Y2K Accounting Bug

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

题目大意：
财政报告一次报五个月的，每次都是亏损，输入s，d 是每个月盈利或亏损的数额，

输出一年最大的盈利数额

思路：由于每5个月的报账都为亏损，所有连续的5个月里至少有1个月为亏损，则可能产生最优解的情况为如下4种 
1 2 3 4 5 6 7 8 9 10 11 12 
s s s s d s s s s d  s  s //每5个月里只有1个月亏损 
s s s d d s s s d d  s  s //每5个月里只有2个月亏损 
s s d d d s s d d d  s  s //每5个月里只有3个月亏损 
s d d d d s d d d d  s  d //每5个月里只有4个月亏损 

当5个月都亏损就不可能有盈利的时候

这四种是保证每五个月都亏损但是还有可能有年盈利的情况

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <cstring>

using namespace std;

int main()
{
    int s,d;
    int ans;
    while(cin>>s>>d)
    {
        ans=0;
        int flag=0;
        if(d>4*s)
            ans=10*s-2*d;
        else if(2*d>3*s)
            ans=8*s-4*d;
        else if(3*d>2*s)
            ans=6*s-6*d;
        else if(4*d>s)
            ans=3*s-9*d;
        else
            ans=-12*d;

        if(ans>=0)
            cout<<ans<<endl;
        else
            printf("Deficit\n");
    }
    return 0;
}