P1021 stamp nominal value design, p1021 stamp nominal value

P1021 stamp nominal value design, p1021 stamp nominal value
Description

Given an envelope, up to N stamps can be pasted, and the number of stamps is calculated (assuming that the number of stamps is sufficient) for a given K (N + K ≤ 15 ), how to Design the face value of a stamp to obtain the maximum value MAX, so that ~ Each postage value between MAX can be obtained.

For example, N = 3, K = 2. If the nominal values are 1 and 4 ~ Each postage value between 6 points can be obtained (of course there are 8 points, 9 points and 12 points); If the face value is 1 point, 3 points, then 1 point ~ Each postage value between 7 points can be obtained. It can be verified that when N = 3, K = 2, 7 is the continuous maximum value of postage, so MAX = 7, the nominal value is 1 minute, 3 points.

Input/Output Format Input Format:

Two integers, N and K.

Output Format:

2 rows. A number in the first row indicates the selected nominal value, which is sorted in ascending order.

In the second row, "MAX = S" is output, and "S" indicates the maximum nominal value.

Input and Output sample Input example #1:

3 2

Output sample #1:

1 3MAX = 7

I wrote a dp and wrote a deep search result, which is as slow as the tortoise ..
However, thanks to the data,
Idea: perform in-depth search and enumeration, and calculate the value of backpack DP

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 const int MAXN = 1001; 7 void read (int & n) 8 {9 char c = '+'; int x = 0; int flag = 0; 10 while (c <'0' | c> '9') 11 {if (c = '-') flag = 1; c = getchar ();} 12 while (c> = '0' & c <= '9') 13 {x = x * 10 + (c-48); c = getchar ();} 14 flag = 1? N =-x: n = x; 15} 16 int n, k; 17 bool vis [MAXN]; 18 int ans [MAXN]; // 19 int dp [MAXN] selected each time; // records whether each number has a solution that can reach 20 int maxnum; 21 int out [MAXN]; 22 int ed = 0; 23 inline int pd () 24 {25 memset (dp, 0x3f, sizeof (dp); 26 dp [0] = 0; 27 int OK = 0; 28 int tot = 0; 29 for (int I = 1; I <= k; I ++) 30 tot = max (ans [I] * n, tot); 31 for (int I = 1; I <= k; I ++) // each item 32 for (int j = ans [I]; j <= n * ans [I]; j ++) 33 if (dp [j-ans [I] + 1 <= n) 34 dp [j] = min (dp [j], dp [j-ans [I] + 1); 35 for (int I = 1; I <= 170; I ++) 36 {37 if (dp [I] <888) 38 {39 OK ++; 40 continue; 41} 42 else break; 43} 44 return OK; 45} 46 inline void dfs (int now, int num) // The number of now selected, the last number of num 47 {48 ans [now] = num; 49 if (now = k) 50 {51 int hh = pd (); 52 // cout