Markdown炸了
```cpp
#include<bits/stdc++.h>
using namespace std;
struct Int{
int len;
short num[500];
int operator[](int a){
return num[a];
}void operator=(int a){
this->len=0;
memset(this->num,0,sizeof(this->num));
while(a){
this->num[++this->len]=a%10;
a/=10;
}
}
};
Int operator+(Int a,Int b){
int m=0;
if(a.len<b.len)swap(a,b);
for(int i=1;i<=a.len;i++){
a.num[i]+=b[i]+m;
m=a[i]/10;a.num[i]%=10;
}if(m>0)a.num[++a.len]=m;
return a;
}Int operator*(Int a,Int b){
Int c;c.len=0;int m=0;
memset(c.num,0,sizeof(c.num));
for(int i=1;i<=a.len;i++){
for(int j=1;j<=b.len;j++){
c.num[i+j-1]+=a[i]*b[j];
if(c[i+j-1])c.len=max(c.len,i+j-1);
}
}for(int i=1;i<=c.len;i++){
c.num[i]+=m;m=c[i]/10;c.num[i]%=10;
}while(m){
c.num[++c.len]=m;
m=c[c.len]/10;
c.num[c.len]%=10;
}return c;
}Int operator+(Int a,int b){
Int c;c=b;
return a+c;
}Int operator*(Int a,int b){
Int c;c=b;
return a*c;
}bool operator<(Int a,Int b){
if(a.len!=b.len)return a.len<b.len;
for(int i=a.len;i>0;i--){
if(a[i]!=b[i])return a[i]<b[i];
}return false;
}bool operator==(Int a,Int b){
if(a.len!=b.len)return false;
for(int i=1;i<=a.len;i++){
if(a[i]!=b[i])return false;
}return true;
}bool operator>(Int a,Int b){
if(a<b)return false;
else if(a==b)return false;
else return true;
}bool operator<=(Int a,Int b){
return !(a>b);
}bool operator>=(Int a,Int b){
return !(a<b);
}istream &operator>>(istream &in,Int &a){
char l[500];cin>>l;a.len=strlen(l);
memset(a.num,0,sizeof(a.num));
int al=a.len,ll=0;
while(al>0&&ll<a.len){
a.num[al]=l[ll]-'0';
al--;ll++;
}return in;
}ostream &operator<<(ostream &out,Int a){
for(int i=a.len;i>0;i--){
cout<<a[i];
}return out;
}Int max(Int a,Int b){
return a>b?a:b;
}Int min(Int a,Int b){
return a<b?a:b;
}
Int dp[50][10],m[50][50];int a[50],n,k;
int main(){
cin>>n>>k;getchar();
for(int i=1;i<=n;i++){
a[i]=getchar()-'0';
dp[i][1]=dp[i-1][1]*10+a[i];
}for(int i=1;i<=n;i++){
m[i][i]=a[i];
for(int j=i+1;j<=n;j++){
m[i][j]=m[i][j-1]*10+a[j];
}
}for(int i=1;i<=n;i++){
for(int j=2;j<=k+1;j++){
if(i<j)break;
for(int l=1;l<i;l++){
if(l<j-1)continue;
dp[i][j]=max(dp[i][j],dp[l][j-1]*m[l+1][i]);
}
}
}cout<<dp[n][k+1];
if(dp[n][k+1].len==0)cout<<0;
return 0;
}
```
by _cyh_ @ 2024-01-25 09:20:45
已AC，输入换一种写法就过了
by _cyh_ @ 2024-01-25 09:29:09