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Python无法通过？？？

发表于 2017-10-10 21:15:22 回复(1)

SKT.RUO

#include<stdio.h>
int gcd(int x, int y)
{
	if (y%x == 0)
		return x;
	return gcd(y%x, x);
}
int main()
{
	int nom1, nom2, den1, den2,
		den, newNom1, newNom2,
		tmpNom, tmpDen, tmpGcd,
		resNom, resDen;
	char op;
	while (~scanf("%d/%d%d/%d %c", &nom1, &den1,&nom2,&den2,&op))
	{
		den = den1*den2 / gcd(den1, den2);
		newNom1 = nom1*den / den1;
		newNom2 = nom2*den / den2;
		switch (op)
		{
		case '+':
			tmpNom = newNom1 + newNom2;
			tmpDen = den;
			break;
		case '-':
			tmpNom = newNom1 - newNom2;
			tmpDen = den;
			break;
		case '*':
			tmpNom = nom1*nom2;
			tmpDen = den1*den2;
			break;
		case '/':
			tmpNom = nom1*den2;
			tmpDen = den1*nom2;
			break;
		}
		tmpGcd = gcd(tmpNom, tmpDen);
		resNom = tmpNom / tmpGcd;
		resDen = tmpDen / tmpGcd;
		resNom = resNom*resDen < 0 ? -abs(resNom) : abs(resNom);
		resDen = abs(resDen);
		printf("%d/%d\n",resNom,resDen);
	}
	return 0;
}

发表于 2017-08-04 13:35:54 回复(0)

MemoryC

实现思路：

1、构造分数类；

2、实现四则运算；

3、利用 “辗转相除法 ”思想构造求最大公约数的算法；

4、利用最大公约数进行月份简化；

5、输出；

另外需要注意的是，尽量减少分数简化（约分）次数，我之前几次超时，原因是构造分数时就进行一次约分。实际上只需要在输出前约分就行；

以下代码仅供参考：

import java.util.Scanner;
/**
 * @author MemoryC
 * */
public class Main {

    public static void main(String[] args) {

        Scanner scanner=new Scanner(System.in);

        while(scanner.hasNext()) {
            String[]aStrings=scanner.next().split("/");
            String[] bString=scanner.next().split("/");
            char operation=scanner.next().charAt(0);

            Fraction a=new Fraction(Integer.parseInt(aStrings[0]), Integer.parseInt(aStrings[1]));
            Fraction b=new Fraction(Integer.parseInt(bString[0]), Integer.parseInt(bString[1]));
            Fraction c=null;
            switch (operation) {
            case '+':
                c=Fraction.add(a, b);
                break;
            case '-':
                c=Fraction.cut(a, b);
                break;
            case '*':
                c=Fraction.multiply(a, b);
                break;
            case '/':
                c=Fraction.divide(a, b);
                break;
            default:
                break;
            }
            c.Simplify();
            System.out.println(c);
        }
        scanner.close();
    }

    static class Fraction{

        int mom=1;
        int son=0;
        boolean negtive=false;

        public Fraction(int son,int mom) {

            if(son<0) {
                son=-son;
                negtive=true;
            }else if(son==0) {
                mom=1;
            }

            this.son=son;
            this.mom=mom;
        }

        public String toString() {
            return negtive?("-"+son+"/"+mom):(son+"/"+mom);
        }

        static int getCommonDivisor(int a,int b) {
            if(a<b) {
                return getCommonDivisor(b,a);
            }
            int aa=a;
            int bb=b;

            while(aa%bb!=0) {
                int r=aa%bb;
                aa=bb;
                bb=r;
            }
            return bb;
        }
        void Simplify() {

            if(son==mom) {
                son=1;
                mom=1;
            }else {
                int common=getCommonDivisor(son, mom);
                son=son/common;
                mom=mom/common;
            }

