poj 2586 Y2K Accounting Bug

                                                                                                                           Y2K Accounting Bug

Time Limit: 1000MS Memory Limit: 65536K   

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题意：一个记账系统遭受到了攻击，导致某些数据丢失，我们已知在一年的十二个月中，连续的五个月（例如1~5,2~6,3~7,4~8,5~9,6~10,7~11,8~12) 必定会亏损（即盈利〈亏损），现已知盈利s,亏损d，让我们求在这一年中，我们能否盈利，如果能，则输出最大的盈利数，否则输出“Deficit”，该题主要是题意很难读懂，其实做法很简单，考得无非就是贪心罢了，我们必须要把亏损的那个月最大利用化，这样才有可能出现最大盈利代码：#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <iomanip>#include <algorithm>#include <iomanip>#include <cmath>using namespace std;int main(){    int a[13],i,sum,j,k;    int s,d;    while(cin>>s>>d)    {        for(i=1;i<=12;i++)        {            a[i]=s;//我们先假设每个月都可以盈利，除去那些亏损的月，这样才可能达到最大盈利        }        for(j=1;j<=8;j++)        {            sum=0;            for(k=j;k<=j+4;k++)            {                sum+=a[k];//求出连续五个月的收入            }            while(sum>=0)//因为题意中，连续的五个月必定亏损，如果连续的五个月的收入大于0，那么我们要把倒着把这五个月中的月份记为亏损，直到收入小于0            {                sum-=(s+d);//我们不仅要减掉亏损的钱，还要减去我们假设这个月能够盈利的钱                a[--k]=d*(-1);//把月份记为亏损            }        }        sum=0;        for(i=1;i<=12;i++)        {            sum+=a[i];            //cout<<a[i]<<" ";        }//这时我们已经把可能出现最大盈利的每个月的收入求出来了，只需要求和        //cout<<endl;        if(sum>=0)//如果和大于等于0，那么可以出去最大盈利        {            cout<<sum<<endl;        }else{//否则，不可能盈利            cout<<"Deficit"<<endl;        }    }    return 0;}