解题思路

完全背包问题

优化思路，对状态转移做等价变形
1. f(i, j) = f(i - 1, j) + f(i - 1, j - a[i]) + f(i - 1, j - 2 * a[i]) ...
2. f(i, j - a[i]) = f(i - 1, j - a[i]) + f(i - 1, j - 2 * a[i]) ...
3. f(i, j) = f(i - 1, j) + f(i, i - a[i]);

代码

原始做法 O(N^3)

#include <iostream>

using namespace std;

const int N = 20, M = 3010;
int a[N];
long f[N][M];

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

    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        f[i][0] = 1;
        for (int j = 1; j <= m; j++) {
            for (int k = 0; j >= k * a[i]; k++) {
                 f[i][j] += f[i - 1][j - k * a[i]];
            }
        }
    }
    cout << f[n][m] << endl;

    return 0;
}

优化时间 O(N^2)

#include <iostream>

using namespace std;

const int N = 20, M = 3010;
int a[N];
long f[N][M];

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

    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        f[i][0] = 1;
        for (int j = 1; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= a[i]) f[i][j] += f[i][j - a[i]];
        }
    }
    cout << f[n][m] << endl;

    return 0;
}

优化空间

#include <iostream>

using namespace std;

const int N = 20, M = 3010;
int a[N];
long f[M];

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

    f[0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = a[i]; j <= m; j++) f[j] += f[j - a[i]];
    }
    cout << f[m] << endl;

    return 0;
}