        }

        static Fraction add(Fraction a,Fraction b) {
            int am=a.mom,as=a.son,
                bm=b.mom,bs=b.son,
                cm=0,cs=1;
            if(a.negtive) {
                as=-as;
            }
            if(b.negtive) {
                bs=-bs;
            }
            cm=am*bm;
            cs=as*bm+bs*am;

            return new Fraction(cs,cm);
        }

        static Fraction cut(Fraction a,Fraction b) {
            int am=a.mom,as=a.son,
                bm=b.mom,bs=-b.son,
                cm=0,cs=1;
            if(a.negtive) {
                as=-as;
            }
            if(b.negtive) {
                bs=-bs;
            }
            cm=am*bm;
            cs=as*bm+bs*am;

            return new Fraction(cs,cm);
        }

        static Fraction multiply(Fraction a,Fraction b) {
            int am=a.mom,as=a.son,
                bm=b.mom,bs=b.son,
                cm=0,cs=1;
            if(a.negtive) {
                as=-as;
            }
            if(b.negtive) {
                bs=-bs;
            }
            cm=am*bm;
            cs=as*bs;

            return new Fraction(cs,cm);
        }

        static Fraction divide(Fraction a,Fraction b) {
            int am=a.mom,as=a.son,
                bm=b.son,bs=b.mom,
                cm=0,cs=1;
            if(a.negtive) {
                as=-as;
            }
            if(b.negtive) {
                bs=-bs;
            }
            cm=am*bm;
            cs=as*bs;

            return new Fraction(cs,cm);
        }
    }
}

编辑于 2018-08-03 16:55:38 回复(3)

思晓

#include <stdio.h>
 
int gcd(int x,int y)  //求最大公约数
{
    if(y==0)
        return x;
    else
        return gcd(y,x%y);
}
 
void add(int a1,int a2,int b1,int b2)
{
    int n1=a1*b2+a2*b1;
    int n2=a2*b2;
    int temp=gcd(n1,n2);
    if(temp<0)
        temp=-temp;
    printf("%d/%d\n",n1/temp,n2/temp);
}
 
void mul(int a1,int a2,int b1,int b2)
{
    int n1=a1*b1;
    int n2=a2*b2;
    int temp=gcd(n1,n2);
    if(temp<0)
        temp=-temp;
    printf("%d/%d\n",n1/temp,n2/temp);
}
 
int main()
{
    int a1,a2,b1,b2;   //第一个数和第二个数得分子和分母
    char ch;
    while(scanf("%d/%d %d/%d %c",&a1,&a2,&b1,&b2,&ch)!=EOF)
    {
        //printf("%d %d %d %d\n",a1,a2,b1,b2);
        switch(ch)
        {
            case '+':
                {
                    add(a1,a2,b1,b2);
                    break;
                }
            case '-':
                {
                    add(a1,a2,-b1,b2);
                    break;
                }
            case '*':
                {
                    mul(a1,a2,b1,b2);
                    break;
                }
            case '/':
                {
                    mul(a1,a2,b2,b1);
                    break;
                }
        }
    }
    return 0;
}

发表于 2020-04-14 22:06:08 回复(0)

小黯

#include<stdio.h>
int gcd(int x,int y)
{
	return !y? x>0?x:-x:gcd(y,x%y);
}
int main()
{
	int a1,b1,a2,b2;
	char c;
	while(scanf("%d/%d %d/%d %c",&a1,&b1,&a2,&b2,&c)!=-1)
	{
		int A,B,C;
		switch(c)
		{
			case '+':
				A=a1*b2+a2*b1;
				B=b1*b2;
				C=gcd(A,B);
				printf("%d/%d\n",A/C,B/C);
				break;
			case '-':
				A=a1*b2-a2*b1;
				B=b1*b2;
				C=gcd(A,B);
				printf("%d/%d\n",A/C,B/C);
				break;
			case '*':
				A=a1*a2;
				B=b1*b2;
				C=gcd(A,B);
				printf("%d/%d\n",A/C,B/C);
				break;
			case '/':
				A=a1*b2;
				B=a2*b1;
				C=gcd(A,B);
				B<0?printf("%d/%d\n",-A/C,-B/C):printf("%d/%d\n",A/C,B/C);
				break;
			default:break;
		}
	}
	return 0;
}

发表于 2020-04-09 16:40:17 回复(0)

熊朝春


	//水题，不过做这感觉比较爽：）



	#include <stdio.h>



	int Simplify(inta, intb);



	int main()



	{



	    int numer0, denom0, numer1, denom1;



	    int numer2, denom2;



	    char oper[1];



	    int res = 0;



	    while(scanf("%d/%d %d/%d %c", &numer0, &denom0, &numer1, &denom1, &oper[0])!=EOF)



