1081 Rational Sum

最新推荐文章于 2021-03-02 14:14:56 发布

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最新推荐文章于 2021-03-02 14:14:56 发布

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分类专栏： PAT甲级题解文章标签： pat 分数的四则运算

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本文解析了1081RationalSum问题，即如何计算多个有理数的和，并提供了一段C++代码实现。输入包括一系列正负有理数，输出为最简形式的整数部分和分数部分。代码中使用了最大公因数和最小公倍数的方法来简化分数。

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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
回头再补充详细解析。
参考代码：

#include<iostream>
#include<vector>
using namespace std;

int gcd(int c1, int c2)   { //最大公因数
	if (c1 < 0)c1 = c1 * (-1);
	if (c2 < 0)c2 = c2 * (-1);
	int a ,b, r;
	a = c1 >= c2 ? c1 : c2;
	b = c1 < c2 ? c1 : c2;
	r = a % b;
	while (r!=0){
		a = b;
		b = r;
		r = a % b;
	}
	return b;
}
int lcm(int a, int b){   //最小公倍数
	int factor = gcd(a, b);
	long long int sum = a * b;
	int temp_lcm = sum / factor;
	return temp_lcm;
}
int main(){
	int n, a, b, temp_gcd, c, d, fenzhi, fenmu;
	cin >> n;
	scanf_s("%d/%d", &a, &b);
	if (a != 0) {
		temp_gcd = gcd(a, b);
		a = a / temp_gcd;
		b = b / temp_gcd;
	}
	for (int i = 1; i < n; i++) {
		scanf_s("%d/%d", &c, &d);
		if (c != 0 && d != 0) {
			temp_gcd = gcd(c, d);
			c = c / temp_gcd;
			d = d / temp_gcd;
		}
		int temp_lcm = lcm(b, d);
		a=fenzhi = temp_lcm / b * a + temp_lcm / d * c;
		b=fenmu = temp_lcm;
		if (fenzhi != 0 && fenmu != 0) {
			temp_gcd = gcd(fenzhi, fenmu);
			a = fenzhi / temp_gcd;
			b = fenmu / temp_gcd;
		}
	}
	int xishu = a / b, flag = 0;
	fenzhi = a % b;
	if (xishu != 0)
	{
		cout << xishu;
		flag = 1;
	}
	if (fenzhi != 0) {
		if (flag)cout << " ";
		cout << fenzhi << "/" << b;
	}
	if (xishu == 0 && fenzhi == 0)
		cout << "0";

	return 0;
}

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