poj 2586 Y2K Accounting Bug

Y2K Accounting Bug

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9769 Accepted: 4869

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

知道一个月的盈利值(s) 和 亏损值(d) ,已知每连续五个月是亏损的，问这一年是否盈利？

连续五个月亏损，所以这五个月不可能是 sssss，

如果 d>4*s   则这一年组成为 ssssdssssdss ,每连续5个月最少1个d，这时如果 10*s > 2*d ,这一年就是盈利的

如果 2*d > 3*s 这一年 为   sssddsssddss ，每连续5个月最少2个d，这时如果 8*s > 4*d  ,这一年就是盈利的

如果 3*d > 2*s 这一年 为   ssdddssdddss ，每连续5个月最少3个d，这时如果 6*s > 6*d  ,这一年就是盈利的
如果 4*d > s   这一年为    sddddsddddsd , 每连续5个月最少4个d，这时如果 3*s > 9*d  ,这一年就是盈利的

#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;int main(){    int s,d;    while(~scanf("%d%d",&s,&d))    {        int b=1;        if((d>4*s)&&(2*d<10*s))           {               printf("%d\n",10*s-2*d);               b=0;           }            else if((2*d>3*s)&&(4*d<8*s))           {               printf("%d\n",8*s-4*d);               b=0;           }            else if((3*d>2*s)&&(6*d<6*s))           {               printf("%d\n",6*s-6*d);               b=0;           }             else if((4*d>s)&&(9*d<3*s))           {               printf("%d\n",3*s-9*d);               b=0;           }        if(b) printf("Deficit\n");    }    return 0;}