原题链接
设(f[i][j])表示在([1, i])中放置(j)个乘号，且第(i)个数字后面放第(j)个乘号时所获得的最大乘积。$ace(1, i)表示将(1 sim i)的数字变为一个数。
有状态转移方程：
[f[i][j] = max { f[i][j], max limits _{suc = j - 1} ^ {suc < i} { ace(suc + 1, i) imes f[suc][j - 1] } }
]
初始化(f[i][1] = ace(1, i))，其余为空。
最后的答案就是(max limits _{i = k} ^ {i < n} { ace(i + 1, n) imes f[i][k] })
因为(n)可达到(40)，所以要用高精。
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 50;
const int K = 10;
struct bigint {
int s[N << 3], l;
void CL() { l = 0; memset(s, 0, sizeof(s)); }
void pr()
{
printf("%d", s[l]);
for (int i = l - 1; i; i--)
printf("%d", s[i]);
}
bigint operator * (bigint &b)
{
bigint c;
int i, j, k, x;
c.CL();
for (i = 1; i <= l; i++)
{
x = 0;
for (j = 1; j <= b.l; j++)
{
x = x + s[i] * b.s[j] + c.s[k = i + j - 1];
c.s[k] = x % 10;
x /= 10;
}
if (x)
c.s[i + b.l] = x;
}
for (c.l = l + b.l; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bool operator < (const bigint &b) const
{
if (l ^ b.l)
return l < b.l;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return s[i] < b.s[i];
return false;
}
};
bigint f[N][K], an;
int a[N], l;
inline void re_l()
{
char c = getchar();
for (; c < '0' || c > '9'; c = getchar());
for (; c >= '0' && c <= '9'; c = getchar())
a[++l] = c - '0';
}
inline bigint maxn(bigint x, bigint y) { return x < y ? y : x; }
bigint ace(int x, int y)
{
bigint o;
o.l = y - x + 1;
for (int i = x; i <= y; i++)
o.s[i - x + 1] = a[y - i + x];
return o;
}
int main()
{
int i, j, k, n, suc;
scanf("%d%d", &n, &k);
re_l();
for (i = 1; i < n; i++)
{
f[i][1] = ace(1, i);
for (j = 2; j <= k; j++)
for (suc = j - 1; suc < i; suc++)
if (f[suc][j - 1].l)
f[i][j] = maxn(f[i][j], ace(suc + 1, i) * f[suc][j - 1]);
if (f[i][k].l)
an = maxn(an, ace(i + 1, n) * f[i][k]);
}
an.pr();
return 0;
}