Go: package main

import (
	"fmt"
)

func main() {
	var n int
	fmt.Scan(&n)
	a := make([]int, n+1)
	for i := 0; i < n; i++ {
		fmt.Scan(&a[i])
		fmt.Scan(&a[i+1])
	}

	m := make([][]int, n+1)
	for i := 0; i <= n; i++ {
		m[i] = make([]int, n+1)
	}

	for d := 2; d <= n; d++ {
		for s := 0; s+d <= n; s++ {
			min := 1 << 30
			for j := s+1; j < s+d; j++ {
				val := m[s][j] + m[j][s+d] + a[s]*a[j]*a[s+d]
				if val < min {
					min = val
				}
			}
			m[s][s+d] = min
		}
	}
	fmt.Println(m[0][n])
}

C++: #include <stdio.h>

int A[101][101];

void chain(int n, int *x){
  int i, j, k, l;
  
  for(i=1; i<n+1; i++)
    A[i][i] = 0;
    
  for(i=2; i<n+1; i++){
    for(j=1; j<n-i+2; j++){
      l = i+j-1;
      A[j][l] = 100000000;
      for(k=j; k<l; k++){
        if(A[j][l] > A[j][k]+A[k+1][l]+x[j-1]*x[k]*x[l])
          A[j][l] = A[j][k]+A[k+1][l]+x[j-1]*x[k]*x[l];
      }
    }
  }
}

int main(){
  int i,n;
  int x[101];
  
  scanf("%d", &n);
  
  for(i=1; i<n+1; i++)
    scanf("%d%d", &x[i-1], &x[i]);
    
  chain(n, x);
  printf("%d\n", A[1][n]);
  
  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 -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>

#define INF(1 << 21)

using namespace std;

int main {

    int i, j, k, l, n, p[101], m[101][101];

    cin >> n;

    for (i = 1; i <= n; i++) {
        cin >> p[i - 1] >> p[i];
    }

    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] = INF;

            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]);
            }
        }
    }

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

    return 0;

}

C++: // https://www.youtube.com/watch?v=prx1psByp7U

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

#define N 101

int dp[N+1][N+1], p[N+1];

// dp[i][j] := [Mi, Mj]におけるminimum number of scalar multiplications 

int main(){
    
    int n;
    cin >> n;
    
    for (int i = 1; i < n+1; i++) cin >> p[i-1] >> p[i];
    
    
    // l はchain lengthを表す変数である.
    // l == 1のとき計算回数は0である.
    for (int l = 2; l <= n; l++){
        
        // 変数iは区間の左端を表す
        // 左端が動ける範囲の上限は, n - l + 1である
        // 具体例を考えてみよ
        for (int i = 1; i <= n-l+1; i++){
            
            // jは右端の上限値を表す
            int j = i + l - 1;
            
            dp[i][j] = (1<<20);
            
            // 今回調べる区間は, [i, j-1]
            for (int k = i; k < j; 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;
}

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

int main()
{
	int n, p[101], dp[101][101];
	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);
			for (int k = i; k < j; ++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<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(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[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j]);
        }
    }
  }
  cout << m[1][n] << endl;
  return 0;
}