	    {



	        if(oper[0]=='+'||oper[0]=='-')



	        {



	            denom2 = denom0*denom1;



	            if(oper[0]=='+')



	            {



	                numer2 = numer0*denom1 + numer1*denom0;



	            }



	            else



	            {



	                numer2 = numer0*denom1 - numer1*denom0;



	            }



	             



	        }



	        elseif(oper[0]=='*'||oper[0]=='/')



	        {



	            if(oper[0]=='*')



	            {



	                numer2 = numer0*numer1;



	                denom2 = denom0*denom1;



	            }



	            else



	            {



	                numer2 = numer0*denom1;



	                denom2 = denom0*numer1;



	            }



	        }



	        res = Simplify(numer2, denom2);



	        denom2 /= res;



	        numer2 /= res;



	        if(denom2<0)



	        {



	            printf("-%d/%d\n", numer2, -denom2);



	        }



	        else



	        {



	            printf("%d/%d\n", numer2, denom2);



	        }



	         



	    }



	    return0;



	}



	int Simplify(int a, int b)



	{



	    int t;



	    while(b!=0)



	    {



	        t = a % b;



	        a = b;



	        b = t;



	    }



	    return a;



	}

发表于 2017-11-27 14:21:41 回复(0)

LuxSun


	//不服，给我一个超时的理由！



	importjava.util.Scanner;



	 



	publicclassMain {



	    publicstaticvoidmain(String[] args) {



	        Scanner in = newScanner(System.in);



	        while(in.hasNext()){



	            String s1=in.next(),s2=in.next(),op=in.next();



	            String[] split1 = s1.split("/"),split2 = s2.split("/");



	            intup1=Integer.valueOf(split1[0]),down1=Integer.valueOf(split1[1]);



	            intup2=Integer.valueOf(split2[0]),down2=Integer.valueOf(split2[1]);



	             



	            intminComm = down1*down2/maxCommon(down1,down2);



	            intup=0,down=0;



	             



	            switch(op) {



	            case"+":



	                inttimes1=minComm/down1,times2=minComm/down2;



	                down=minComm;



	                up=up1*times1+up2*times2;



	                intmaxComm = maxCommon(down,up);



	                up/=maxComm;



	                down/=maxComm;



	                break;



	            case"-":



	                times1=minComm/down1;



	                times2=minComm/down2;



	                down=minComm;



	                up=up1*times1-up2*times2;



	                if(up<0){



	                    maxComm = maxCommon(down,up*=-1);



	                    up/=maxComm;



	                    down/=maxComm;



	                    up*=-1;



	                }



	                elseif(up==0){



	                    System.out.println("ERROR");



	                }



	                else{



	                    maxComm = maxCommon(down,up);



	                    up/=maxComm;



	                    down/=maxComm;



	                }



	                break;



	            case"*":



	                up=up1*up2;



	                down=down1*down2;



	                maxComm = maxCommon(down,up);



	                up/=maxComm;



	                down/=maxComm;



	                break;



	            case"/":



	                up=up1*down2;



	                down=down1*up2;



	                maxComm = maxCommon(down,up);



	                up/=maxComm;



	                down/=maxComm;



	                break;



	            }



	             



	            System.out.printf("%d/%d\n",up,down);



	        }



	    }



	     



	    publicstaticintmaxCommon(intm,intn){



	        if(m<n){



	            inttemp=n;



	            n=m;



	            m=temp;



	        }



	        if(m%n==0)



	            returnn;



	        else



	            returnmaxCommon(n,m%n);



	    }



	}

发表于 2017-07-17 12:33:03 回复(1)

