Go: package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

var cache [100][100]int

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	sc.Scan()
	num, _ := strconv.Atoi(sc.Text())
	tables := []table{}
	for i := 0; i < num; i++ {
		sc.Scan()
		num1, _ := strconv.Atoi(sc.Text())
		sc.Scan()
		num2, _ := strconv.Atoi(sc.Text())
		t := table{num1, num2}
		tables = append(tables, t)
	}

	fmt.Println(minimum(0, len(tables)-1, tables))
}

type table struct {
	v int
	h int
}

func minimum(i int, j int, t []table) int {
	if i == j {
		return 0
	}

	// fmt.Printf("i:%v j:%v\n", i, j)
	if cache[i][j] != 0 {
		return cache[i][j]
	}

	min := math.MaxInt64
	for k := i; k < j; k++ {
		p := t[i].v
		n := t[k].h
		q := t[j].h
		val := minimum(i, k, t) + minimum(k+1, j, t) + p*n*q
		if val < min {
			min = val
		}
	}

	cache[i][j] = min
	return min
}

C++: #include<iostream>
#include<algorithm>
using namescape std;

static const int N = 100;

int main(){
  int n,p[N+1],m[N+1][N+1];
  cin >> n;
  for (int i=1;i<=n;i++){
    cin >> p[i-1] >> p[i];
  }

  for( int i=1;i<=n;i++)m[i][i]=0;
  for( int l=2;l<=n;l++){
    for( int i=1;i<=n-l+1;i++){
      int j=i+l-1;
      m[i][j]=(1<<21);
      for( int k=i;k<=j-1;k++){
        m[i][j]=min(m[i][j], m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j]);
        }
      }
    }

  cout << m[1][n] << endl;
  
  return 0;
}

C++: //矩阵连乘问题，递归方程采用课本P47的递归方程。
//与课本P47的程序相比，两者都是填m[][]的上三角，
//但本程序是横向填表，而课本程序是斜向填表。
//本程序按照 自底向上，自左向右 的顺序来计算m[][]的上三角。
//程序中有记录相应断开位置到s[][]。

#include <iostream>
using namespace std;

void MatrixChain(int, int[], int **, int **);
void print(int, int, int **);
void CalculateMin(int i, int j, int p[], int **m, int **s);

int main()
{
    int n;
    cin >> n;
    int *tempP = new int[2 * n]; //p[0]为A1的行数；p[i]为Ai的列数

    int *p = new int[n + 1];    //p[0]为A1的行数；p[i]为Ai的列数
    int **m = new int *[n + 1]; //m[i][j]为Ai连乘到Aj的最少乘次数
    int **s = new int *[n + 1]; //s[i][j]为 Ai连乘到Aj的最优解的第一层断开位置
    for (int i = 1; i <= n; i++)
    {
        m[i] = new int[n + 1];
        s[i] = new int[n + 1];
    }

    for (int i = 0; i < 2 * n; i++)
    {
        cin >> tempP[i];
    }

    for (int i = 0; i <= n; i++)
    {
        p[i] = i == 0 ? tempP[0] : tempP[i * 2 - 1];
    }

    MatrixChain(n, p, m, s);
    cout << m[1][n] << endl; //最优值存储在m[1][n]
    //print(1, n, s); //打印最优计算方式

    return 0;
}

void MatrixChain(int n, int p[], int **m, int **s)
{
    //以下双重循环对m[][]的上三角进行填表
    //填表顺序为自底向上，自左向右
    for (int i = n; i >= 1; i--) //i表示行
    {
        for (int j = i; j <= n; j++) //j表示列，因为只填上三角，所以j的初值为i
        {
            //以下按照矩阵连乘问题的递归公式来求每一个m[i][j]
            if (i == j)
            {
                m[i][j] = 0;
            }
            else
            {
                CalculateMin(i, j, p, m, s);
            }
        }
    }
}

