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  • poj 2586

    题意:每个月或者亏损s或者盈利d   调查8次连续的5个月  问若能盈利  则最大能盈利多少

    贪心算法  :先求出能使五个月亏损的最小亏亏损月数  把那些亏损月数向后靠边 

    #include<iostream>
    using namespace std;
    int main()
    {
    	int s,d,i;
    	while(cin>>s>>d)
    	{
    		for(i=1;i<=5;i++)
    			if(d*i>(5-i)*s)
    				break;
    		if(i>=5)
    		{
    			cout<<"Deficit"<<endl;
    			continue;
    		}
    		switch (i)
    		{
    		case 1:if(d*2<s*10) 
    			    cout<<s*10-d*2<<endl;
    			else cout<<"Deficit"<<endl;break;
    		case 2:if(d*4<s*8) 
    			    cout<<s*8-d*4<<endl;
    			else cout<<"Deficit"<<endl;break;
    		case 3:if(d*6<s*6) 
    			    cout<<s*6-d*6<<endl;
    			else cout<<"Deficit"<<endl;break;
    		case 4:if(d*9<s*3) 
    			    cout<<s*3-d*9<<endl;
    			else cout<<"Deficit"<<endl;break;
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zhangdashuai/p/3753701.html
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