Go: package main

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

//util
var sc = bufio.NewScanner(os.Stdin)

func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func printList(list []int) {
	buf := bufio.NewWriter(os.Stdout)
	for _, v := range list {
		buf.Write([]byte(fmt.Sprintf(" %d", v)))
	}
	buf.Write([]byte("\n"))
	buf.Flush()
}

//solve
//単純な場合
func matrixChainMultiplication() {
	//行列が1つだけのとき
	for i := 1; i <= n; i++ {
		m[i][i] = 0
	}
	for l := 2; l <= n; l++ {
		for i := 1; i <= n-l+1; i++ {
			j := i + l - 1
			m[i][j] = 1 << 21
			//kまでで区切る
			for 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])
			}
		}
	}
}

func max(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func min(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

const MAX = 100

var n int
var p [MAX + 1]int

//m[i][j]: Mi * Mi+1 * ... * Mjを計算するときに必要なスカラー乗算の最小回数
var m [MAX + 1][MAX + 1]int

func main() {
	sc.Split(bufio.ScanWords)
	n = nextInt()
	for i := 1; i <= n; i++ {

		p[i-1] = nextInt() //行数
		p[i] = nextInt()   //列数
	}

	matrixChainMultiplication()
	fmt.Println(m[1][n])
}

C++: #include <iostream>
#include <algorithm>
using namespace std;
#define MAX_N 100
#define INF (1 << 21)
int cost[MAX_N][MAX_N];
int rc[MAX_N+1];

int main() {
  int n; cin >> n;
  for(int i=0;i<n;i++) {
    int tmp;
    if(i==0) {
      cin >> rc[i] >> rc[i+1];
    }
    else {
      cin >> tmp >> rc[i+1];
    }
  }

  // cost[i][i] = 0
  /*
  for(int i=0;i<n;i++) {
    cost[i][i] = 0;
  }
  */
  // cost[i][i+1] = rc[i]*rc[i+1]*rc[i+2]
  for(int i=0;i<n-1;i++) {
    cost[i][i+1] = rc[i]*rc[i+1]*rc[i+2];
  }
  // calculation cost[i][i+j]
  for(int j=2;j<n;j++) {
    for(int i=0;i+j<n;i++) {
      cost[i][i+j] = INF;
      for(int k=i+1;k<=i+j;k++) {
        cost[i][i+j] = min(cost[i][i+j], cost[i][k-1] + cost[k][i+j] + rc[i]*rc[k]*rc[i+j+1]);
      }
    }
  }

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

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

static const int N=100;

//求解矩阵链乘法的最优解
//p 用于存储矩阵的行和列的
//dp[i][j] 表示从矩阵i乘到矩阵j的最小乘法次数

int main(){
    int n;
    int p[N+1],dp[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++){
        dp[i][i]=0;
    }
    for(int l=2;l<=n;l++){
        for(int i=1;i<=n-l+1;i++){
            int j=i+l-1;
            dp[i][j]=(1<<21);//取一个很大的值
            for(int k=i;k<=j-1;k++){
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+p[i-1]*p[k]*p[j]);
            }
        }
    }

    cout<<dp[1][n]<<endl;

    return 0;
}

C++: #define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <utility>
#include <numeric>

using namespace std;
typedef long long llong;
typedef unsigned long long ullong;

const ullong MOD = 1000000000 + 7;

const int N = 101;

int main()
{
	int n;
	cin >> n;

	vector<int> p(N);
	for (int i = 1; i <= n; i++)
	{
		cin >> p.at(i - 1) >> p.at(i);
	}

	vector<vector<int>> m(N, vector<int>(N));

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

	cout << m.at(1).at(n) << endl;
}

C++: #include <bits/stdc++.h>
using namespace std;