C/C++

cantin

#include <stdio.h>
int zdgys(int a,int b)
{
    int r;
    if(a==0)
    {
    	return a;
	}
	else
	{
    while(b!=0)
    {
        r=a%b;
        a=b;
        b=r;
    }
    return a;
    }
}
int zxgbs(int a,int b)
{
    return a*b/zdgys(a,b);
}
int main()
{
    char a[6],b[6],c,tab;
    int afz,afm,bfz,bfm,i,j,m,n,temp;
    while(scanf("%s%s%c%c",a,b,&tab,&c)!=EOF)
    {
		afz=0;afm=0;bfz=0;bfm=0;
        for(i=0;a[i]!='/';i++)
        {
            afz=afz*10+a[i]-'0';
        }
        for(;a[i]!='\0';i++)
        {
            if(a[i]>='0'&&a[i]<='9')
            {
                afm=afm*10+a[i]-'0';
            }
        }
        for(i=0;b[i]!='/';i++)
        {
            bfz=bfz*10+b[i]-'0';
        }
        for(;b[i]!='\0';i++)
        {
            if(b[i]>='0'&&b[i]<='9')
            {
                bfm=bfm*10+b[i]-'0';
            }
        }
        if(c=='+')
        {
            m=afz*zxgbs(afm,bfm)/afm+bfz*zxgbs(afm,bfm)/bfm;
            n=zxgbs(afm,bfm);
            printf("%d/%d\n",m/zdgys(m,n),n/zdgys(m,n));
        }
        else if(c=='-')
        {
            m=afz*zxgbs(afm,bfm)/afm-bfz*zxgbs(afm,bfm)/bfm;
            n=zxgbs(afm,bfm);
				if(m>0)
				{
					printf("%d/%d\n",m/zdgys(m,n),n/zdgys(m,n));
				}
				else if(m==0)
				{
					printf("0\n");
				}
				else if(m<0)
				{
					printf("-%d/%d\n",-m/zdgys(-m,n),n/zdgys(-m,n));
				}
        }
        else if(c=='*')
        {
            m=afz*bfz;n=afm*bfm;
				printf("%d/%d\n",m/zdgys(m,n),n/zdgys(m,n));
        }
        else if(c=='/')
        {
            temp=bfz;bfz=bfm;bfm=temp;
            m=afz*bfz;n=afm*bfm;
				printf("%d/%d\n",m/zdgys(m,n),n/zdgys(m,n));
        }
    }
    return 0;
}

发表于 2020-04-17 19:00:28 回复(0)

zzxds

有点长，但是很好理解

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a1,b1,a2,b2,g;
	char _1,_2,op;
	while(cin>>a1>>_1>>b1>>a2>>_2>>b2>>op)
		{
			switch(op)
			{
				case'+':a1=a1*b2+b1*a2;
						b1*=b2;
						g=abs(__gcd(a1,b1));
						cout<<to_string(a1/g)+"/"<<to_string(b1/g)<<endl;
						break;
				case'-':a1=a1*b2-b1*a2;
						b1*=b2;
						g=abs(__gcd(a1,b1));
						cout<<to_string(a1/g)+"/"<<to_string(b1/g)<<endl;
						break;	
				case'*':a1*=a2;
						b1*=b2;
						g=abs(__gcd(a1,b1));
						cout<<to_string(a1/g)+"/"<<to_string(b1/g)<<endl;
						break;
				case'/':
						a1*=b2;
						b1*=a2;
						g=abs(__gcd(a1,b1));
						cout<<to_string(a1/g)+"/"<<to_string(b1/g)<<endl;
						break;
				default:break;
			}	
		}
	return 0;
}

发表于 2020-04-13 10:50:09 回复(0)

pphkaa

本题感觉主要知识点在于***()函数求最大公约数，可以直接调用C++中的__***(a,b),

也可以自己写***()函数

#include<bits/stdc++.h>
using namespace std;
inline int stringint(string x)
{
	long long temp=0;
	int a=1;
	int length=x.size();
	for(int i=length-1;i>=0;i--)
	{
		temp+=a*(x[i]-'0');
		a*=10;
	}
	return temp;
	
