POJ 286 Y2K Accounting Bug【简单暴力】

链接：

http://poj.org/problem?id=2586

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26733#problem/C

Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8472		Accepted: 4185

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

Source

Waterloo local 2000.01.29

题目：

题意很坑，题目很水，我等弱菜+英语白痴怎么受得了

有个公司丢失了一些数据，他们的数据【盈利s,亏损d】都是每5个月报一次,所以一年当然是报 8 次了咯。

现在给你一组 s 和 d 。要你求出全年最大可能的盈利值，如果不肯能盈利则输出 Deficit

思路：

给你的数据有盈利 S 和亏损 D

那么上报一次数据的 5 个月，只可能有 5 种情况：

盈利四个月，亏损一个月：ssssd -> ssssdssssdss 【保证了对应每次报数据都是盈利四个月，亏损一月，下面同】

盈利三个月，亏损二个月：sssdd -> sssddsssddss

盈利二个月，亏损三个月：ssddd -> ssdddssdddss

盈利一个月，亏损四个月：sdddd -> sddddsddddsd

连续亏损 5 个月必定亏损= =

记得去年看过MMchen的博客Orz

code：

#include<stdio.h>
int main()
{
    int s, d;
    while(scanf("%d%d", &s,&d) != EOF)
    {
        int ans = -1;
        if(4*s < d) ans = 10*s-2*d; //ssssd
        else if(3*s < 2*d) ans = 8*s-4*d; //sssdd
        else if(2*s < 3*d) ans = 6*s-6*d; //ssddd
        else if(s < 4*d) ans = 3*s-9*d; //sdddd
        if(ans < 0) printf("Deficit\n");
        else printf("%d\n", ans);
        //for(int i = 1; i <= 12; i++) printf("%d\n", i*s-(12-i)*d);
    }

    return 0;
}