区间DP，将(maxn[i][j][k])表示为i到j区间内分为k个区间所得到的最大值，(minn)表示最小值。
然后可以得到状态转移方程：
[maxn[i][j][k]= max(maxn[i][j][k],maxn[i][l][k-1]*(l到j的\%10后的和))$$。
然后判断一下边界就能得到最终答案。
```c++
#include <bits/stdc++.h>
using namespace std;
int n, m, data[1001010];
int sum[111], ha[111][111];//每一段区间的和
int maxn[111][111][111], minn[111][111][111];
int ans1, ans2 = 2147483647;
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
{
scanf("%d", &data[i]);
data[i + n] = data[i];
}
n *= 2;
for (int i = 1; i <= n; i++)
sum[i] = sum[i - 1] + data[i];
memset(minn, 12, sizeof(minn));
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
ha[i][j] = (sum[j] - sum[i] + 100000) % 10, maxn[i][j][1] = minn[i][j][1] = ha[i][j];
for (int k = 2; k <= m; k++)
for (int i = 1; i <= n; i++)
for (int j = i + k - 1; j <= n; j++)
{
minn[i][j][k] = 214748367;
for (int l = i + k - 2; l < j; l++)
{
maxn[i][j][k] = max(maxn[i][l][k - 1] * ha[l][j], maxn[i][j][k]);
minn[i][j][k] = min(minn[i][l][k - 1] * ha[l][j], minn[i][j][k]);
}
}
for (int i = 1; i <= n / 2; i++)
ans1 = max(maxn[i][i + n / 2][m], ans1), ans2 = min(minn[i][i + n / 2][m], ans2);
printf("%d
%d", ans2, ans1);
return 0;
}
```]