POJ2586Y2K Accounting Bug

Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15922		Accepted: 7983

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：有一个公司赢亏数据丢失，但该公司每月的 赢亏是一个固定的数字，要么一个月赢利s，要么一月亏d。现在只知道该公司每五个月有一个赢亏报表，而且报表每次都为亏损。在一年中这样的报表总共有8次（1月到5月，2月到6月，…，8月到12月），求当盈利为s和亏损为d时，并使每张报表为亏损的情况下，全年公司最高可赢利多少，输出盈利金额，若亏损，输出"Deficit"。

思路：连续五个月总额亏损，有以下几种情况（尽可能使亏损的月份数降到最低）

1）ssssd，ssssd，ss  4s<d 年利润=10s-2d；

2）sssdd，sssdd，ss  3s<2d 年利润=8s-4d；

3）ssddd，ssddd，ss  2s<3d 年利润=6s-6d；

4）sdddd，sdddd，sd  s<4d  年利润=3s-9d；

5）ddddd，ddddd，dd  无盈利。

在写程序时充分体现贪心思路，从盈利月份最大的假设开始讨论。

#include<stdio.h>
int main()
{
    int s,d;
    while(scanf("%d %d",&s,&d)!=EOF)
    {
        int sum=0;
        if(4*s<d)
            sum=10*s-2*d;
        else if(3*s<2*d)
            sum=8*s-4*d;
        else if(2*s<3*d)
            sum=6*s-6*d;
        else if(s<4*d)
            sum=3*s-9*d;
        else
            sum=0;
        if(sum<0)
            printf("Deficit\n");
        else
            printf("%d\n",sum);
    }
    return 0;
}