1081. Rational Sum (20)

最新推荐文章于 2019-02-06 16:53:54 发布

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最新推荐文章于 2019-02-06 16:53:54 发布

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该博客介绍了PAT竞赛中1081题——有理数求和的问题，要求计算给定不超过100个有理数的和，并以最简形式输出。内容包括问题解析、解题策略，特别是如何找到两个整数的最大公约数（GCD）以简化结果，提供了一个C++解决方案，包含递归和非递归两种求GCD的方法。

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1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

这道题是简单题，目的是求有理数的加法，并且按照要求输出结果，题目的关键是求两个整数的最大公约数和最小公倍数，代码如下，同时附上链接递归及非递归求最大公约数

#include<iostream>
#include <vector>

using namespace std;

//最大公约数
long long gcd( long long x , long long y){
	long long max,min,temp;
	max = x > y ? x : y ;
	min = x < y ? x : y ;
	while( max % min ){

		temp = max % min;
		max = min;
		min = temp;
	}
	return min;
}
//最小公倍数
long long lcm( long long x , long long y ){
	return x*y/gcd(x,y);
}

int main(){

	vector<long long> numerator,denominator,factor;
	char c;
	long long a,b,temp,sum = 0;
	int n;
	cin>>n;

	for ( int i = 0; i < n ; i++ ){
		cin>>a>>c>>b;
		numerator.push_back(a);
		denominator.push_back(b);
	}
	temp = denominator[0];
	for ( int i = 1; i < denominator.size() ; i++ ){
		temp = lcm( temp , denominator[i] );
	}
	for ( int i = 0; i < denominator.size() ; i++ ){
		factor.push_back(temp/denominator[i]);
	}
	for ( int i = 0; i < numerator.size() ; i++ ){
		numerator[i]=numerator[i]*factor[i];
		sum += numerator[i];
	}
	if( sum%temp && sum/temp )
		cout<<(sum/temp)<<" "<<((sum%temp)/gcd((sum%temp),temp))<<"/"<<(temp/gcd((sum%temp),temp));
	else if( sum%temp && !(sum/temp) )
		cout<<((sum%temp)/gcd((sum%temp),temp))<<"/"<<(temp/gcd((sum%temp),temp));
	else
		cout<<(sum/temp);

	return 0;
}