POJ 2586 Y2K Accounting Bug

题目理解比较麻烦，用贪心求解，让亏损的有效覆盖距离最远即可，所有可能的情况有5种:

ssssd ssssd ss 满足4 * s - d < 0

sssdd sssdd ss 满足3 * s - 2 * d < 0

ssddd ssddd ss 满足2 * s - 3 * d < 0

sdddd sdddd sd 满足s - 4 * d < 0

ddddd ddddd dd 满足s - 4 * d >= 0

Code:

#include <iostream>
 using namespace std;

int main()
{
    int a, b;
    while (scanf("%d%d", &a, &b)!=EOF)
    {
        int sum = 0;
        if (4 * b - a <= 0)
            sum = -1;
        else if(3 * b - 2 * a <= 0)
            sum = 3 * a - 9 * b;
        else if(2 * b - 3 * a <= 0)
            sum = 6 * (a - b);
        else if(b - 4 * a <= 0)
            sum = 8 * a - 4 * b;
        else sum = 10 * a - 2 * b;
        sum < 0 ? printf("Deficit\n") : printf("%d\n",sum);
    }
    return 0;
}