POJ2586 Y2K Accounting Bug

POJ 2586  Y2K Accounting Bug

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：

输入两个数s和d，代表某公司的全年每月平均盈利或亏损，即如该月盈利则为s亏损则为d。现已知该公司每五个月都是亏损的，现在问你全年是否盈利，如果能盈利输出盈利的最大值。

题解：

仔细分析一下题目，抓住每五个月都是亏损的，可以得到一下分析：

1. 若s>=4d，则说明五个月必须都为亏损。
2. 若2s>=3d，则说明每五个月中至多有一个月盈利。
3. 若3s>=2d，则说明每五个月中至多有两个月盈利。、
4. 若4s>=d，则说明每五个月中至多有三个月盈利。
5. 否则，每五个月至多有四个月盈利。

然后得考虑最终盈利值最大，而考虑的没五个月一循环，因此将盈利月安排在最前面几个月可以保证最后的11，12月能更多盈利（见下面演示）

1. 若s>=4d，dddddddddddd
2. 若2s>=3d，sddddsddddsd
3. 若3s>=2d，ssdddssdddss
4. 若4s>=d，sssddsssddsss
5. 否则，ssssdssssdss

有以上分析，结果就很明了了。

代码

#include<iostream>
using namespace std;

int main()
{
    int s,d;
    while(cin>>s>>d)
    {
        int sum = -1;
        if(s>=4*d);        //dddddddddddd
        else if(2*s>=3*d)  //sddddsddddsd
            sum = 3*s-9*d;
        else if(3*s>=2*d)  //ssdddssdddss
            sum = 6*s-6*d;
        else if(4*s>=d)  //sssddsssddss
            sum = 8*s-4*d;
        else               //ssssdssssdss
        {
            sum = 10*s-2*d;
        }
        if(sum>0)
            cout<<sum<<endl;
        else cout<<"Deficit"<<endl;

    }
    return 0;
}