PAT.A1081 Rational Sum

最新推荐文章于 2022-01-22 14:34:45 发布

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本文介绍了一个用于计算多个有理数之和的算法。输入包括一系列形式为“分子/分母”的有理数，并通过该算法计算这些有理数的总和。输出结果以最简形式给出，包括整数部分及分数部分。

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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

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <iostream>
using namespace std;
struct frac {
	int up, down;
}a[110],sum;
int gcd(int a, int b) {
	if (b == 0) return a;
	else return gcd(b, a%b);
}
frac huajian(frac f) {
	if (f.down < 0) {
		f.up = -f.up;
		f.down = -f.down;
	}
	if (f.up == 0) {
		f.down = 1;
	}
	else {
		int d = gcd(abs(f.up), abs(f.down));
		f.up /= d;
		f.down /= d;
	}
	return f;
}
int main(){
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d/%d", &a[i].up, &a[i].down);
		if (i == 0) {
			sum.up = a[i].up;
			sum.down = a[i].down;
		}
		else {
			sum.up = sum.up*a[i].down + a[i].up*sum.down;
			sum.down = a[i].down*sum.down;
		}
       }
	//printf("%d/%d", sumup, sumdo);
	sum=huajian(sum);
	if (sum.down == 1) printf("%lld", sum.up);
	else if (abs(sum.up) > sum.down) {
		printf("%d %d/%d", sum.up / sum.down, abs(sum.up)%sum.down, sum.down);
	}
	else printf("%d/%d", sum.up, sum.down);
	return 0;
	
}