POJ2586,Y2K Accounting Bug,简单是简单，无法理解此类叙述的思维

最新推荐文章于 2021-06-23 18:41:37 发布

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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：

看了半天discuss也没理解给出的s,d是什么，每季度的还是每个月的还是一年的。。。

大意是一个公司在12个月中，或固定盈余s，或固定亏损d.
但记不得哪些月盈余，哪些月亏损，只能记得连续5个月的代数和总是亏损(<0为亏损)，而一年中只有8个连续的5个月，分别为1~5，2~6，…，8~12
问全年是否可能盈利？若可能，输出可能最大盈利金额，否则输出“Deficit".

实际上;只要讨论5种情况即可；(任一月固定盈余s，或固定亏损d).

SSSSDSSSSDSS 4s<d 保证“连续5个月必亏损”，每连续5个月种至少1个月D，保证可能有全年最大盈余，每连续5个月中至多4个月S

SSSDDSSSDDSS 3s<2d 保证“连续5个月必亏损”，每连续5个月种至少2个月D，保证可能有全年最大盈余，每连续5个月中至多3个月S

SSDDDSSDDDSS 2s<3d 保证“连续5个月必亏损”，每连续5个月种至少3个月D，保证可能有全年最大盈余，每连续5个月中至多2个月S

SDDDDSDDDDSD s<4d 保证“连续5个月必亏损”，每连续5个月种至少4个月D，保证可能有全年最大盈余，每连续5个月中至多1个月S

DDDDDDDDDDDD s>=4d 保证“连续5个月必亏损”，每连续5个月种至少5个月D，每月亏损，此情况全年必亏损

code:

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