// https://dashoj.com/p/114
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int main() {
	int n, m;
	cin >> n >> m;
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	sort(a.begin(), a.end());
	vector<int> dp(m + 1, 1e5);
	dp[0] = 0;
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= n; j++)
			if (a[j] <= i) dp[i] = min(dp[i], dp[i - a[j]] + 1);
	cout << dp[m] << endl;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;

int a[1010];
int f[10010];

int main(){
    int n, cost;
    cin>>n>>cost;
    for(int i = 1; i <= n; i++){
        cin>>a[i];
        f[a[i]] = 1;
    }

    for(int i = 1; i <= cost; i++){
        if(f[i]) continue;
        int mi = cost;
        for(int j = 1; j < n; j++){
            if(a[j] > i) continue;
            mi = min(f[i - a[j]] + 1 , mi);
        }
        f[i] = mi;
    }
    cout<<f[cost];
}

#include<bits/stdc++.h>
using namespace std;
int dp[100001],n,a[101010],cost,w;
int main(){
	cin>>n>>w;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1;i <= w;i++){
		cost = 0x3f3f3f3f;
		for(int j = 1; j <= n; j++){
           //特判一下，防止RE
		    if(i - a[j] >= 0) cost = min(cost, dp[i - a[j]]);
           //求之前用的纸币数最小值
		}
		dp[i] = cost + 1;//加一是这次会多用1张纸币
	} 
	cout << dp[w];
	return 0;
}

时间复杂度$O(nm)$

空间复杂度$O(m)$

#include <bits/stdc++.h>

using namespace std;

const int N = 1010 , M = 1e4 + 10 ; 

int f[M] ;
// f[i , j] 表示前 i 种纸币凑出金额为 j 的最少数量
int n , m , a[N] ;

int main(int argc, char const *argv[])
{
	cin.tie(nullptr)->ios::sync_with_stdio(false) ;
	cin >> n >> m ;
	for(int i = 1 ; i <= n ; ++ i) cin >> a[i] ;
	memset(f , 0x3f , sizeof f) ; 
	f[0] = 0 ; 
	for(int i = 1 ; i <= n ; ++ i) 
		for(int j = a[i] ; j <= m ; ++ j)
			f[j] = min(f[j] , f[j - a[i]] + 1) ;

	cout << f[m] << '\n' ;
	return 0;
}

麻烦看看有什么可以优化👀️

#include <bits/stdc++.h>
using namespace std;

int main()
{
    //输入n张纸币，凑成m元，n赋给cir
    int n, m, cir;
    cin >> n >> m;
    cir = n;
    //初始化money数组来接收面额
    int money[n] = {0};
    while(cir)
    {
        cin >> money[n-cir];
        cir--;
    }
    //简单排个序
    sort(money, money+n);
    //从最高面额往下凑
    int i = 1, num[n] = {0}, l = n, com;
    //大循环是怕3*5这样的例子出现
    //不然从最高往下凑会有11的影响
    for(;l;l--)
    {
        com = m;
        i = 1;
        for(int j = 0; j < l; j++)
        {
            while(com/money[l-i])
            {
                num[n-l]++;
                com -= money[l-i];
            }
            i++;
        }
    }
    //找最小值
    int* max1 = min_element(num, num+n);
    cout << *max1;

}