PTA(Advanced Level)1081.Rational Sum

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

思路

• 分数化简的思路——求分母和分子的最大公约数，同时÷最大公约数
• 分数相加的思路——求两个分母的最大公倍数，分子乘以对应的因子，相加即可
• 设(a,b)的最大公约数为(d)，有性质：(a*b/d = 最大公倍数)

代码

#include<bits/stdc++.h>
using namespace std;
long int a[101];
long int b[101];
long int gcd(long int a, long int b)
{
	if(b == 0)	return a;
	else return gcd(b, a%b);
}	//求最大公约数
void reduction(long int a[], long int b[], int i)
{
	int d = gcd(abs(a[i]), abs(b[i]));
	a[i] /= d;
	b[i] /= d;
}	//根据最大公约数化简分子分母
int main()
{
	int n;
	scanf("%d", &n);
	for(int i=1;i<=n;i++)
	{
		scanf("%ld/%ld", &a[i], &b[i]);
		if(a[i] == 0)	b[i] = 1;	//为了不让求最大公约数出错，让实际为0的分数的分母弄为1——1和任何数的最大公约数都是1
		reduction(a, b, i);
	}
	a[0] = 0; b[0] = 1;   //对第i位的处理是让第i位和第i-1位来求和，所以让第0位表示为0
	for(int i=1;i<=n;i++)
	{
		long int gcd_value = gcd(b[i], b[i-1]);
		long int lcm_value = b[i] / gcd_value * b[i-1];		//先/后*减少溢出可能
		a[i-1] *= lcm_value / b[i-1];
		a[i] *= lcm_value /b[i];
		a[i] = a[i] + a[i-1];
		b[i] = lcm_value;
		reduction(a,b,i);
	}
	if(abs(a[n]) >= abs(b[n]))		//这里要用绝对值是因为会有负数的情况出现，比如-1/1
	{
		int t = a[n] / b[n];  
		int up = a[n] - t * b[n];
		if(up == 0)
			printf("%ld
", t);
		else
			printf("%ld %ld/%ld
", t, up, b[n]);
	}else
    {
        if(a[n] == 0)
            printf("0
");		//这里不单独处理的话，是会输出0/n这种类型的数字的
        else
            printf("%ld/%ld
", a[n], b[n]);
    }


	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805386161274880