Stamp face design-Deep Search + dp pruning

Continuous postage
Time Limit: 1500 MS memory limit: 10000 KB
Total submit: 59 (25 users) accepted submit: 22 (16 users) Page View: 3748

 

Country g issues n stamps with different denominations and requires that at most M stamps can be pasted on each envelope. The continuous postage problem requires the best design of the stamp face value for the given n and M values so that the postage starts from 1 on one envelope, the maximum continuous postage interval with an increment of 1. For example, when n = 5 and M = 4, five stamps with a nominal value of (, 32) can be pasted with a maximum continuous postage interval of 1 to 70. Programming task: Calculate the optimal design of the stamp face value for the given positive integers m and n.

Input

Each row of the input data has two positive integers, M and N (1 <= n, m <= 9). Finally, 0 indicates the end of the file.

Output

For the positive integers m and n in each row in the input, the maximum continuous postage interval is output.

Sample Input

4 5

0 0

Sample output

70

Http://acm.nankai.edu.cn/p1428.html

Analysis: This is a deep search and DP pruning question. However, this is still inefficient. Of course, I submitted it here.

 

Code

  1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 int n,m,f[1000000]={0},max,ma,s[100],limit,flag[1000000]={0};
 5 int dp(int k){
 6     int i,j;
 7     for(j=1;j<=m*limit;j++)
 8     f[j]=m+1;
 9     for(j=1;j<=m*limit;j++)
10     {
11      for(i=1;i<=k;i++)
12      if(j>=s[i])
13      if(f[j]>f[j-s[i]]+1)f[j]=f[j-s[i]]+1;
14      if(f[j]>m)
15      { 
16        if(k==n)ma=j-1;
17        return j;
18      }
19     }  
20     return j; 
21             }
22      
23 void dfs(int x,int k){
24      int i,j,z;
25      if(k<=n)
26       for(i=2;i<=x;i++)
27          if(flag[i]!=1)
28          {
29          s[k]=i;
30          flag[i]=1;
31          if(i>limit)limit=i;
32          z=dp(k);
33          if(k==n&&max<ma)
34           max=ma;
35          dfs(z,k+1);
36          flag[i]=0;
37          }
38                           }
39 
40 int main(){
41  scanf("%d%d",&m,&n);
42  while(m+n!=0)
43  {
44  max=0;limit=1;
45  int i,j,x;
46  s[1]=1;flag[1]=1; 
47  ds(m+1,2);
48  printf("%d\n",max); 
49  scanf("%d%d",&m,&n);
50  }
51 return 0;
52           }