[Dynamic Programming] Complete Knapsack: Integer Division (Number of Solutions)

It can be transformed into a complete knapsack problem 
. Select several numbers from 1 to n so that their sum is exactly n. How many solutions are there in total?

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dp[i][j] represents the number of solutions selected from the previous i numbers so that their sum is exactly j

Consider how many times the i-th number is selected 
dp[i][j]=dp[i-1][j]+dp[i-1][ji]+dp[i-1][j-2*i]+ ...

Time optimization:
note that dp[i][ji]=dp[i-1][ji]+dp[i-1][j-2*i]+dp[i-1][j-3*i] +....
so dp[i][j]=dp[i-1][j]+dp[i][ji]

Space optimization:
rolling array 
for i 1~n:
  for ji~n: dp[j]=dp[j]+dp[ji]  

Code:

	int n;
	cin>>n;
	memset(dp,0,sizeof(dp));
	dp[0]=1;
    for(int i=1;i<=n;i++)
    {
    	for(int j=i;j<=n;j++)
    	{
    		dp[j]=(dp[j]+dp[j-i])%mod;
		}
	}
	cout<<dp[n];