2755: The Magic of pocket (recursive, dynamic programming)

 

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Total time limit: 

10000ms

 

Memory Limit: 

65536kB

description

There is a magic pocket, the total volume was 40, with some variations of this article pockets, the total volume of these items must be 40. John now there are n items to be obtained, a volume of each item are A _{. 1} , A ₂ ...... A _n . John may be selected from some of these items, if the total volume of the object 40 is selected, then use the magic pocket, John these items can be obtained. The question now is, John How many different ways of selecting items.

Entry

The first input line is a positive integer n (1 <= n <= 20), indicates the number of different items. The next n lines, each line has a positive integer between 1 and 40, are given A _{. 1} A, ₂ ...... A _n values.

Export

The number of different items of output selection mode.

Sample input

3
20
20
20

Sample Output

3

1  // Method a: the number of recursive items and item volume n array a [100] to global variables;
 2  // COUNT (i, SUM) represents the number of combinations from the start of the i-th array back to statistics and sum of the number of types,
 . 3  // sum and number combination is: COUT (I, sum) = COUT (I +. 1, sum-a [I]) + COUT (I +. 1, sum),
 . 4  // wherein cout (i + 1, sum- a [i]) represents contains a [i], that is, from the first i + 1 next count number
 5  // number and combination of sum-a [i] is the type number, and cout (i + 1, sum) indicates not comprise a [i], that is, from the number i + 1 onward statistical combination of numbers and the number of types of ******** sum *************************** 
. 6 #include <the iostream>
 . 7  the using  namespace STD;
 . 8  
. 9  int A [ 100 ];
 10  int n-= . 1 ;
 . 11  int COUNT (int I, int sum) {
 12 is      IF (sum == 0 ) {
 13 is          return  . 1 ;    // find a set of sum and number combination; 
14      }
 15      IF (I == n-sum || < 0 ) return  0 ; // I == no other n-number of instructions are combined, sum <0 indicate that a combination not; 
16      return COUNT (I + . 1 , a-SUM [I]) + COUNT (I + . 1 , SUM); // from the array i is first started, comprising A [i], and does not include; 
. 17  }
 18 is  
. 19  int main () {
 20 is      the while (CIN >> n-) {
 21 is         for(int i=0; i<n; i++)
22             cin>>a[i];
23         cout<<count(0,40)<<endl;
24     }
25     return 0;
26 }

1  // Method two: dynamic programming 
2 #include <the iostream>
 . 3  the using  namespace STD;
 . 4  #define N 100
 . 5  int n-, A [N];
 . 6  int main () {
 . 7      the while (CIN >> n-) {
 . 8          int (DP *) [ 50 ] = new new  int [N] [ 50 ]; // DP [i] [J] indicates the i-th item in the front volume Couchu J; 
. 9          for ( int i = . 1 ; i <= n-; ++ I ) {
 10              CIN >> A [I];
 . 11              DP [I] [0]=1; //初始边界
12         }
13         dp[0][0]=1;
14         for(int i=1; i<=n; i++)
15             for(int j=1; j<=40; j++) {
16                 dp[i][j]=dp[i-1][j];
17                 if(a[i]<=j)
18                     dp[i][j]+=dp[i-1][j-a[i]];
19             }
20         cout<<dp[n][40]<<endl;
21         delete []dp;
22     }
23     return 0;
24 }