Go: package main

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

var sc = bufio.NewScanner(os.Stdin)

func nextInt() int {
	sc.Scan()
	i, _ := strconv.Atoi(sc.Text())
	return i
}
func main() {
	sc.Split(bufio.ScanWords)
	var n int
	var M [101][101]int

	n = nextInt()

	P := make([]int, n*2)

	for i := 1; i <= n; i++ {
		P[i-1] = nextInt()
		P[i] = nextInt()
	}
	for l := 2; l <= n; l++ {
		for i := 1; i <= n-l+1; i++ {
			j := i + l - 1
			M[i][j] = 1000000
			for k := i; k <= j-1; k++ {
				M[i][j] = int(math.Min(float64(M[i][j]), float64(M[i][k]+M[k+1][j]+P[i-1]*P[k]*P[j])))
			}
		}
	}
	fmt.Println(M[1][n])
}

C++: /**
 * 	matrix chain multiplication
 * 	f(n,m) = multiply Mn*M(n+1)*...*M(m-1)*M(m)
 * 	f(n,n) = 1
 * 	r(n)=c(n-1)
 * 	f(n,n+1) = C(n) * R(n) * C(n+1) = c(n) * c(n-1) * c(n+1)
 * 	f(n,k,m) = f(n,k)*f(k+1,m)*C(n)*r(n)*c(k)
 * 	f(n,m) = min(f(n,k,m)
 */

#include <iostream>
#include <cmath>
#include <climits>

using namespace std;

int r[45] = {0};
int a[100][100] = {0};

int f(int n, int m);
int f(int n, int k, int m);


int f(int n, int k, int m)
{
	return f(n,k) + f(k+1,m) + (r[n] * r[k+1] * r[m+1]);
}

int f(int n, int m)
{
	if (n==m) return 0;
	int & ra = a[n][m];
	if (ra) return ra;
	ra = INT_MAX;
	for (int k=n;k<m;k++)
	{
		int ans = f(n,k,m);
		if (ans < ra) ra = ans;
	}
	return ra;
}

int main()
{
	int rr = 0;
	int cc = 0;
	int n = 0;
	cin >> n;
	for (int i=0;i<n;i++)
	{
		cin >> rr >> cc;
		r[i]=rr;
		r[i+1]=cc;
	}
	for (int i=0;i<n-1;i++)
	{
		a[i][i+1] = r[i] * r[i+1] * r[i+2];
	}
	int ans = f(0,n-1);
	cout << ans << endl;
}

C++: //  AOJ Algorithms and Data Structures I - Dynamic Programming - Matrix Chain Multiplication
// s 2013 0925 1640
// e 

/* 2014 0329 1621

イメージ的には
---------------------
|M11|M12|M13|   |   | 
---------------------
|   |M22|M23|   |   | 
---------------------
|   |   |M33|   |   |  
---------------------
|   |   |   |M44|   |
---------------------
|   |   |   |   |M55|
---------------------

-------------------
|r1|r2|r3|r4|r5|r6|
-------------------

この二つを作って使う

*/


#include<iostream>
#include<cstdio>
#include<vector>

using namespace std;


//コストの計算 costを参照渡しで持ってきて、直接入れる
// s eはstart end の略でcostの配列添え字
void cal_cost(vector< vector<int> >& cost, vector<int> r, int s, int e){
	int ans;
	int k; //切る場所 (k=2 なら、　1|234 )

	if(s==e){ cost[s][e]=0; } //Miiのコストは0
	else{
		cost[s][e]=2000000000; //最小値求めるためのセンチネル　数値は適当
		for(k=s+1; k<=e; k++){
			ans = cost[s][k-1] + cost[k][e] + r[s]*r[k]*r[e+1];
			if(ans<cost[s][e]) cost[s][e] = ans;
		}
	}
}


//http://algorithms.blog55.fc2.com/blog-entry-66.html
//このホームページ見て正解っぽく動的計画法で書いてみる版
int DP_myself( vector< vector<int> > A, int n){
	int i, diff;

	//最終的な出力は最小コストだけでいいので、かけ算の順番は保存しない。するとしたら、cost配列を構造体にして、ルーツの保存しとけば、それでたどれると思う
	// (cost[][]の使い方)  M_ijのコストをcost[i][j]に入れる (M11, M22等はコスト0)
	vector< vector<int> > cost(n,vector<int>(n)); //Mijにかかるコスト（最少）

	//行とか列の長さを入れる配列
	vector<int> r(n+1);

	//rを埋めてく
	for(i=0;i<n;i++){
		r[i]=A[i][0];
	}
	r[n]=A[n-1][1];

	//costを埋めていく(右斜め下に進んで。ピラミッド状に)
	for(diff=0; diff<n; diff++){
		for(i=0; i+diff<n; i++){
			// デバッグ用printf("%d,%d,%d\n", diff,i,i+diff);
			cal_cost(cost, r, i, i+diff); //コストの計算
		}
		
