AtCoder Beginner Contest 099

C

由于可以装的是有限个，因此可以全部都装进去，紧接着用一个完全背包进行枚举即可。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
// LUOGU_RID: 174639729
#include <bits/stdc++.h>
using namespace std;
int n, cnt, a[233], f[1000010], p;
int main()
{
    cin >> n;
    a[++cnt] = 1;
    p = 1;
    for (int i = 1; i <= 8; i++)
    {
        p *= 6;
        a[++cnt] = p;
    }
    p = 1;
    for (int i = 1; i <= 6; i++)
    {
        p *= 9;
        a[++cnt] = p;
    }
    sort(a + 1, a + cnt + 1);
    memset(f, 0x3f, sizeof(f)); 
    f[0] = 0;
    for (int i = 1; i <= cnt; i++)
        for (int j = a[i]; j <= n; j++)
            f[j] = min(f[j], f[j - a[i]] + 1);
    cout << f[n];
    return 0;
}

D

分成3组，统一计算，最后暴力枚举即可，颜色数量是满足题意的。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
// LUOGU_RID: 174644485
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 50, M = 3, inf = 2e18;
long long n, m, s, x, i, j, k, ans = inf, c[N][N], f[M][N];
int main()
{
    cin >> n >> m;
    for (i = 1; i <= m; i++)
        for (j = 1; j <= m; j++)
            cin >> c[i][j];
    for (i = 1; i <= n; i++)
        for (j = 1; j <= n; j++)
        {
            s = (i + j) % 3;
            cin >> x;
            for (k = 1; k <= m; k++)
                f[s][k] += c[x][k];
        }
    for (i = 1; i <= m; i++)
        for (j = 1; j <= m; j++)
            for (k = 1; k <= m; k++)
                if (i != j && i != k && j != k)
                    ans = min(ans, f[0][i] + f[1][j] + f[2][k]);
    cout << ans << '\n';
}