解题思路

代码

原始做法 O(N^3)

#include <iostream>

using namespace std;

const int N = 1010;
int a[5] = {0, 10, 20, 50, 100};
int f[5][N];

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

    f[0][0] = 1;
    for (int i = 1; i <= 4; 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[4][m] << endl;

    return 0;
}

优化时间 O(N^2)

#include <iostream>

using namespace std;

const int N = 1010;
int a[5] = {0, 10, 20, 50, 100};
int f[5][N];

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

    f[0][0] = 1;
    for (int i = 1; i <= 4; 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[4][m] << endl;

    return 0;
}

优化空间

#include <iostream>

using namespace std;

const int N = 1010;
int a[5] = {0, 10, 20, 50, 100};
int f[N];

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

    f[0] = 1;
    for (int i = 1; i <= 4; i++) {
        for (int j = a[i]; j <= m; j++) {
            f[j] += f[j - a[i]];
        }
    }
    cout << f[m] << endl;

    return 0;
}