PAT 1034. 有理数四则运算

PAT 1034. 有理数四则运算

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

分析

自己写的程序如下，测试点2未过，如有大神路过，请指点一二，万分感谢

自己的代码

#include<iostream>
#include<math.h>
using namespace std;
long long int inf=0,flag3=1,int3,son3,mom3;
void print(long long int flag,long long int Int,long long int son,long long int mom){
	if(inf==1){
	cout<<"Inf"; return;
	}
	if(flag==0) 
	cout<<0;
	else if(flag<0){
	if(Int==0) 
	cout<<"("<<-son<<"/"<<mom<<")";
	else if(Int!=0&&son!=0)
	cout<<"("<<-Int<<" "<<son<<"/"<<mom<<")";
	else 
	cout<<"("<<-Int<<")";
	}
	else if(flag>0){
	if(Int==0) cout<<son<<"/"<<mom;
	else if(Int!=0&&son!=0)cout<<Int<<" "<<son<<"/"<<mom;
	else cout<<Int; 
	}	
	flag3=1;
}
long long int gcd(long long int t1,long long int t2) {
    return t2 == 0 ? t1 : gcd(t2, t1 % t2);
}
void huajian(long long int &son,long long int &mom,long long int &Int,long long int &flag){
	long long int s=abs(son);
	long long int temp=gcd(s,mom);
	son/=temp; mom/=temp;
	if(son<0) {flag=-1; son=-son;}
	else if(son==0) flag=0;
	Int=son/mom; son=son%mom;
} 
void jia(long long int s1,long long int m1,long long int s2,long long int m2){
	mom3=m1*m2;
	son3=m1*s2+m2*s1;
}
void jian(long long int s1,long long int m1,long long int s2,long long int m2){
	mom3=m1*m2;
	son3=s1*m2-s2*m1;
}
void chen(long long int s1,long long int m1,long long int s2,long long int m2){
	mom3=m1*m2;
	son3=s1*s2;
}
void chu(long long int s1,long long int m1,long long int s2,long long int m2){
	if(s2==0) inf=1;
	else{
		son3=s1*m2;
		mom3=s2*m1;
	}
	if(son3*mom3>0) {son3=abs(son3); mom3=abs(mom3); }
	else if(son3*mom3<0){son3=-abs(son3); mom3=abs(mom3);} 
}
int main(){
	long long int mom1,son1,int1,mom2,son2,int2,flag1=1,flag2=1;
	scanf("%lld/%lld %lld/%lld",&son1,&mom1,&son2,&mom2);
	char fuhao[4]={'+','-','*','/'};
	long long int m1=mom1,s1=son1,m2=mom2,s2=son2;
	huajian(son1,mom1,int1,flag1);
	huajian(son2,mom2,int2,flag2);
	for(int i=0;i<4;i++){
	print(flag1,int1,son1,mom1);
	cout<<" "<<fuhao[i]<<" ";
	print(flag2,int2,son2,mom2);
	cout<<" = ";
	switch(i){
		case 0:jia(s1,m1,s2,m2); break;
		case 1:jian(s1,m1,s2,m2); break;
		case 2:chen(s1,m1,s2,m2); break;
		case 3:chu(s1,m1,s2,m2); break;
		default:;
	}
	huajian(son3,mom3,int3,flag3);
	print(flag3,int3,son3,mom3);
	cout<<endl;
	}
	return 0;
} 

目前看过的本题最好的代码

#include<iostream>
#include<math.h>
using namespace std;
long long int getgcd(long long int s,long long int m){
	int r;
	while(r=s%m){
		s=m;
		m=r;
	}
	return m; 
}
void print(long long int s,long long int m){
	int sign=1;
	if(m==0){
	cout<<"Inf";
	return;	
	}
	if(s<0){
	  s=abs(s);
	  sign*=-1;
	}
	if(m<0){
	  m=abs(m);
	  sign*=-1;
	}
	int gcd=getgcd(s,m);
	s/=gcd;
	m/=gcd;
	if(sign<0) cout<<"(-";
	if(m==1) cout<<s;
	else if(s>m) cout<<s/m<<" "<<s%m<<"/"<<m;
	else cout<<s<<"/"<<m;
	if(sign<0) cout<<")";
}
int main(){
	long long int s1,m1,s2,m2;
	char op[4]={'+','-','*','/'};
	scanf("%lld/%lld %lld/%lld",&s1,&m1,&s2,&m2);
	for(int i=0;i<4;i++){
		print(s1,m1);
		cout<<" "<<op[i]<<" ";
		print(s2,m2);
		cout<<" = ";
		switch(i){
			case 0:print(s1*m2+s2*m1,m1*m2); break;
			case 1:print(s1*m2-s2*m1,m1*m2); break;
			case 2:print(s1*s2,m1*m2); break;
			case 3:print(s1*m2,m1*s2); break;
		}
		cout<<endl;
	}
	return 0;
}