HDU acm1028 Integer Division recursion problem (recursive)

We use recursive + memory to solve the problem of ordinary integer partitioning: Define f (n,m) to divide the integer n into a series of integers, where addend

Maximum no more than M.

Get the following recursive relationship:

When N==1 | | M==1 has only one division, that is, 1 or 1+1+1......+1.

When M>n is obviously equivalent to F (n,n)

When m==n at this time: I consider two cases where addend contains m or not:

1) Division does not contain m, i.e. f (n,m-1)---all m-1

2) division contains M, at this time only one is M

So when m==n, there is f (n,m) =f (n,m-1) +1

When M<n,

1) contains M, {M,X1,X2,X3....XI} is now equivalent to F (n-m,m)

2) does not contain m when obviously F (n,m-1)

So f (n,m) =f (n,m-1) +f (n-m,m)

Using memory technology to optimize complexity:

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 #defineMAXN 1306 intNUM[MAXN][MAXN];7 intdpintNintk)8 {9     if(n==1|| k==1)Ten         return 1; One     if(k>N) A     { -k=N; -         if(num[n][n]==-1) the             returnnum[n][n]=DP (n,n); -         Else -             returnNum[n][n]; -     } +     Else if(k==N) -     { +         if(num[n][k]==-1) A             returnNUM[N][K]=DP (n,k-1)+1; at         Else -             returnNum[n][k]; -     } -     Else -     { -         if(num[n][k]==-1) in             returnNUM[N][K]=DP (n,k-1) +DP (nk,k); -         Else to             returnNum[n][k]; +     } - } the intMain () * { $     intN;Panax Notoginseng      while(cin>>N) -     { thememset (num,-1,sizeof(num)); +COUT&LT;&LT;DP (n,n) <<Endl; A     } the     return 0; +}

HDU acm1028 Integer Division recursion problem (recursive)