邮票面值设计 （动态规划+DFS）

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n, k, res, ans[105], f[2005], curans[105];
int solve(int dep, int sum){
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i <= dep; ++i){
        for (int j = curans[i]; j <= n*sum; ++j)            //完全背包，f[i]记录的是在已知的面值中选择最少数量
            f[j] = min(f[j], f[j - curans[i]] + 1);            //如果最少数量大于n时，说明i这个数字不能表示，但是
    }                                                        //i-1表示可表示的最大值。
    for (int i = 1; i <= n*sum; ++i){
        if (f[i] > n){
            return i - 1;
        }
    }
    return n*sum;
}

void dfs(int dep, int last, int maxn, int sum){
    if (dep > k){                                        //dep作为递归结束，maxn是一个估计上界（因为不能无限
        if (res < maxn){                                //枚举下去，但是面值肯定小于n*sum)
            res = maxn;                                    //而枚举的下界就是i+1
            for (int i = 1; i <= k; ++i){
                ans[i] = curans[i];
            }
        }
        return;
    }
    for (int i = last + 1; i <= maxn + 1; ++i){
        curans[dep] = i;
        int x = solve(dep, sum + i);
        cout << "dep=" << dep<<endl;
        cout << "maxn=" << maxn << "   sum+i=" << sum + i << endl;
        cout << "x=" << x << endl;
        for (int i = 1; i <= dep; ++i)
            cout << "cur[" << i << "]="<<curans[i] << endl;
        cout << endl << endl;
        dfs(dep + 1, i, x, sum + i);
    }
}

int main(){
    cin >> n >> k;
    dfs(1, 0, 0, 0);
    for (int i = 1; i <= k; ++i)
        cout << ans[i] << " ";
    cout << endl << "MAX=" << res << endl;
}