题目链接:点这里!!!!
题解:
因为是环形,我们就把长度扩大一倍。
然后我们设dp[i][j]表示消除i~j区间剩余a[i]的最大值为多少?
则转移方程:dp[i][j]=max(dp[i][k],dp[k+1][j]+a[i]*a[k+1]*a[j])。
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<vector>
#include<bitset>
#include<set>
#include<queue>
#include<stack>
#include<map>
#include<cstdlib>
#include<cmath>
#define LL long long
#define pb push_back
#define pa pair<int,int>
#define clr(a,b) memset(a,b,sizeof(a))
#define lson lr<<1,l,mid
#define rson lr<<1|1,mid+1,r
#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)
#define key_value ch[ch[root][1]][0]
#pragma comment(linker, "/STACK:102400000000,102400000000")
const LL MOD = 1000000007;
const int N = 200+15;
const int maxn = 1e5+15;
const int letter = 130;
const LL INF = 1e7;
const double pi=acos(-1.0);
const double eps=1e-10;
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,dp[N][N],a[N];
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",a+i),a[i+n]=a[i];
for(int len=2;len<=2*n;len++){
for(int i=1;i<=2*n;i++){
int j=i+len-1;
if(j>2*n) break;
for(int k=i;k<j;k++){
dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]+a[i]*a[k+1]*a[j+1]);
}
}
}
int max1=0;
for(int i=1;i<=n;i++) max1=max(max1,dp[i][i+n-1]);
printf("%d\n",max1);
return 0;
}