	}

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

	return 0;
}


int main(){
	int i;
	int n;
	int num_multi=0;
	
	cin >> n;
	//dequeやlistでなくても、途中の要素の削除できるんでvectorでいい。
	// C++で2次元配列を動的に確保する方法 http://samidarehetima.blog9.fc2.com/blog-entry-95.html
	//連鎖行列はA[0][] ~ A[n-1][] まで。
	vector<vector<int> > A(n,vector<int>(2)); //[][0]=r, [][1]=c
	for(i=0;i<n;i++){
		cin >> A[i][0];
		cin >> A[i][1];
	}


	DP_myself(A,n);

    return 0;
}

C++: // ALDS1_10_B.cpp
// Dynamic Programming - Matrix Chain Multiplication

#include <iostream>
#include <algorithm>
#include <vector>
#include <climits>
using namespace std;
static const int N = 100;
int main()
{
    int n;
    int P[N + 1];
    int m[N + 1][N + 1];
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> P[i] >> P[i + 1];
    }

    for (int k = 0; k < n; k++) {
        for (int i = 1; i + k <= n; i++) {
            if (k == 0) m[i][i+k] = 0;
//            else if (k == 1) m[i][i+1] = P[i - 1] * P[i] * P[i + 1];
            else {
                int res = INT_MAX;
                for (int j = 0; j < k; j++) {
                    res  = min(res, m[i][i+j] + m[i+j+1][i+k] + P[i-1] * P[i+j] * P[i+k]);
                }
                m[i][i+k] = res;
            }
        }
    }

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

C++: #include <stdio.h>
#include <stdlib.h>
#define N 100

int order(int, int *);

int main(){

  int i;
  int n, p[N + 1];

  scanf("%d", &n);

  for(i = 0; i < n; i++){
    scanf("%d%d", &p[i], &p[i + 1]);
  }

  //for(i = 0; i <= n; i++) printf("%d\n", p[i]);

  printf("%d\n", order(n, p));

  return 0;
}

int order(int n, int *p){

  int i, j, k;
  int t, q, m;
  int a[N][N];

  for(i = 1; i <= n; i++){
    a[i][i] = 0;
  }

  for(i = 2; i <= n; i++){
    for(j = 1; j <= n - i + 1; j++){
      t = i + j - 1;
      a[j][t] = 114514;
      for(k = j; k < t; k++){
	q = a[j][k] + a[k + 1][t] + p[j - 1] * p[k] * p[t];
	if(j == k) m = q;
	  else if(m > q) m = q;
      }
      a[j][t] = m;
    }
  }
  
  return a[1][n]; 
}

C++: #include <stdio.h>
#include <iostream>
#include <string>
using namespace std;
#define NUM 100
#define INF 1000000000

int X_dim[NUM];
int Y_dim[NUM];
int M[NUM][NUM];

void init(){
    for(int i = 0; i < NUM; i++){
        for(int j = 0; j < NUM; j++){
            if(i == j){
                M[i][j] = 0;
            }else{
                M[i][j] = INF;
            }
        }
    }

}

int main(void){
    init();

    int n;
    scanf("%d", &n);

    for(int i = 0; i < n; i++){
        scanf("%d %d", &X_dim[i], &Y_dim[i]);
    }

    for(int k = 1; k <= n - 1; k++){
        for(int x = 0; x + k <= n - 1; x++){
            for(int t = x; t <= x + k - 1; t++){
                M[x][x + k] = min(M[x][x + k], M[x][t] + M[t + 1][x + k] + X_dim[x] * Y_dim[t] * Y_dim[x + k]);
                // printf("M[%d][%d]: M[%d][%d] + M[%d][%d] + %d * %d * %d\n", x, x + k, x,t,t+1,x+k,X_dim[x],Y_dim[t],Y_dim[x+k]);
            }
        }
    }

    printf("%d\n", M[0][n - 1]);
}

C++: #include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;

int dp[100][100];

int main(void)
{   
    int n;
    cin >> n;
    pair<int, int> rc[n];
    rep(i, n) cin >> rc[i].first >> rc[i].second;

    vector<vector<int> > dp(n, vector<int>(n, 1e9));
    
    rep(i, n) {
        dp[i][i] = 0;
    }

    for(int i = 1; i <= n-1; i++) {
        for(int j = 0; j <= n - 1 - i; j++) {
            for(int k = j; k < j + i; k++) {
                dp[j][j+i] = min(dp[j][j+i], dp[j][k] + dp[k+1][j+i] + rc[j].first*rc[k].second*rc[j+i].second);
            }
        }
    }

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

C++: #include "bits/stdc++.h"
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;
}