E

 

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

59 237
375 743
200000 849694
2500000 8000000

Sample Output

116
28
300612
Deficit

//没理解题的意思
看到别人的题解才明白

某公司要统计全年盈利状况，对于每一个月来说，如果盈利则盈利S，如果亏空则亏空D。

公司每五个月进行一次统计，全年共统计8次(1-5、2-6、3-7、4-8、5-9、6-10、7-11、8-12)，已知这8次统计的结果全部是亏空(盈利-亏空<0)。

四种能盈利的情况：

1、若SSSSD亏空，那么全年最优情况为SSSSDSSSSDSS

2、若SSSDD亏空，那么全年最优情况为SSSDDSSSDDSS

3、若SSDDD亏空，那么全年最优情况为SSDDDSSDDDSS

4、若SDDDD亏空，那么全年最优情况为SDDDDSDDDDSD

#include <iostream>
 using namespace std;
 int main ()
 {
     int n,m;
     int i,j;
     int sum;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         int max=-12*m;
         if(4*n < m)
         {
             sum=10*n-2*m;
             if(sum>max)
                 max=sum;
         }
         else
             if(3*n < 2*m)
             {
                 sum=8*n-4*m;
                 if(sum>max)
                     max=sum;
             }
             else     if(2*n < 3*m)
             {
                 sum=6*n-6*m;
                 if(sum>max)
                     max=sum;
             }
             else     if(n < 4*m)
             {
                 sum=3*n-9*m;
                 if(sum>max)
                     max=sum;
             }
             if(max>0)
             {
                 printf("%d
",max);
             }
             else
             {
                 printf("Deficit
");
             }
     }
     return 0;
 }