Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10674		Accepted: 5344

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

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.

59 237

375 743

200000 849694

2500000 8000000

116

28

300612

Deficit

题意： 有一个公司因为某个病毒使公司赢亏数据丢失，但该公司每月的 赢亏是一个定数，要么一个月赢利s，要么一月亏d。

如今ACM仅仅知道该公司每五个月有一个赢亏报表。并且每次报表赢利情况都为亏。

在一年中这种报表总共同拥有8次（1到5，2到6，…，8到12），如今要编一个程序确定当赢s和亏d给出。并满足每张报表为亏的情况下，全年公司最高可赢利多少，若存在。则输出多多额，若不存在，输出"Deficit"。

分析：

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

代码例如以下：

#include <iostream>

using namespace std;

int main()

{

	int s,d;

	int res;

	while(cin>>s && cin>>d)

	{

		if(d>4*s)res=10*s-2*d;

		else if(2*d>3*s)res=8*s-4*d;

		else if(3*d>2*s)res=6*(s-d);

		else if(4*d>s)res=3*(s-3*d);

		else res=-1;

		if(res<0)cout<<"Deficit"<<endl;

		else cout<<res<<endl;

	}

	return 0;

}