poj 2586 Y2K Accounting Bug

http://poj.org/problem?id=2586

 

大意是一个公司在12个月中，或固定盈余s，或固定亏损d.

但记不得哪些月盈余，哪些月亏损，只能记得连续5个月的代数和总是亏损(<0为亏损)，而一年中只有8个连续的5个月，分别为1~5，2~6，…，8~12

问全年是否可能盈利？若可能，输出可能最大盈利金额，否则输出“Deficit".

贪心，符合最优子结构性质。5个月统计一次都亏空，那么有5种情况：
      1、若SSSSD亏空，那么全年可能最大盈利情况为: SSSSDSSSSDSS
      2、若SSSDD亏空，那么全年可能最大盈利情况为：SSSDDSSSDDSS
      3、若SSDDD亏空，那么全年可能最大盈利情况为: SSDDDSSDDDSS
      4、若SDDDD亏空，那么全年可能最大盈利情况为: SDDDDSDDDDSD
      5、若DDDDD亏空，那么全年可能最大盈利情况为: DDDDDDDDDDDD
                                                       Y2K Accounting Bug
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10761   Accepted: 5391

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
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 int main()
 5 {
 6     int s,d;
 7     while(~scanf("%d%d",&s,&d))
 8     {
 9            int flag=0;
10           int ans=0;
11         if(4*s-d<0)
12              {
13                  ans=10*s-2*d;
14                   if(ans<=0)
15                      flag=1;
16             }
17         else if(3*s-2*d<0)
18         {
19             ans=8*s-4*d;
20                 if(ans<=0)
21                      flag=1;
22          }
23           else if(2*s-3*d<0)
24           {
25              ans=6*s-6*d;
26               if(ans<=0)
27                      flag=1;
28              
29           }
30          else if(s-4*d<0)
31          {
32              ans=3*s-9*d;
33               if(ans<=0)
34                      flag=1;
35           
36          }
37           else
38             {
39                 printf("Deficit\n");
40                  continue;
41             }
42              if(flag)
43                  printf("Deficit\n");
44              else
45             printf("%d\n",ans);
46 
47     }
48     return 0;
49 }

 

posted @ 2014-12-03 21:50  疯狂的癫子  阅读(159)  评论(0)  编辑  收藏  举报