The number of combinations related to the number of combinations of multiple sets

1. The number of division (% MOD) requirements of $ A_n ^ m $. There recurrence equation: $ dp [i] [j] = \ sum_ {k = 0} ^ {j} dp [i-1] [jk] $, where $ dp [i] [j] $ to $ $ J the total number of $ i $ divided.
2. The number of division (% MOD) required $ C_n ^ m $, there is recurrence equation: $ dp [i] [j ] = dp [i] [j-1] + dp [i-1] [j] $.
   code show as below:
   

    1 int n, m;
    2 int dp[MAX_M + 1][MAX_N + 1];  // DP数组
    3 
    4 int solve() {
    5     dp[0][0] = 1;
    6     for (int i = 1; i <= m; i++) {
    7         for (int j = 0; j <= n; j++) {
    8             if (j - i >= 0) {
    9                 dp[i][j] = (dp[i - 1][j] + dp[i][j - i]) % MOD;
   10             } else {
   11                 dp[i][j] = dp[i - 1][j];
   12             }
   13         }
   14     }
   15     return dp[m][n];
   16 }

    

3. Multiple sets the number of combinations, there are n items, there are i-th $ a $ a_i. Different types of items can be distinguished from each other but can not distinguish the same species. From these items of m number of species in a borrowing.
   Recurrence equation: $ dp [i + 1] [j] = dp [i + 1] [j-1] + dp [i] [j] -dp [i] [j-1-a_i] $
   code is as follows:
   

   . 1  int n-, m;
    2  int A [MAX_N];
    . 3  int DP [MAX_N + . 1 ] [M + + mAX . 1 ];
    . 4  
   . 5  int Solve ({)
    . 6      // method does not always take only one 
   . 7      for ( int I = 0 ; I <= n-; I ++ ) {
    . 8          DP [I] [ 0 ] = . 1 ;
    . 9      }
    10      for ( int I = 0 ; I <n-; I ++ ) {
    . 11          for ( int J =1; j <= m; j++) {
   12             if (j - 1 - a[i] >= 0) {
   13                 dp[i + 1][j] =
   14                     (dp[i + 1][j - 1] + dp[i][j] - dp[i][j - 1 - a[i]] + MOD) %
   15                     MOD;
   16             } else {
   17                 dp[i + 1][j] = (dp[i + 1][j - 1] + dp[i][j]) % MOD;
   18             }
   19         }
   20     }
   21     return dp[n][m];
   22 }