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			A DP Problem poj 2295
		
            
		
            WA了N多次,猛然發(fā)現(xiàn)一處少寫一符號,總算A掉了.這種題目就是要細心細心再細心.
            

            #include <iostream>
            #include            <cstring>
            #include            <cmath>
            using namespace std;
            int main()
            {
                            char str[1000]={0};
                            int i=0;
                            int n=0;
                cin            >>n;
                            while (n--)
                            {
                                int len=0;
                    cin            >>str;
                    len            =strlen(str);
                                int sign=1;
                                int xx[2]={0};
                                int cc[2]={0};
                                int cnt=0;//
                    int stack=0;
                                int empty=1;
                                for(i=0;i<len;i++)
                                {
                                    switch (str[i])
                                    {
                                    case '=':
                                        if(!empty)
                                        {
                                cc[cnt]            +=stack*sign;
                            }
                            empty            =1;
                            stack            =0;
                            cnt            ++;
                            sign            =1;
                                        break;
                            
                                    case '-':
                            
                                        if(!empty)
                                        {
                                cc[cnt]            +=stack*sign;
                                stack            =0;
                                empty            =1;
                                sign            =-1;
                            }
                                        else
                                sign            *=-1;
                            
                            
                                        break;
                                    case '+':
                                        if(!empty)
                                        {
                                cc[cnt]            +=stack*sign;
                                stack            =0;
                                sign            =1;
                                empty            =1;
                            }
                                        break;
                                    case 'x':
                                        if(!empty)
                                        {
                                xx[cnt]            +=stack*sign;
                                stack            =0;
                                empty            =1;
                            }
                                        else
                                xx[cnt]            +=sign;
                            sign            =1;
                                        break;
                            
                                    default:
                            stack            =stack*10+str[i]-'0';
                            empty            =0;
                                        break;
                            
                        }
                    }
                                if(!empty)
                                {
                        cc[cnt]            +=sign*stack;
                    }
                                if(xx[0]==xx[1])
                                {
                                    if(cc[0]==cc[1])
                                    {
                            cout            <<"IDENTITY"<<endl;
                        }
                                    else
                            cout            <<"IMPOSSIBLE"<<endl;
                    }
                                else 
                                if((cc[1]-cc[0])%(xx[0]-xx[1])==0)
                                {
                        cout            <<(cc[1]-cc[0])/(xx[0]-xx[1])<<endl;
                    }
                                else 
                                {
                                    int flag1,flag2;
                                    if(cc[1]>cc[0])
                            flag1            =1;
                                    else 
                            flag1            =-1;
                                    if(xx[0]>xx[1])
                            flag2            =1;
                                    else flag2=-1;
                                    if(flag1*flag2>0)
                            cout            <<(cc[1]-cc[0])/(xx[0]-xx[1])<<endl;
                                    else
                            cout            <<(cc[1]-cc[0])/(xx[0]-xx[1])-1<<endl;
                    }
                    
                }
                
                            return 0;
            }
            

            

            
		
            
			posted on 2010-08-21 17:24 若余 閱讀(226) 評論(0)  編輯 收藏 引用  
		
            
	
            
	
	


	


            
	    
    




            






            

				
    


            

            
    
        
        

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