2586 Y2K Accounting Bug 贪心_acmy2k accounting bug-优快云博客

本文探讨了一种基于Y2K会计漏洞的问题，通过给定的月度盈亏信息，设计算法来判断公司全年是否出现赤字，并计算可能的最大盈余。文章提供了具体的输入输出示例及源代码。

Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3874		Accepted: 1795

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

s s s s d s s s s d s s d>4*s

s s s d d s s s d d s s d<=4*s && 2*d>3*s

s s d d d s s d d d s s 2*d<=3*s && 3*d>2*s

s d d d d s d d d d s d 3*d<=2*s && 4*d>s

上面情况还要满足利润>0

d d d d d d d d d d d d else

#include<iostream>

using namespace std;

int main()

{

__int64 s,d;

while(cin>>s>>d)

{

if(d>4*s && 10*s-2*d>0 ) cout<<10*s-2*d<<endl;

else

if(2*d>3*s && 8*s-4*d>0 ) cout<<8*s-4*d<<endl;

else

if(3*d>2*s && 6*s-6*d>0) cout<<6*s-6*d<<endl;

else

if(4*d>s && 3*s-9*d>0) cout<<3*s-9*d<<endl;

else

cout<<"Deficit"<<endl;

}

return 0;

}