}
inline void Print(int x1,int x2,int y1,int y2,string h3)
{
	if(h3=="+"||h3=="-")
	{
		while(x2!=y2)
		{
			int temp=x2;
			x2*=y2;
			x1*=y2;
			y2*=temp;
			y1*=temp;
		}
		if(h3=="+")
		{
			int g;
			int res=x1+y1;
			while((g=__***(res,x2))!=1)
			{
				res/=g;
				x2/=g;
			}
			cout<<res<<"/"<<x2<<endl;
		}
		else
		{
			int g,d=1;
			int res=x1-y1;
			if(res<0)
			{
				d=0;
				res=-res;
			}
			while((g=__***(res,x2))!=1)
			{
				res/=g;
				x2/=g;
			}
			if(d)
			{
				cout<<res<<"/"<<x2<<endl;
			}
			else
			{
				cout<<"-"<<res<<"/"<<x2<<endl;
			}	
		}
	}
	else if(h3=="*")
	{
		int g;
		int x3=x1*y1;
		int x4=x2*y2;
		while((g=__***(x3,x4))!=1)
		{
			x3/=g;
			x4/=g;
		}
		cout<<x3<<"/"<<x4<<endl;
	}
	else
	{
		int g;
		int x3=x1*y2;
		int x4=x2*y1;
		while((g=__***(x3,x4))!=1)
		{
			x3/=g;
			x4/=g;
		}
		cout<<x3<<"/"<<x4<<endl;
		
	}

}
int main()
{
	string h1,h2,h3;
	int i,j;
	 while(cin>>h1>>h2>>h3)
	 {
	 	for(i=0;;i++)
	 	{
	 		if(h1[i]=='/') break; 
	 		
		}
		int x1=stringint(h1.substr(0,i));
		int x2=stringint(h1.substr(i+1));

		for(j=0;;j++)
		{
			if(h2[j]=='/') break;
		} 
		int y1=stringint(h2.substr(0,j));
		int y2=stringint(h2.substr(j+1));
		Print(x1,x2,y1,y2,h3);
	 }
 
}

发表于 2020-04-02 11:40:31 回复(0)

Slashyouth

#include<cstdio>
(802)#include<cstdlib>
int a,b,c,d;
char x;
void ans()
{
    if(x=='+')  
    {
        int k=b*d,z=a*d+b*c;
        a=z,b=k;
    }
    else if(x=='-')
    {
        int k=b*d,z=a*d-b*c;
        a=z,b=k;      
    }
    else if(x=='*')
    {
        int k=b*d,z=a*c;
        a=z,b=k; 
        
    }
    else if(x=='/')
    {
        int k=b*c,z=a*d;
        a=z,b=k;        
    }
    int k=abs(a)>abs(b)?abs(b):abs(a);
    if(abs(a)==1 || abs(b)==1) return ;
    for(int i=k;i>1;i--)
    {
        if(abs(a)%i==0 && abs(b)%i==0) {a=a/i;b=b/i;return;}
    }
    return ;
}
int main()
{
    while(scanf("%d/%d %d/%d %c",&a,&b,&c,&d,&x)!=EOF)
    {
        ans();
        printf("%d/%d\n",a,b);     
        
    }
    
    
    return 0;
}

发表于 2020-03-14 23:12:50 回复(0)

hestyle

$\color{blue}解题思路：$
本题的关键是最大公因数、最小公倍数的求解。

两数的最大公因数求解常用辗转相除法，比如a，b两数（假设a >= b）

while (a % b != 0) {
    b’ 赋值为 a / b 余数 (即 a % b)
    a‘ 赋值为 b

    a = a’, b = b'
}

两数的最小公倍数求解有个数学规律

a、b最小公倍数 = a * b / (a、b最大公因数)

$\color{blue}代码实现：$

#include <iostream>
(720)#include <math.h>
using namespace std;

// 求number1、number2的最大公因数
int gcd(int number1, int number2) {
    int tempNum = 0;
    //保持number1不小于number2
    if (number1 < number2) {
        //借助tempNum，将number1、number2进行交换
        tempNum = number1;
        number1 = number2;
        number2 = tempNum;
    }
    //判断number1是否能整除number2
    while (number1 % number2 != 0) {
        //ba余数赋给tempNum
        tempNum = number1 % number2;
        //把number2赋给number1，tempNum赋给number2
        //由于进行求余时，number2是除数，tempNum是余数，所以number2 > tempNum
        //赋值操作后继续维持了 number1 > number2
        number1 = number2;
        number2 = tempNum;
    }
    return number2;
}

