POJ 2586 贪心＋枚举

Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15626		Accepted: 7843

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

题意：一个公司每一个月的情况为：盈利s或亏损d。 每五个月进行一次统计，共统计八次（1-5月一次，2-6月一次.......） 统计的结果是这八次都是亏空。

问题：判断全年是否能盈利，如果能则求出最大的盈利。 如果不能盈利则输出Deficit

思路：按d从大到小枚举情况，最开始每五个月里最多4个s

eg：若d>4s,则对于答案的排列里每五个月最多只有4个s，这样存在最多s的排列为:ssssdssdssss 10s+2d 

　　若2d>3s，则最多3个s 排列：sssddssddsss 8s+4d...

　　...

代码：

#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define db double
#define inf 0x3f3f3f
#define mj
typedef long long ll;
using  namespace std;
const int N=1e5+5;
int profit(int s, int d)
{
    if (d > 4 * s)
        return -2*d+10*s;
    else if (2 * d >3 * s)
        return -4*d+8*s;
    else if (3 * d > 2 * s)
        return -6*d+6*s;
    else if (4 * d > s)
        return -9*d+3*s;
    else
        return -1;
}

int main() {
    int s, d;
    while (cin >> s >> d)
    {
        int sum = profit(s, d);
        if (sum >= 0)
        {
            cout << sum << endl;
        }
        else {
            cout << "Deficit" << endl;
        }
    }
    return 0;
}