POJ 2586 Y2K Accounting Bug （贪心）

题目链接： http://poj.org/problem?id=2586

Y2K Accounting Bug

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15286 Accepted: 7653

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

题目大意：  佶屈聱牙，晦涩难懂 -  -   没5 个月份报一次表，  连续报8次并且 每次都是亏损，     

1-5

2-6

3-7

4-8

5-9

6-10

7-11

8-12

共八次报表：

  解法一：  画一画 

5 10  月份 是一个   重要点，  可以照顾前后；   因此 有

#include <iostream>#include <stdio.h>#include <queue>#include <cmath>#include <cstring>using namespace std;int main(){    int s,d;    int ans;    while(cin>>s>>d)    {        ans=0;        int flag=0;        if(d>4*s)            ans=10*s-2*d;        else if(2*d>3*s)            ans=8*s-4*d;        else if(3*d>2*s)            ans=6*s-6*d;        else if(4*d>s)            ans=3*s-9*d;        else            ans=-12*d;        if(ans>=0)            cout<<ans<<endl;        else            printf("Deficit\n");    }    return 0;}

 注意： 4d >s 时  注意 此时12 月份必须为亏损

解法二 贪心思路：