TOJ-1335 Y2K Accounting Bug

最新推荐文章于 2021-01-12 17:03:50 发布

转载最新推荐文章于 2021-01-12 17:03:50 发布 · 103 阅读

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原文链接：http://www.cnblogs.com/shenchuguimo/p/6366100.html

本文探讨了一个因Y2K错误导致数据丢失的案例，涉及一家公司在1999年每月盈亏情况的数据恢复问题。通过分析连续五个月的亏损记录，文章提出了一种算法来确定公司是否在该年度遭受亏损，或可能的最大盈余额。代码示例展示了如何在限制条件下计算年度最大盈余。

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 is a sequence of lines, each containing two positive integers s and d. 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

Output for Sample Input

116
28
300612
Deficit

Source: Waterloo Local Contest Jan. 29, 2000

开始没读懂题一头雾水，后来知道题意了一脸蒙比。

已知某公司一年的每连续五个月的盈利状况均为亏损。又已知每个月的盈利值或亏损值，求全年可能的最大盈利。

求解方法即保证连续五个月的盈利状况为亏损情况下使亏损月数最少，求最后的盈利值。

每五个月中，亏损约束可为1~5，依次最高盈利状况序列如下：

每五个月亏损月数　　最高盈利状况序列　　满足条件　　总盈利

　　　1　　　　　　ssssd ssssd ss　　　d>4s　　 10s-2d

　　 2　　　　　　sssdd sssdd ss　　 2d>3s　　 8s-4d

　　 3　　　　　　ssddd ssddd ss　　 3d>2s　　 6s-6d

　　 4　　　　　　sdddd sdddd sd　　 4d>s　　 3-9d

　　 5　　　　　 ddddd ddddd dd　　4d>s　　 -12d

#include <iostream>
using namespace std;
int main(){
int s,d,a;
while(cin>>s>>d){
        if(d>4*s)a=10*s-2*d;
        else if(2*d>3*s)a=8*s-4*d;
        else if(3*d>2*s)a=6*s-6*d; 
        else if(4*d>s)a=3*s-9*d; 
        else a=-12*d;
        if(a<0)cout<<"Deficit"<<endl;  
        else cout<<a<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/shenchuguimo/p/6366100.html