// 求number1、number2的最小公倍数
int lcm(int number1, int number2) {
    //最小公倍数 = 两数相乘 / 最大公因数
    return number1 * number2 / gcd(number1, number2);
}


int main(int argc, const char * argv[]) {
    char op = '\0';
    // 分数numerator1/denominator1，参数1   分数numerator2/denominator2  参数2
    int numerator1 = 1, denominator1 = 1, numerator2 = 1, denominator2 = 1;
    //scanf返回值为正确输入数据的变量个数，当一个变量都没有成功获取数据时，此时返回-1
    while (scanf("%d/%d %d/%d %c", &numerator1, &denominator1, &numerator2, &denominator2, &op) != - 1) {
        //记录结果是否为负数
        bool flag = false;
        //分数numerator3/denominator3,结果
        int numerator3 = 0, denominator3 = 0;
        if (op == '+') {
            //处理分数加法，先分母同分，然后分子与分母交叉相乘
            denominator3 = denominator1 * denominator2;
            numerator3 = numerator1 * denominator2 + numerator2 * denominator1;
        }
        if (op == '-') {
            //处理分数减法，先分母同分，然后分子与分母交叉相乘
            denominator3 = denominator1 * denominator2;
            numerator3 = numerator1 * denominator2 - numerator2 * denominator1;
            if (numerator3 < 0) {
                //这里单独记录负号，是因为求解公因数、公倍数是不能有负数
                flag = true;
                numerator3 = -numerator3;
            }
        }
        if (op == '*') {
            //处理分数乘法，分母相乘，分子相乘
            denominator3 = denominator1 * denominator2;
            numerator3 = numerator1 * numerator2;
        }
        if (op == '/') {
            //处理分数除法，分母1 分子2 相乘作分母，分子1 分母2 相乘 作分子
            denominator3 = denominator1 * numerator2;
            numerator3 = numerator1 * denominator2;
        }
        if (numerator3 == 0) {
            printf("0\n");
        } else {
            //分子、分母分别除最大公因数进行约分
            int tempNum = gcd(numerator3, denominator3);
            numerator3 /= tempNum;
            denominator3 /= tempNum;
            //如果是减法可能产生负数，因此需要输出负号
            printf((flag ? "-%d/%d\n" : "%d/%d\n"), numerator3, denominator3);
        }
    }
    return 0;
}

发表于 2020-03-04 11:51:58 回复(0)

C/C++

不偷不抢安度因_

//30以内的质数只有2，3，5，7，11，13，17，19，23，29
#include <stdio.h>
#include <stdlib.h>
 
int prime[10] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
 
int is_relative(int a, int b)
{
    for(int i = 0; i < 10; i++)
    {
        if(a % prime[i] == 0 && b % prime[i] == 0)
                return prime[i];//不互质,有因数prime[i]
    }
    return 1;//互质
}
 
int main()
{
    int a1, b1, a2, b2;
    char sign;
    int temp1, temp2;
    while(~scanf("%d/%d %d/%d %c", &a1, &b1, &a2, &b2, &sign))
    {
        int x;
        if(sign == '+')
        {
            temp2 = b1 * b2;
            temp1 = a1*b2 + a2*b1;
        }
        else if(sign == '-')
        {
            temp2 = b1 * b2;
            temp1 = a1*b2 - a2*b1;
        }
        else if(sign == '*')
        {
            temp2 = b1 * b2;
            temp1 = a1 * a2;
        }
        else
        {
            temp2 = a2 * b1;
            temp1 = a1 * b2;
        }
        if(temp1 == temp2)
            temp1 = temp2 = 1;
        while((x = is_relative(temp1, temp2)) != 1)
        {
            temp1 = temp1/x;
            temp2 = temp2/x;
        }
         
        printf("%d/%d\n", temp1, temp2);
    }
}

编辑于 2020-03-02 19:47:49 回复(0)

C/C++

我是个小白菜

//可能不是你的程序真的超时，而是你的输入没有加入!=EOF，真的牛皮
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
int MinCommonMultiple(int a, int b)
{
	int minnum = 0;
	for (int i = max(a,b);;i++)
	{
		if (i%a == 0 && i%b == 0)
		{
			minnum = i;
			break;
		}
	}
	return minnum;
}
int MaxConventionNumber(int a, int b)
{
	int maxnum = 0;
	for (int i = min(a, b);i >= 2;i--)
	{
		if (a%i == 0 && b%i == 0)
		{
			maxnum = i;
			break;
		}
	}
	return maxnum;
}
//约分
void reduction(int &a0, int &a1)
{
	if (a0 < 0)
	{
		a0 = -a0;
		cout << '-';
	}
	if (a0 == 0)
	{
		cout << a0 << endl;
		return;
	}
	//得到最大公约数
	int maxnum = MaxConventionNumber(a0, a1);
	if (maxnum != 0)
	{
		//约分
		a0 /= maxnum;
		a1 /= maxnum;
	}
	cout << a0 << '/' << a1 << endl;


