C语言 深入理解动态规划之计数类DP

// f[i][j] = f[i - 1][j] + f[i][j - i]
#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 7, mod = 1e9 + 7;

int f[N][N];

int main() {
    int n;
    cin >> n;

    for (int i = 0; i <= n; i ++) {
        f[i][0] = 1; // 容量为0时，前 i 个物品全不选也是一种方案
    }

    for (int i = 1; i <= n; i ++) {
        for (int j = 0; j <= n; j ++) {
            f[i][j] = f[i - 1][j] % mod; // 特殊 f[0][0] = 1
            if (j >= i) f[i][j] = (f[i - 1][j] + f[i][j - i]) % mod;
        }
    }

    cout << f[n][n] << endl;
}

// f[i][j] = f[i - 1][j] + f[i][j - i]
#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 7, mod = 1e9 + 7;

int f[N];

int main() {
    int n;
    cin >> n;


    f[0] = 1; // 容量为0时，前 i 个物品全不选也是一种方案

    for (int i = 1; i <= n; i ++) {
        for (int j = i; j <= n; j ++) {
            f[j] = (f[j] + f[j - i]) % mod;
        }
    }

    cout << f[n] << endl;
}

//非背包做法
//状态表示:f[i][j] 所有总和是i，并且恰好可以表示成j个数的和的方案
#include <bits/stdc++.h>

using namespace std;

const int N = 1010, mod = 1e9 + 7;

int n;
int f[N][N];

int main()
{
    cin >> n;

    f[0][0] = 1;
    for (int i = 1; i <= n; i ++ )
        //i最多表示成i个数的和，因此j<=i
        for (int j = 1; j <= i; j ++ )
            f[i][j] = (f[i - 1][j - 1] + f[i - j][j]) % mod;

    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = (res + f[n][i]) % mod;

    cout << res << endl;

    return 0;
}