PAT乙级考试-1034 有理数四则运算 (20分)

题目


思路
本题的关键是要知道怎么求最大公约数来进行化简，这里采用辗转相除法来计算。知道如何化简后，依次录入，化简，计算，打印即可

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

typedef struct
{
	long long int fz;
	long long int fm;
	long long int zs;
}number;

long long int gy(long long int i, long long int j)
{
	long long int k;
	while (i % j != 0)
	{
		k = i % j;
		i = j;
		j = k;
	}
	return abs(j);
}

void trans(char temp[], number* a)
{
	char* flag = strchr(temp, '-');
	long long int i = 0, j = 0;
	long long int k = flag ? 1 : 0;
	for (; temp[k] != '/'; k++)
		i = i * 10 + temp[k] - '0';
	for (k = k + 1; temp[k] != '\0'; k++)
		j = j * 10 + temp[k] - '0';
	long long int e = gy(i, j);
	k = flag ? -1 : 1;
	i = (i * k) / e;
	j = j / e;
	a->zs = i / j;
	a->fz = i % j;
	a->fm = j;
}

void print(number a)
{
	long long int flag = 0;
	if (a.zs == 0 && a.fz == 0)
	{
		printf("0");
		return;
	}
	if (a.fm == 0)
	{
		printf("Inf");
		return;
	}
	if (a.fz < 0 || a.zs < 0)
	{
		flag = 1;
		printf("(-");
		a.zs = -a.zs;
		a.fz = -a.fz;
	}
	if (a.zs != 0)
		printf("%d", a.zs);
	if (a.fz != 0)
	{
		if (a.zs != 0)
			printf(" ");
		if (a.fm == 1)
			printf("%d", a.fz);
		else
			printf("%d/%d", a.fz, a.fm);
	}
	if (flag)
		printf(")");
}

number compute(number a, number b, char ch)
{
	long long int z, m, l, e, temp;
	a.fz = a.zs * a.fm + a.fz;
	b.fz = b.zs * b.fm + b.fz;
	number c;
	if (ch == '+')
	{
		m = a.fm * b.fm;
		z = a.fz * b.fm + b.fz * a.fm;
	}
	else if (ch == '-')
	{
		m = a.fm * b.fm;
		z = a.fz * b.fm - b.fz * a.fm;
	}
	else if (ch == '*')
	{
		m = a.fm * b.fm;
		z = a.fz * b.fz;
	}
	else
	{
		if (b.fz == 0)
			m = 0;
		else
		{
			if (b.fz > 0)
			{
				temp = b.fz;
				b.fz = b.fm;
				b.fm = temp;
			}
			else
			{
				temp = b.fz;
				b.fz = -b.fm;
				b.fm = -temp;
			}
			m = a.fm * b.fm;
			z = a.fz * b.fz;
		}
	}
	if (m != 0)
	{
		e = gy(z, m);
		z = z / e;
		m = m / e;
		c.zs = z / m;
		c.fz = z % m;
		c.fm = m;
	}
	else
		c.fm = 0;
	return c;
}

 int main(void)
{
	char temp1[50], temp2[50];
	scanf("%s%s", temp1, temp2);
	number a, b, c;
	trans(temp1, &a);
	trans(temp2, &b);
	char op[4] = { '+','-','*','/' };
	for (long long int i = 0; i < 4; i++)
	{
		print(a);
		printf(" %c ", op[i]);
		print(b);
		c = compute(a, b, op[i]);
		printf(" = ");
		print(c);
		printf("\n");
	}
	return 0;
}