}
//通分
void reverse_red(int &a0, int &a1, int &b0, int &b1)
{
	if (a1 != b1)
	{
		int minnum = MinCommonMultiple(a1, b1);
		a0 = minnum / a1 * a0;
		b0 = minnum / b1 * b0;
		a1 = b1 = minnum;
	}
}

int main()
{
	string str0, str1;
	char ch4;
	int a0, a1, b0, b1;
	while (scanf_s("%d/%d %d/%d %c",&a0,&a1,&b0,&b1,&ch4)!=EOF)
	{
		if (ch4 == '+')
		{
			reverse_red(a0, a1, b0, b1);  //通分
			a0 += b0;
			reduction(a0, a1);        //约分
		}
		else if (ch4 == '-')
		{
			reverse_red(a0, a1, b0, b1);
			a0 -= b0;
			reduction(a0, a1);
		}
		else if (ch4 == '*')
		{
			a0 *= b0;
			a1 *= b1;
			reduction(a0, a1);
		}
		else if (ch4 == '/')
		{
			a0 *= b1;
			a1 *= b0;
			reduction(a0, a1);
		}
		else
		{
			cout << "The operation you inputed was failed." << endl;
		}
	}
	return 0;
}

发表于 2019-09-06 13:56:09 回复(0)

寂之夜谁与

#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;
int fun(int a,int b) //求最大公约数然后化简
{
int i,j;
if(abs(a)>b)
i=b;
else
i=abs(a);
for(j=i;j>0;j--)
if(a%j==0&&b%j==0)
{
cout<<a/j<<'/'<<b/j<<endl;
break;
}
return 0; //没有这句会显示编译错误，我也不知道为什么
}
int main()
{
int n1,n2,n3,n4;
char s,s1,s2;
while(~scanf("%d/%d %d/%d %c", &n1, &n2, &n3, &n4, &s))

//while(cin>>n1>>s1>>n2>>n3>>s2>>n4>>s)

//最开始告诉我段错误，后来提交就好了，也不知什么毛病

{
switch(s)
{
case '+':
fun(n1*n4+n2*n3,n2*n4);
break;
case '-':
fun(n1*n4-n2*n3,n2*n4);
break;
case '*':
fun(n1*n3,n2*n4);
break;
case '/':
fun(n1*n4,n2*n3);

break;

default：

break;

}
}

}

发表于 2019-07-13 13:57:33 回复(0)

bajiSuan

掉坑了，一直超时，反复检查自己的***算法，结果最后发现原来是循环里的scanf()忘记写!=EOF了

发表于 2019-06-28 15:56:16 回复(0)

只是看一看

#include<stdio.h>//除法转换为乘法，分情况计算，化简的时候注意负数
void Abb(int *a,int *b);//化简
int main()
{
char S[3];
int a,b,i,f[4];
while(scanf("%d%c%d %d%c%d %c",&f[0],&S[0],&f[1],&f[2],&S[1],&f[3],&S[2])!=EOF)
{
if(S[2]=='+')
{
if(f[1]==f[3])
{
a=f[0]+f[2];
b=f[1];
Abb(&a,&b);

}
else
{
f[2]=f[2]*f[1]; ;//通分
f[0]=f[0]*f[3];
f[1]=f[1]*f[3];
f[3]=f[1];
a=f[0]+f[2];
b=f[1];
Abb(&a,&b);

}
}
if(S[2]=='-')
{
if(f[1]==f[3])
{
a=f[0]-f[2];
b=f[1];
Abb(&a,&b);

}
else
{
f[2]=f[2]*f[1]; ;//通分
f[0]=f[0]*f[3];
f[1]=f[1]*f[3];
f[3]=f[1];
a=f[0]-f[2];
b=f[1];
Abb(&a,&b);

}
}
if(S[2]=='*')
{
a=f[0]*f[2];
b=f[1]*f[3];
Abb(&a,&b);

