poj 2586 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

大致题意就是有一个年度报告，每五个月为一个季度往上提交报告，一年共上交八次报告。分别为1月-5月，2月-6月，3月-7月，4月-8月，5-9，6-10,7-11,8-12.
每个季度都是亏损，问一年有可能盈余吗，求最大盈余，若不能盈余则输出“Deficit”。主要采用贪心的思想,把亏损放在每季度的后面。
输入s和d,s为每月的盈余，d为每月的亏损。

样例1:  59 237
　　     SSSSD SSSSD SS 
样例2:  375 743
            SSSDD SSSDD SS

一共有这么几种情况：

 SSSSD SSSSD SS    一年有2个月为亏损
 SSSDD SSSDD SS   一年有4个月为亏损
 SSDDD SSDDD SS   一年有6个月为亏损

 SDDDD SDDDD SD  一年有9个月为亏损

 DDDDD DDDDD DD 一年有12个月为亏损

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int s,d;
    int a[10]={0,2,4,6,9,12};
    while(cin>>s>>d)
    {
        int k=1;
        while(s*(5-k)>d*k)
            k++;
        int ans=s*(12-a[k])-d*a[k];
        if(ans<0)
            cout<<"Deficit"<<endl;
        else
        printf("%d\n",ans);
    }
    return 0;
}