PAT 1034 有理数四则运算 C语言实现

本题要求编写程序，计算 2 个有理数的和、差、积、商。

输入格式：

输入在一行中按照 a1/b1 a2/b2 的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为 0。

输出格式：

分别在 4 行中按照 有理数1 运算符 有理数2 = 结果 的格式顺序输出 2 个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式 k a/b，其中 k 是整数部分，a/b 是最简分数部分；若为负数，则须加括号；若除法分母为 0，则输出 Inf。题目保证正确的输出中没有超过整型范围的整数。

输入样例 1：

2/3 -4/2

输出样例 1：

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

输入样例 2：

5/3 0/6

输出样例 2：

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

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

void yue_fen(long long int *fen_zi,long long int *fen_mu,long long int *k)		
{
	*k = *fen_zi / *fen_mu;
	if(*k != 0)
	{
		*fen_zi = llabs(*fen_zi % *fen_mu);
	}
	else
	{
		*fen_zi = *fen_zi % *fen_mu;	
	}
		*fen_mu = llabs(*fen_mu);
	for(int i = 10 ; i >= 2; i--)
	{
		if(0 == (*fen_zi % i) && 0 == (*fen_mu % i))
		{
				*fen_zi = *fen_zi / i;
				*fen_mu = *fen_mu / i;
				i = 11;
		}
	}
}

int main()
{
	long long int a1 = 0, b1 = 0, a2 = 0, b2 = 0,k1 = 0,k2 = 0;
	long long int resu_k[4]= {},resu_a[4] = {},resu_b[4] = {};
	scanf("%lld/%lld %lld/%lld",&a1,&b1,&a2,&b2);
	
	char symbol[4] = {'+','-','*','/'};
	
	//和、差
	if(0 == a1)
	{
		resu_a[0] = a2;
		resu_b[0] = b2;
		resu_a[1] = a2;
		resu_b[1] = b2;
	}
	else if(0 == a2)
	{
		resu_a[0] = a1;
		resu_b[0] = b1;
		resu_a[1] = a1;
		resu_b[1] = b1;
	}
	else
	{
		resu_a[0] = a1 * b2 + a2 * b1;
		resu_b[0] = b1 * b2;
		resu_a[1] = a1 * b2 - a2 * b1;
		resu_b[1] = b1 *b2;
	}
	yue_fen(&resu_a[0],&resu_b[0],&resu_k[0]);
	yue_fen(&resu_a[1],&resu_b[1],&resu_k[1]);
	
	//积
	if(0 == a1 || 0 == a2)
	{
		resu_a[2] = 0;
		resu_k[2] = 0;
	}
	else
	{
		resu_a[2] = a1 * a2;
		resu_b[2] = b1 * b2;
		yue_fen(&resu_a[2],&resu_b[2],&resu_k[2]);
	}
	
	//商
	int mark = 0;
	if(0 == a1 || 0 == a2)
	{
		mark = 1;	
	}
	else if(a1 < 0 && a2 >0 || (a2 < 0 && a1 > 0))
	{
		resu_a[3] = -1*llabs(a1) * llabs(b2);
		resu_b[3] = llabs(a2)* llabs(b1);
		yue_fen(&resu_a[3],&resu_b[3],&resu_k[3]);
	}
	else
	{
		resu_a[3] = -1 * a1 * b2;
		resu_b[3] = -1 * a2 * b1;
		yue_fen(&resu_a[3],&resu_b[3],&resu_k[3]);
	}
	
	//分子分母有理化
	yue_fen(&a1,&b1,&k1);
	yue_fen(&a2,&b2,&k2);
	
	
	for(int i = 0 ; i < 4 ; i++)				//分别输出四则运算
	{
		if(0 == a1)								//判断第一个有理数
		{
			if(k1 < 0)
				printf("(%lld)",k1);
			else
				printf("%lld",k1);
		}
		else
		{
			if(0 == k1)
			{
				if(a1 > 0)
					printf("%lld/%lld",a1,b1);
				else
					printf("(%lld/%lld)",a1,b1);
			}
			else if(k1 < 0)
				printf("(%lld %lld/%lld)",k1,a1,b1);
			else
				printf("%lld %lld/%lld",k1,a1,b1);		
		}
		printf(" %c ",symbol[i]);

		if(0 == a2)								//判断第二个有理数
		{
			if(k2 < 0)
				printf("(%lld)",k2);
			else
				printf("%lld",k2);
		}
		else
		{
			if(0 == k2)
			{
				if(a2 > 0)
					printf("%lld/%lld",a2,b2);
				else
					printf("(%lld/%lld)",a2,b2);
			}
			else if(k2 < 0)
				printf("(%lld %lld/%lld)",k2,a2,b2);
			else
				printf("%lld %lld/%lld",k2,a2,b2);
		}
		printf(" = ");
		
		if(3 == i && 1 == mark)
		{
			printf("Inf\n");	
		}
		else if(0 == resu_a[i])								//判断结果
		{
			if(resu_k[i] < 0)
				printf("(%lld)\n",resu_k[i]);
			else
				printf("%lld\n",resu_k[i]);
		}
		else
		{
			if(resu_k[i] < 0)
				printf("(%lld %lld/%lld)\n",resu_k[i],resu_a[i],resu_b[i]);
			else if(0 == resu_k[i] && resu_a[i] > 0)
				printf("%lld/%lld\n",resu_a[i],resu_b[i]);
			else if(0 == resu_k[i] && resu_a[i] < 0)
				printf("(%lld/%lld)\n",resu_a[i],llabs(resu_b[i]));
			else
				printf("%lld %lld/%lld\n",resu_k[i],resu_a[i],resu_b[i]);
		}
	}

	return 0;	
}

标签：PAT,fen,resu,a1,else,a2,C语言,1034,lld
来源： https://blog.csdn.net/weixin_45393375/article/details/99601125