PAT (Advanced Level) Practise 1081 Rational Sum(20)

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本文介绍了一个算法，用于计算多个分数的总和，并将其简化为最简形式。输入包括一系列分数，输出则是这些分数的和，以最简形式表示，包括整数部分和分数部分。

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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

题意：给你n个分数，求出它们的和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

LL a[105], b[105];
int n;

LL gcd(LL x, LL y)
{
	if (x > y) swap(x, y);
	while (y%x)
	{
		LL k = y%x;
		y = x;
		x = k;
	}
	return x;
}

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++) scanf("%lld/%lld", &a[i], &b[i]);
		LL ans1=0,ans2=1;
		for (int i = 1; i <= n; i++)
		{
			LL k = gcd(ans2, b[i]);
			ans2 = b[i] * ans2 / k;
		}
		for (int i = 1; i <= n; i++) ans1 += ans2 / b[i] * a[i];
		if (!ans1) { printf("0\n"); continue; }
		LL x = abs(ans1);
		LL k = gcd(x, ans2);
		ans1 /= k, ans2 /= k;
		if (ans1%ans2 == 0) printf("%lld\n", ans1 / ans2);
		else
		{
			if (abs(ans1) < ans2) printf("%lld/%lld\n", ans1, ans2);
			else
			{
				if (ans1 > 0) printf("%lld %lld/%lld\n", ans1 / ans2, ans1%ans2, ans2);
				else printf("%lld -%lld/%lld\n",(LL)(ans1 / ans2), abs(ans1)%ans2, ans2);
			}
		}
	}
	return 0;
}