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

題意：

給n-1個矩陣排成的序列，現在讓你求這n-1個矩陣相乘的最小計算量。

思路：

區間dp

n*m的矩陣和m*p的矩陣相乘的計算量是n*m*p，，這題可以把整個序列分成更小的問題，即每次都是計算第i個矩陣乘到第j個矩陣的最小計算量，記憶化搜尋即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn=105;
int dp[maxn][maxn];
int a[maxn];
int n;
int work(int x, int y)   //計算x到y的最小計算量
{
    if(dp[x][y]!=INF)
        return dp[x][y];
    if(x==y)
        return dp[x][y]=0;
    else
    {
        for(int i=x; i<=y; i++)    //再次分成更小的區間
            dp[x][y]=min(dp[x][y], work(x, i)+work(i+1, y)+a[x]*a[i+1]*a[y+1]);
        return dp[x][y];
    }
}
int main()
{
    while(~scanf("%d", &n))
    {
        memset(a, 0, sizeof(a));
        memset(dp, 0x3f, sizeof(dp));
        for(int i=1; i<=n; i++)
            scanf("%d", &a[i]);
        printf("%d\n", work(1, n-1));
    }
    return 0;
}