python/TangDou/AcWing/JiShuDP/900_2.cpp

#include <bits/stdc++.h>

using namespace std;

const int N = 1010;
const int MOD = 1e9 + 7;
int n;
int f[N][N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) f[i][0] = 1;

    for (int i = 1; i <= n; i++)       // 枚举每个物品，物品的序号与物品的体积是相等的，都是i
        for (int j = 1; j <= n; j++) { // 枚举背包容量j
            if (j >= i)                // ① 背包容量j>=当前体积i,可以选择当前数字
                f[i][j] = (f[i][j - i] + f[i - 1][j]) % MOD;
            else
                f[i][j] = f[i - 1][j] % MOD; // ② 放弃当前数字
        }

    cout << f[n][n] << endl;
    return 0;
}