// m[i][j]はM_i...M_jの最小コスト
vector<vector<int> > m(101, vector<int>(101));
// M_iはP[i-1]xP[i]行列
vector<int> P(101);

int n;

void solve()
{
  // M_iの最小コストをメモ
  for (int i = 1; i <= n; i++)
  {
    m.at(i).at(i) = 0;
  }
  for (int l = 2; l <= n; l++)
  {
    // 長さlの行列積の最小コストをメモ
    for (int i = 1; i <= n - l + 1; i++)
    {
      int j = i + l - 1;
      // M_i...M_jの最小コストを、M_i...M_kの最小コストとM_k...M_jの最小コストを参照して計算する
      m.at(i).at(j) = (1 << 21); // 2の21乗
      for (int k = i; k <= j - 1; k++)
      {
        m.at(i).at(j) = min(m.at(i).at(j), m.at(i).at(k) + m.at(k + 1).at(j) + P.at(i - 1) * P.at(k) * P.at(j));
      }
    }
  }
}

int main()
{
  cin >> n;
  for (int i = 0; i < n; i++)
  {
    int r, c;
    if (i == 0)
    {
      cin >> r >> c;
      P.at(0) = r;
      P.at(1) = c;
    }
    else
    {
      cin >> r >> c;
      P.at(i + 1) = c;
    }
  }

  solve();

  cout << m.at(1).at(n) << endl;
}

C++: #include <iostream>
#include<climits>
#include<vector>
using namespace std;
 
int MCM(int p[],int n) {
    vector<vector<int>> m(n,vector<int>(n));
    for (int i = 0; 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] = INT_MAX;
            for (int k = i; k <= j-1; k++) {
                int q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (m[i][j] > q) {
                    m[i][j] = q;
                }
 
            }
        }
 
    }
    return m[1][n];
}
 
int main() {
    int n, p[200];
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> p[i] >> p[i + 1];
    }
    cout << MCM(p, n) << endl;
    return 0;
}

C++: #include<stdio.h>

#define N 100
int A[N + 1][N + 1];
int p[N + 1];

int solve(int i, int j)
{
    if(!(A[i][j] < 0)){ return A[i][j]; }
    if(j == i){ A[i][i] = 0; return 0; }
    if(j == i + 1){
        A[i][j] = p[i - 1] * p[i] * p[i + 1];
        return A[i][j];
    }
    int min = -1, x;
    for(int k = i; k < j; k++){
        x = solve(i, k) + solve(k + 1, j) + p[i - 1] * p[k] * p[j];
        if(min < 0){ min = x; }
        else{ if(min > x) min = x; }
    }
    A[i][j] = min;
    return min;
}

int main()
{
    int i, j, n;
    scanf("%d", &n);
    for(i = 1; i <= n; i++){
        scanf("%d %d", &p[i - 1], &p[i]);
        for(j = 1; j <= n; j++){
            A[i][j] = -1;
        }
    }
    printf("%d\n", solve(1, n));
    return 0;
}

C++: // http://algorithms.blog55.fc2.com/blog-entry-66.html

#include <iostream>
using namespace std;
#include <stdio.h>

#define MAX 100
#define INFTY (1 << 21)

int SIZE;
int R[MAX + 2];
int C[MAX+1][MAX+1]; // cost table
int B[MAX+1][MAX+1]; // best table

void compute(){
    for ( int i = 1; i <= SIZE; i++ )
    for ( int j = 1; j <= SIZE; j++ ) C[i][j] = INFTY;
    
    for ( int i = 1; i <= SIZE; i++ ) C[i][i] = 0;
    
    for ( int j = 1; j <= SIZE - 1; j++ ){
        for ( int i = 1; i <= SIZE - j; i++ ){
            for ( int k = i + 1; k <= i + j; k++ ){
                // C[i][i+j] = min( C[i][i+j], 
                //                  C[i][k-1] + C[k][i+j] + R[i]*R[k]*R[i+j+1]);
                int cost = C[i][k-1] + C[k][i+j] + R[i]*R[k]*R[i+j+1];
                if ( cost < C[i][i+j] ){
                    C[i][i+j] =  cost;
                    B[i][i+j] = k;
                }
                
            }
        }
    }
}

void order( int i, int j ){
    if ( i == j ) cout << "M" << i;
    else{
        cout << "(";
        order( i, B[i][j]-1 ); order(B[i][j], j);
        cout << ")";
    }
}

void input(){
    cin >> SIZE;
	int z,i=1;
    for (; i <= SIZE; i++ ) cin >> R[i]>>z;
	R[i]=z;
	//SIZE++;
}

void output(){
    cout <<  C[1][SIZE] << endl;
    //order(1, SIZE);
    //cout << endl;
}

main(){
    input();
    compute();
    output();
}