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题目链接：http://poj.org/problem?id=2586
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Language: Default 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 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 |
题意比较难懂，其实只要读懂题意，就很简单了。
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".
思路：贪心思想，每连续5个月中，在保证这5个月经营之和为亏损的情况下，亏损的月数肯定应尽量往后选，盈利的月数应尽量往前选。
一共五种情况：
把5种情况可以归纳为关于s的判定条件：
0 <= s <1/4d 每连续5个月种至少1个月D
1/4d <= s < 2/3d 每连续5个月种至少2个月D
2/3d <= s < 3/2d 每连续5个月种至少3个月D
3/2d <= s < 4d 每连续5个月种至少4个月D
4d <= s 全年各月必亏损
代码如下：
#include <iostream>
using namespace std;
int main()
{
int s, d, flag;
while(cin>>s>>d)
{
int sum = 0;
flag = 0;
if(0<=s && s<(1.0/4)*d)
{
sum = 10*s - 2*d;
}
else if((1.0/4)*d<=s && s<(2.0/3)*d)
{
sum = 8*s-4*d;
}
else if((2.0/3)*d<=s && s<(3.0/2)*d)
{
sum = 6*s-6*d;
}
else if((3.0/2)*d<=s && s<4*d)
{
sum = 3*s-9*d;
}
else if(4*d <= s)
{
flag = 1;
}
if( sum < 0)
flag = 1;
if(flag)
cout<<"Deficit"<<endl;
else
cout<<sum<<endl;
}
return 0;
}
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