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 <cstdlib>
#include <climits>

/*
 * http://ttrsq.exblog.jp/24318687/
 */

const int MAX_N = 100;
int DP[MAX_N + 1][MAX_N + 1] = {0};
int *p;

int best(int r, int c) {
  int min = INT_MAX;
  int num = c - r;

  for (int i = 0; i < num; i++) {
    min = std::min(min, DP[r][r + i] + DP[r + i + 1][c] + p[r - 1] * p[r + i] * p[c]);
  }
  
  return min;
}

int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);

  int n = 0;
  std::cin >> n;

  if (n == 1) {
    std::cout << "0" << std::endl;
  }
  
  p = (int *) std::malloc(sizeof(int) * n + 1);

  int bb = 0;

  std::cin >> p[0];
  std::cin >> p[1];
  
  for (int i = 2; i <= n; i++) {
    std::cin >> bb;
    std::cin >> p[i];
  }

  for (int i = 1; i < n; i++) {
    DP[i][i + 1] = p[i - 1] * p[i] * p[i + 1];
  }

  for (int j = 0; j <= n - 3; j++) {
    for (int i = j + 3; i <= n; i++) {
      DP[i - 2 - j][i] = best(i - 2 - j, i);
    }
  }

  std::cout << DP[1][n] << std::endl;

  return 0;
}

C++: //?????????B 
#include <iostream>
#include <algorithm>
using namespace 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++: #include <iostream>
#include <algorithm>
using namespace std;
#define INF (1 << 20)
int main()
{
    int n;
    cin >> n;
    int p[101];

    // i番目の行列は p[i]*p[i+1]
    for (int i = 0; i < n; i++)
    {
        cin >> p[i] >> p[i + 1];
    }

    // m[i][j] : i番目からj番目までの部分積でのスカラー乗算の最小回数
    //      i = 0,1, ... , n-1
    // m[i][i] := 0
    // m[i][j] = min_{i <= k < j} m[i][k] + m[k+1][j] + p[i]*p[k+1]*p[j+1]
    //      l = 2,3, ... , n
    //      i = 0,1, ... , n-l
    //          j = i+l-1
    //      k = i, i+1, ... , j-1
    // l = j-i
    int m[101][101];
    for (int i = 0; i < n; i++)
    {
        m[i][i] = 0;
    }

    for (int l = 2; l <= n; l++)
    {
        for (int i = 0; i <= n-l; i++)
        {
            int j = i + l - 1;
            int minv=INF;
            for (int k = i; k < j; k++)
            {
                minv = min(minv, m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1]);
            }
            m[i][j] = minv;
            
            // cout << "m[ " << i<<" ][ "<<j<< " ] = " << minv << endl;
        }
    }
    cout << m[0][n-1] << endl;
    return 0;
}

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++: 
#include<algorithm>
#include<iostream>

using namespace std;
typedef long long ll;

const int MAX = 100;



int main() {


	int n, p[MAX + 1], m[MAX + 1][MAX + 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 calcLen = 2;calcLen <= n;calcLen++) {

		for (int st = 1;st <= n - calcLen + 1;st++) {
			int ed = st + calcLen - 1;
			m[st][ed] = (1 << 21);

			for (int divP = st;divP <= ed-1;divP++) {
				m[st][ed] = min(m[st][ed], m[st][divP] + m[divP + 1][ed] + p[st - 1] * p[divP] * p[ed]);
			}
		}
	}
	cout << m[1][n] << endl;

	return 0;
}

C++: #include<iostream>
#define MAX 99999999;
using namespace std;
int p[102];
int mcm[102][102];
int n;
int min(int x,int y) {
	int m = MAX;
	int cost = 0;
	for (int j = x; j < y; j++) {
		cost = mcm[x][j] + mcm[j + 1][y] + p[x-1]*p[j]*p[y];
		if (cost < m) {
			m = cost;
		}
	}
	return m;
}
int calMcm() {
	for (int j = 2; j <= n; j++) {
		for (int i = 1; i <= n-j+1; i++) {
			mcm[i][j+i-1] = min(i, j+i-1);
		}
	}
	return mcm[1][n];
}
int main() {
	int t;

	cin >> n;
	cin >> p[0];
	cin >> p[1];
	for (int i = 2; i <= n; i++) {
		cin >> t >> p[i];
	}

	/*for (int i = 0; i <= n; i++) {
		cout << p[i] << " ";
	}
	cout << endl;*/

	cout << calMcm() << endl;
}

C++: #include <iostream>
#include <bits/stdc++.h>
#define endl '\n'
#define MAXN 1000000000
#define vi vector<int>
#define pii pair<int, int>
#define vii vector<pii>
using namespace std;


long long dp[101][101];
int s[101][101];
void order(int i,int j){
    int k;
    if(i==j) printf("A%d",i);
    else{
        k=s[i][j];
        printf("(");
        order(i,k);
        order(k+1,j);
        printf(")");
    }
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, n1; cin >> n; n1 = n;
    vi dim;
//    dim.push_back(0);
    int r, c;
    cin >> r >> c;
    dim.push_back(r);
    dim.push_back(c);
    n1--;
    while(n1--){
        cin >> r >> c;
        dim.push_back(c);
    }
    n++;
    for (int i = 0; i < n; ++i) {
        dp[i][i] = 0;
    }
    long long cost = 0;
//    n = dim.size() - 1;
    long long mincost = -MAXN;
    for (int len = 2; len < n; ++len) {
        for (int i = 1; i < n - len + 1; ++i) {
            int j = i + len - 1;
            dp[i][j] = MAXN;
            for (int k = i; k < j; ++k) {
                cost = dp[i][k] + dp[k + 1][j] + dim[i - 1] * dim[j] * dim[k];
                if(cost < dp[i][j]){
                    mincost = cost;
                    dp[i][j] = cost;
                    s[i][j] = k;
                }
            }
        }
    }
//    for (int l = 0; l <= n; ++l) {
//        for (int i = 0; i <= n; ++i) {
//            cout << dp[l][i] << " ";
//        }
//        cout << endl;
//    }
//    printf("Minimum Multiplication : %Ld\n",dp[1][n]);
//    printf("Order : ");
//    order(1,n);
//    printf("\n\n");
    cout << dp[1][n - 1] << endl;
//    cout << dp[2][n] << endl;
    return 0;
}

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