法一：完全背包问题

#include <iostream>
using namespace std;

const int maxn = 1010, mod = 1e9 + 7;
int f[maxn];

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

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

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

    cout << f[n] << endl;

    return 0;
}

法二：总和是i，恰好表示成j个数的和

#include <iostream>
using namespace std;

const int maxn = 1010, mod = 1e9 + 7;
int f[maxn][maxn];

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

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

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

    cout << ans << endl;

    return 0;
}