POJ 2586 貪心＋枚舉
阿新 • • 發佈：2017-07-20
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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 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 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 |
題意：一個公司每一個月的情況為：盈利s或虧損d。 每五個月進行一次統計，共統計八次（1-5月一次，2-6月一次.......） 統計的結果是這八次都是虧空。
問題：判斷全年是否能盈利，如果能則求出最大的盈利。 如果不能盈利則輸出Deficit
思路：按d從大到小枚舉情況，最開始每五個月裏最多4個s
eg：若d>4s,則對於答案的排列裏每五個月最多只有4個s，這樣存在最多s的排列為:ssssdssdssss 10s+2d
若2d>3s，則最多3個s 排列：sssddssddsss 8s+4d...
...
代碼：
1 #include "cstdio" 2 #include "stdlib.h" 3 #include "iostream" 4 #include "algorithm" 5 #include "string" 6 #include "cstring" 7 #include "queue" 8 #include "cmath" 9 #include "vector" 10 #include "map" 11 #include "set" 12 #define db double 13 #define inf 0x3f3f3f 14 #define mj 15 typedef long long ll; 16 using namespace std; 17 const int N=1e5+5; 18 int profit(int s, int d) 19 { 20 if (d > 4 * s) 21 return -2*d+10*s; 22 else if (2 * d >3 * s) 23 return -4*d+8*s; 24 else if (3 * d > 2 * s) 25 return -6*d+6*s; 26 else if (4 * d > s) 27 return -9*d+3*s; 28 else 29 return -1; 30 } 31 32 int main() { 33 int s, d; 34 while (cin >> s >> d) 35 { 36 int sum = profit(s, d); 37 if (sum >= 0) 38 { 39 cout << sum << endl; 40 } 41 else { 42 cout << "Deficit" << endl; 43 } 44 } 45 return 0; 46 }
POJ 2586 貪心＋枚舉