原题链接
又是一道挺水的类区间(DP)。
因为题目给定的是一个环，所以先断环成链再(DP)即可。
设(f[i][j][l])表示(i sim j)之间的数分成(l)段的最大值，(g[i][j][l])为最小值，(mod(x))为((x mod 10 + 10) mod 10)，(s[])为断环后的链上数据的前缀和。
则有状态转移方程：
[f[i][j][l] = max limits _{k = i + l - 2} ^ {j - 1} { f[i][k][l - 1] imes mod(s[j] - s[k]) }
]
[g[i][j][l] = min limits _{k = i + l - 2} ^ {j - 1} { f[i][k][l - 1] imes mod(s[j] - s[k]) }
]
初始化(f[i][j][1] = g[i][j][1] = mod(s[j] - s[i - 1]))，其余(g)数组为无穷大，(f)数组为(0)。
最后答案就是(max limits _{i = 1} ^ {n} f[i][i + n - 1][m], min limits _{i = 1} ^ {n} g[i][i + n - 1][m])。
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 110;
const int M = 15;
int f[N][N][M], g[N][N][M], a[N];
inline int re()
{
int x = 0;
char c = getchar();
bool p = 0;
for (; c < '0' || c > '9'; c = getchar())
p |= c == '-';
for (; c >= '0' && c <= '9'; c = getchar())
x = x * 10 + c - '0';
return p ? -x : x;
}
inline int maxn(int x, int y) { return x > y ? x : y; }
inline int minn(int x, int y) { return x < y ? x : y; }
inline int mod(int x) { return (x % 10 + 10) % 10; }
int main()
{
int i, j, k, l, n, m, o;
n = re(); m = re();
o = n << 1;
for (i = 1; i <= n; i++)
a[i + n] = a[i] = re();
for (i = 1; i <= o; i++)
a[i] += a[i - 1];
memset(g, 60, sizeof(g));
for (i = 1; i <= o; i++)
for (j = i; j <= o; j++)
f[i][j][1] = g[i][j][1] = mod(a[j] - a[i - 1]);
for (l = 2; l <= m; l++)
for (i = 1; i + l - 1 <= o; i++)
for (j = i + l - 1; j <= o; j++)
for (k = i + l - 2; k < j; k++)
{
f[i][j][l] = maxn(f[i][j][l], f[i][k][l - 1] * mod(a[j] - a[k]));
g[i][j][l] = minn(g[i][j][l], g[i][k][l - 1] * mod(a[j] - a[k]));
}
int mi = 1e9, ma = 0;
for (i = 1; i <= n; i++)
{
mi = minn(mi, g[i][i + n - 1][m]);
ma = maxn(ma, f[i][i + n - 1][m]);
}
printf("%d
%d", mi, ma);
return 0;
}