(2.18) The number of multiple set combinations

Title (DP)

Code

In order not to count repeatedly, it is best to handle the same kind of items at once.
dp[i+1][j] := The total number of combinations of taking out j items from the first i kinds of items O(nm ² )
Taking out j items from the previous i kinds of items, you can first take out jk items from the first i-1 kinds of items, and then Take out k items from the i-th species and add them in

O(nm)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int MAX_N = 1000; 
const int MAX_M = 1000;
int dp[MAX_M+1][MAX_N+1]; // DP数组 

//输入 
int n = 3;   
int m = 3;
int a[3] = {
    
    1, 2, 3};
int M = 10000;

void solve(){
    
    
	// 一个都不取的方法总是只有一种
	for(int i = 0 ; i <= n ; i ++)
		dp[i][0] = 1;
	for(int i = 0 ; i < n ; i ++)
		for(int j = 1 ; j <= m ; j ++){
    
    
			if(j-1-a[i] >= 0) // 在有取余的情况下，避免减法运算的结果出现负数 
				dp[i+1][j] = (dp[i+1][j-1] + dp[i][j] - dp[i][j-1-a[i]] + M) % M;
			else
				dp[i+1][j] = (dp[i+1][j-1] + dp[i][j]) % M;
		}
	printf("%d\n", dp[n][m]);					 
} 	


int main(){
    
    	
	solve(); 
	return 0;
}