den;
yue(i + 1);
}
int y = ration[N].num / ration[N].den;
ration[N].num = ration[N].num % ration[N].den;
if (y > 0)
{
printf("%d", y);
}
if (ration[N].num != 0)
{
if (y != 0)
{
printf(" ");
}
printf("%d/%d\n", ration[N].num, ration[N].den);
}
else
{
if (y == 0)
{
printf("0\n");
}
}
}