poj日记（2586）java

Description

Accounting for Computer Machinists (ACM) has sufferred from the Y2K bug and lost some vital data for preparing annual report for MS Inc. 
All what they remember is that MS Inc. posted a surplus or a deficit each month of 1999 and each month when MS Inc. posted surplus, the amount of surplus was s and each month when MS Inc. posted deficit, the deficit was d. They do not remember which or how many months posted surplus or deficit. MS Inc., unlike other companies, posts their earnings for each consecutive 5 months during a year. ACM knows that each of these 8 postings reported a deficit but they do not know how much. The chief accountant is almost sure that MS Inc. was about to post surplus for the entire year of 1999. Almost but not quite. 

Write a program, which decides whether MS Inc. suffered a deficit during 1999, or if a surplus for 1999 was possible, what is the maximum amount of surplus that they can post.

Input

Input is a sequence of lines, each containing two positive integers s and d.

Output

For each line of input, output one line containing either a single integer giving the amount of surplus for the entire year, or output Deficit if it is impossible.

Sample Input

59 237375 743200000 8496942500000 8000000

Sample Output

11628300612Deficit

题目大意：公司每个月要么盈利s元，要么亏损d元，一年之中任意连续的五个月是亏损的，最后问一年的总收入是多少，如果盈利即输出数额，如果亏损，则输出Deficit。

解题思路：枚举。
              在保证连续5个月都亏损的前提下，使得每5个月中亏损的月数最少。
              x=1:  ssssd,ssssd,ss        d>4s      赢利10个月  10s-2d
              x=2:  sssdd,sssdd,ss       2d>3s    赢利8个月     8s-4d
              x=3:  ssddd,ssddd,ss      3d>2s    赢利6个月     6s-6d
              x=4:  sdddd,sdddd,sd     4d>s      赢利3个月     3s-9d
              x=5:  ddddd,ddddd,dd    4d<s      无赢利

import java.util.Scanner;public class Main {int ans;        void greedy(int s, int d)    {                if( 4 * s < d)        {            ans = 10 * s - 2 * d;        }        else if(3 * s < 2 * d)        {            ans = 8 * s - 4 * d;        }        else if( 2 * s < 3 * d)        {            ans = 6 * s - 6 * d;        }        else if(s < 4 * d)        {            ans = 3 * s - 9 * d;        }        else        {            ans = -1;        }    }        void output()    {        if(ans < 0)            System.out.println("Deficit");        else            System.out.println(ans);    }        public static void main(String args[])    {        Scanner cin = new Scanner(System.in);        Main ct = new Main();        int s;        int d;                while(cin.hasNext())        {            s = cin.nextInt();            d = cin.nextInt();            if(s > 0 && d > 0)            {                 ct.greedy(s, d);                ct.output();            }            else                break;        }                cin.close();    }}