poj2586 典型贪心

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Y2K Accounting Bug

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7846 Accepted: 3885

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 237375 743200000 8496942500000 8000000

Sample Output

11628300612Deficit

Source

Waterloo local 2000.01.29

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".

根据经验，贪心选择往往都在极端处（临界点）选择。（其实这题不用贪心，单纯枚举也可以AC，因为不同情况实在太少呐。。。。

不难证明，每连续5个月中，在保证这5个月经营之和为亏损的情况下，亏损的月数肯定应尽量往后选，盈利的月数应尽量往前选。证明省略。

先处理处理完1~5月后，剩下的月份可以根据“连续5个月经营之和为亏损”这个条件进行确定亏损还是盈利。

本题的贪心选择每次仅仅选取其中一种情况（1~5月），因为之后月份无需再选择，所以每次总共只做了一次贪心选择。

实际上;只要讨论5种情况即可；(任一月固定盈余s，或固定亏损d).

SSSSDSSSSDSS 4s<d 保证“连续5个月必亏损”，每连续5个月种至少1个月D，

保证可能有全年最大盈余，每连续5个月中至多4个月S

SSSDDSSSDDSS 3s<2d 保证“连续5个月必亏损”，每连续5个月种至少2个月D，

保证可能有全年最大盈余，每连续5个月中至多3个月S

SSDDDSSDDDSS 2s<3d 保证“连续5个月必亏损”，每连续5个月种至少3个月D，

保证可能有全年最大盈余，每连续5个月中至多2个月S

SDDDDSDDDDSD s<4d 保证“连续5个月必亏损”，每连续5个月种至少4个月D，

保证可能有全年最大盈余，每连续5个月中至多1个月S

DDDDDDDDDDDD s>=4d 保证“连续5个月必亏损”，每连续5个月种至少5个月D，

每月亏损，此情况全年必亏损

要注意的是，前4种情况都仅仅是“可能有全年的盈余”，而不是“一定有全年的盈余”。

但是若果一旦有盈余，必定是最大盈余

把5种情况可以归纳为关于s的判定条件：

0 <= s <1/4d 每连续5个月种至少1个月D

1/4d <= s < 2/3d 每连续5个月种至少2个月D

2/3d <= s < 3/2d 每连续5个月种至少3个月D

3/2d <= s < 4d 每连续5个月种至少4个月D

4d <= s 全年各月必亏损

正确代码：

#include<iostream>#include<stdio.h>#include<iomanip>using namespace std;int a,b;int main(){while(scanf("%d%d",&a,&b)!=EOF){if (a>=4*b){cout<<"Deficit"<<endl;}else if (a<4*b&&a>=3.0/2*b){if (3*a-9*b<0){cout<<"Deficit"<<endl;}else cout<<3*a-9*b<<endl;}else if (a<3.0/2*b&&a>=2.0/3*b){if (a<b){cout<<"Deficit"<<endl;}else cout<<6*(a-b)<<endl;}else if (a<2.0/3*b&&a>=1.0/4*b){if (8*a-4*b<0){cout<<"Deficit"<<endl;}else cout<<8*a-4*b<<endl;}else if (a<1.0/4*b){if (10*a-2*b<0){cout<<"Deficit"<<endl;}else cout<<10*a-2*b<<endl;}}}