Programming Contest challenges: the number of divisions

Subject to the effect

Problem-solving ideas

Such problems are generally referred to as $n$ the $m$ division. Note that this is classified asless than m equalparts, the use of dynamic programming solution.

• definition: $dp[i][j]$: $j$ 's $i$ total number of splits.
• Target state: $dp[m][n]$.
• State transition:
  • Wrong idea: that the demand $j$ 's $i$ When the division, from $j$ in Naqu $k$ a, then seek $j-k$ 's $i-1$ division, that is,
    $dp[i][j] = \sum_{k=1}^{j-1}dp[i-1][j-k]$
  • Error point: will $\{1,1,2 \}$ and $\{1,2,1\}$ considered as different divisions.
  • The right idea: for $j$ 's $i$ division, there are two cases: (1) In order to exactly divide $i$ parts. (2): For less than partitioning $i$ parts.
    • In the first case: We want to take away from every which are a result of the original and the result is one to one. Therefore, a total of $dp[i][j-i]$
    • Second case: direct correspondence is $dp[i-1][j]$ . Thus eventually
      $dp[i][j] = dp[i-1][j]+dp[i][j-i]$
• Update Policy: $dp[i][j]$ dependent on the left of and above $dp[i'][j']$, According to line by line so the update cycle.
• Initial state: $dp[0][*] = 0, dp[0][0] = 1$ : i.e., when no article is divided into only a method that 0 parts.
• Note: In the transfer process, there will be $dp[2][0] = 1$ such a situation, such as if it is not realistic, because he represents $0$ 2-division two items actually there is a. This is because we have takentake the equivalentof strategy, this status will only be $dp[2][2]$ cited. Therefore must $1$ without affecting other values.
• the complexity: $O (mn)$

Code

#include<iostream>
using namespace std;
const int MAX = 1005;
int dp[2][MAX];
int main()
{
    int m, n;
    cin >> n >> m;
    int M;
    cin >> M;
    dp[0][0] = 1;
    for(int i=1; i<=m; i++)
    {
        for(int j=0; j<=n; j++)
        {
            if(j-1 >= 0)
                dp[i%2][j] = (dp[1-i%2][j] + dp[i%2][j-i]) % M; 
            else
                dp[i%2][j] = dp[1-i%2][j];// 当总数j小于划分数i时，不可能每一份都大于0
        }
    }
    cout << dp[m%2][n] << endl;
}