POJ 2586 Y2K Accounting Bug

Posted 2021-02-09 bravewtz

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8282		Accepted: 4106

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

 

 

某公司要统计全年盈利状况，对于每一个月来说，如果盈利则盈利S，如果亏空则亏空D。

公司每五个月进行一次统计，全年共统计8次(1-5、2-6、3-7、4-8、5-9、6-10、7-11、8-12)，已知这8次统计的结果全部是亏空(盈利-亏空<0)。

题目给出S和D，判断全年是否能盈利，如果能则求出盈利的最大值，如果不能盈利则输出Deficit

分析：利用贪心策略，每个月都只能是盈利s或者亏损d，总共12个月，按照s与d的关系，总共分为5种情况，挨个遍历就好了。

 

 

 1 #include<iostream>
 2 using namespace std;
 3 int main(){
 4     int s,d,flag,sum;
 5     while(~scanf("%d%d",&s,&d)){
 6         sum=0;
 7         flag=0;
 8         if(s>=0&&s<(1.0/4)*d){
 9             sum=10*s-2*d;
10         }
11         else if((s>=(1.0/4)*d)&&(s<(2.0/3)*d)){
12             sum=8*s-4*d;
13         }
14         else if((s>=(2.0/3)*d)&&(s<(3.0/2)*d)){
15             sum=6*s-6*d;
16         }
17         else if((s>=(3.0/2)*d)&&(s<4*d)){
18             sum=3*s-9*d;
19         }
20         else if(s>=4*d){
21             flag=1;
22         }
23         if(sum<0) flag=1;
24         if(flag){
25             cout<<"Deficit"<<endl;
26         }
27         else{
28             cout<<sum<<endl;
29         }
30     }
31     return 0;
32 }