}
if(S[2]=='/')
{
int t;
t=f[2];
f[2]=f[3];//除法转换为乘法
f[3]=t;
a=f[0]*f[2];
b=f[1]*f[3];
Abb(&a,&b);

}
printf("%d/%d\n",a,b);
}

}
void Abb(int *a,int *b)
{
int i,t;
if(*a>=0)
{
for(i=2;i<=*a;i++)
{
t=0;
while(t==0)
{
if(*a%i==0&&*b%i==0)
{
*a=*a/i;
*b=*b/i;
}
else
{
t=1;
}
}
}
}
else
{
*a=-(*a);
for(i=2;i<=*a;i++)
{
t=0;
while(t==0)
{
if(*a%i==0&&*b%i==0)
{
*a=*a/i;
*b=*b/i;
}
else
{
t=1;
}
}
}
*a=-(*a);
}

}

发表于 2019-04-13 14:01:19 回复(0)

执剑人96

//超时 感觉是死循环另外呀 貌似java全超时了

import java.io.*;
import java.util.*;

public class Test1012Caculate {
public static void main(String[] args) {
	Scanner in= new Scanner(new BufferedInputStream(System.in));
	while (in.hasNext()) {
	
	String []a=in.next().split("/");
	String []b=in.next().split("/");
	int a1=Integer.parseInt(a[0]);
	int b1=Integer.parseInt(a[1]);
	int a2=Integer.parseInt(b[0]);
	int b2=Integer.parseInt(b[1]);
	char op=in.next().charAt(0);
	switch (op) {
	case '+':
	{
		int x=a1*b2+a2*b1;
		int y=b1*b2;
		int gcd=getGcd(x, y);
		x=x/gcd;
		y=y/gcd;
		System.out.printf("%d/%d\n",x,y);
		break;
	}
	case '-':
	{
		int x=a1*b2-a2*b1;
		int y=b1*b2;
		int gcd=Math.abs(getGcd(x, y));
		x=x/gcd;
		y=y/gcd;
		System.out.printf("%d/%d\n",x,y);
		break;
	}
	case '*':
	{
		int x=a1*a2;
		int y=b1*b2;
		int gcd=getGcd(x, y);
		x=x/gcd;
		y=y/gcd;
		System.out.printf("%d/%d\n",x,y);
	
		break;
	}
	case '/':
	{
		int x=a1*b2;
		int y=b1*a2;
		int gcd=getGcd(x, y);
		x=x/gcd;
		y=y/gcd;
		System.out.printf("%d/%d\n",x,y);
		break;
	}
	default:
		break;
	}
		
	}
}
public static int getGcd(int p,int q)  {
	if(p%q==0){
		return q;
	}
	int r= p%q;
	return getGcd(q,r);
}
}

发表于 2018-12-04 14:35:20 回复(0)

Disconnect

想问一下，为啥代码一样的，我的却显示超时啊

发表于 2018-11-30 17:43:22 回复(0)

C/C++

BlackCarDriver

#include<iostream>
#include<stdio.h>
#include<math.h>
#define fun(a) a*a+a+41
using namespace std;

int gcb(int a,int b) {
    if(a<b) swap(a,b);
if(b==0) return a;
return gcb(b ,a%b);
}

struct node{
    int a;
    int b;
    void ans(int a,int b){
        this->a=a;
        this->b=b;
        bool fu = false;
        if(a*b<0) fu=true;
        a=abs(a);
        b=abs(b);
        int m = gcb(a,b);
        if(m!=1) a/=m,b/=m;
        if(fu) a=-a;
        printf("%d/%d\n",a,b);
    }
};
int main() {
    node f,s,ans;
    char c;
    while(scanf("%d/%d %d/%d %c",&f.a , &f.b , &s.a , &s.b , &c)!=EOF){
        switch(c){
            case '+': ans.ans(f.a*s.b + s.a*f.b , f.b*s.b); break;
            case '-': ans.ans(f.a*s.b - s.a*f.b , f.b*s.b ); break;
            case '*': ans.ans(f.a*s.a , f.b*s.b); break;
            case '/': ans.ans(f.a*s.b , s.a*f.b ); break;
        }
    }
return 0;
}

发表于 2018-11-20 15:52:24 回复(0)

分数运算

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