做法1: 认为价值等于体积

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 30;
const int MAXM = 20000;
int n, m;
int v[MAXN + 5];
int dp[MAXM + 5];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> m >> n;
    for (int i = 1; i <= n; i++)
        cin >> v[i];
    for (int i = 1; i <= n; i++)
        for (int j = m; j >= v[i]; j--)
            dp[j] = max(dp[j], dp[j - v[i]] + v[i]);
    cout << m - dp[m] << "\n";
    return 0;
}

做法2

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 30;
const int MAXM = 20000;
int n, m;
int v[MAXN + 5];
bool dp[MAXM + 5]; // dp[i]: 能否装满 i 的体积
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> m >> n;
    for (int i = 1; i <= n; i++)
        cin >> v[i];
    dp[0] = true;
    for (int i = 1; i <= n; i++)
        for (int j = m; j >= v[i]; j--)
        {
            // 考虑j的体积能否装满
            //    要不然是原来就能装满
            //    要不然是用了第i件物品后就能装满了
            //        前i-1件物品能装满剩余的空间 j-v[i]
            dp[j] = dp[j] || dp[j - v[i]];
        }
    for (int i = m; i >= 0; i--)
        if (dp[i])
        {
            cout << m - i;
            break;
        }
    return 0;
}