算法1

时间复杂度 $O(n^2m)$

C++ 代码

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110, M = 11;

int n, m;
int f[N][M];

int main() {
    while(cin >> n >> m) {
        for(int i = 1; i <= n; ++i) f[i][1] = i;
        for(int i = 1; i <= m; ++i) f[1][i] = 1;
        for(int i = 2; i <= n; ++i)
            for(int j = 2; j <= m; ++j) {
                f[i][j] = f[i][j - 1];
                for(int k = 1; k <= i; ++k) 
                    f[i][j] = min(f[i][j], max(f[k - 1][j - 1], f[i - k][j]) + 1);
            }
        cout << f[n][m] << endl;
    }
    return 0;
}