2016SDAU课程练习一1013 Problem N

最新推荐文章于 2016-05-26 17:23:51 发布

原创最新推荐文章于 2016-05-26 17:23:51 发布 · 293 阅读

0 ·

CC 4.0 BY-SA版权

贪心算法专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种算法，用于解决公司MSInc在特定条件下如何计算全年盈亏的问题。通过给定每月可能的最大盈余和最大赤字，该算法能够判断全年是否会出现赤字，并计算可能的最大盈余。

Problem N

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 43 Accepted Submission(s) : 21

Problem 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

这道题最难的不是思路也不是编码而是理解题意！！！

找了无数个版本的翻译后总算是明白了，盈利运算，给出了盈利与亏损的s，d，每五个月结算一次，共8次，让你计算盈利值。

思维量相当小，直接硬做就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<algorithm>
using namespace std;
int main(){
int s, d, sum;
while (scanf("%d%d",&s,&d)==2){
  if (d>4*s)
            sum=10*s-2*d;
  else if(2*d>3*s)
            sum=8*s-4*d;
  else if(3*d>2*s)
            sum=6*s-6*d;
  else if(4*d>s)
            sum=3*s-9*d;
  else
            sum=-12*d;
        if(sum<=0)
            cout<<"Deficit"<<endl;
  else
            cout<<sum<<endl;
}
return 0;
}