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

分析：ms公司由于某病毒导致数据流失，该公司的盈(s)亏(d)是一定的，每五个月记录一次财务报表，这五个月总额为亏，问一年中是否盈利，如果盈利的话，求最大盈利值，反正输出Deficit。

    亏一个月(d > 4*s): ssssd ssssd ss 

    亏两月(2*d > 3*s): sssdd sssdd ss

    亏三月(3*d > 2*s): ssddd ssddd ss

    亏四个月(4*d > s): sdddd sdddd sd

    亏

参考代码：

#include<cstdio>#include<cmath>#include<cstdlib>#include<cctype>#include<cstring>#include<string>#include<sstream>#include<algorithm>#include<stack>#include<queue>#include<vector>#include<set>#include<map>#include<iostream>using namespace std;int s,d;int main(){    while( ~scanf("%d%d",&s,&d))    {        int ans;        if( d > 4*s)//每五个月亏一个月            ans = 10*s-2*d;        else if( 2*d > 3*s)//每五个月亏两月            ans = 8*s-4*d;        else if( 3*d > 2*s)//每五个月亏三月            ans = 6*s-6*d;        else if( 4*d > s)//每五个月亏四月            ans = 3*s-9*d;        else        {            printf("Deficit\n");            continue;        }        if( ans < 0)            printf("Deficit\n");        else            printf("%d\n",ans);    }    return 0;}