Description
The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.
Input
The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.
Output
Output must contain a single integer - the minimal score.
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
题意：
一系列的数字，除了头尾不能动，每次取出一个数字，这个数字与左右相邻数字的乘积为其价值，最后将所有价值加起来，要求最小
思路：
设定dp【i】【j】表示已经拿完：【i，j-1】区间的所有数之后的最小总价值。
状态转移方程：
dp【i】【j】=min（dp【i】【j】，dp【i】【k】+dp【k+1】【j】+a【k】*a【j】*a【i-1】）；
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 105;
const ll mod = 1e9+7;
const ll INF = 1e18;
const double eps = 1e-9;
ll a[maxn];
ll dp[maxn][maxn];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{cin>>a[i];
}
for(int len=2;len<n;len++) //阶段
for(int i=1;i<=n;i++) //状态
{
int j=i+len-1;
if(j>n) break;
dp[i][j]=1e9;
for(int k=i;k<j;k++) //决策
{
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+a[i-1]*a[j]*a[k]);
}
}
printf("%d\n",dp[2][n]);
}
return 0;
}