void CalculateMin(int i, int j, int p[], int **m, int **s)
{
    m[i][j] = m[i][i] + m[i + 1][j] + p[i - 1] * p[i] * p[j]; //若i与j不相等，m[i][j]的初值为断开位置为i时的最优值

    s[i][j] = i;                    //记录当前断开位置位置为i
    for (int k = i + 1; k < j; k++) //k用于尝试不同的断开位置
    {
        int curr = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
        if (curr < m[i][j]) //当前k值作为Ai连乘到Aj的断开位置能获得更少计算量
        {
            m[i][j] = curr; //刷新最优值
            s[i][j] = k;   //刷新相应断开位置
        }
    }
}

void print(int i, int j, int **s)
{
    if (i == j)
        cout << "A" << i;
    else
    {
        cout << "(";
        print(i, s[i][j], s);
        print(s[i][j] + 1, j, s);
        cout << ")";
    }
}

C++: #include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <complex>
#include <ctime>
#include <cstdlib>

using namespace std;

inline int to_int(string s) {int v; istringstream sin(s); sin >> v; return v;}
template<class T> inline string to_str(T x) {ostringstream sout; sout << x; return sout.str();}

typedef long long ll;
typedef pair<int, int> P;
const int INF = 1<<22;

// dp[i][j] := [i , j]??????????????????????????§???????????????????°???¨????????°
// |(4*5   (5*3)  3*2)|  2*1
int main()
{
	int n, dp[128][128], r[128];
	P T[128];

	cin >> n;

	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
		 	dp[i][j] = INF;
		}
	}
	/*
	for(int i = 0; i < n; i++)
	{
		dp[i][i] = 0;
	}
	*/
	for(int i = 0; i < n; i++)
	{
		cin >> T[i].first >> T[i].second;
		dp[i][i] = 0;
	}
	/*
	r[0] = T[0].first;
	for(int i = 0; i < n; i++)
	{
		r[i+1] = T[i].second;
	}
	*/
	/*
	for(int i = 0; i <= n; i++)
	{
		cout << r[i] << endl
	}
	*/
	for(int l = 1; l < n; l++)
	{
		for(int i = 0; i < n-l; i++)
		{ 
			int j = i + l;
			for(int k = 1; k <= l; k++)
			{
				dp[i][j] = std::min(dp[i][j], dp[i][i + k - 1] + dp[i + k][j] + T[i].first*T[i + k].first*T[j].second); //r[j]*r[k]*r[i+j+1]);
			}
		}
	}

	/*for(int i = 1; i < n; i++)
	  {
	  for(int j = 0; j < n-i; j++)
	  { 
	  for(int k = 1; k <= i; k++)
	  {
	  dp[i][j] = std::min(dp[i][j], dp[k-1][j] + dp[i-k][j+k] + T[j].first*T[j + k].first*T[j + i].second); //r[j]*r[k]*r[i+j+1]);
	  printf("%3d : %3d : %3d => %3d\n", i, j, k, dp[i][j]);
	//puts("run");
	}
	}
	}*/

	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			printf("%10d ",dp[i][j]);
		}
		puts("");
	}

	cout << dp[0][n - 1] << endl;
}

C++: #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX 100
int min(int,int);
int p[MAX]={},m[MAX][MAX]={};

int min(int n1,int n2){
  if(n1>=n2) return n2;
  else return n1;
}

int main()
{
  int n,r,l,i,j,k;
  
  scanf("%d",&n);
  
  for(i=1;i<=n;i++){
    scanf("%d %d",&p[i-1],&p[i]);
  }
  
  for(l=2;l<=n;l++){
    for(i=1;i<=n-l+1;i++){
      j=i+l-1;
      m[i][j]=(1<<21);
      for(k=i;k<=j-1;k++){
	m[i][j]=min(m[i][j],m[i][k]+m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j]);
      }
    }
  }
  
  printf("%d\n",m[i][n]);
  
  return 0;
}