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 <bits/stdc++.h>
#define LOOP(n) for (int _i = 0; _i < (n); _i++)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define RREP(i, n) for (int i = (n); i >= 0; --i)
#define FOR(i, r, n) for (int i = (r); i < (n); ++i)
#define ALL(obj) begin(obj), end(obj)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;
using Graph = vector<vector<int>>;

const int INF = 2100100100;
const int MOD = 1e9 + 7;

// 多次元 vector 生成
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}

int N, M, K, Q, W, H, R, C;
int dp[200][200];

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

    cin >> N;
    vector<int> a(N), b(N);
    REP(i, N) cin >> a.at(i) >> b.at(i);

    REP(i, N + 1) {
        REP(j, N + 1) {
            dp[i][j] = INF;
            if (i + 1 == j) dp[i][j] = 0;
        }
    }

    FOR(len, 1, N) {
        REP(l, N) {
            int r = l + len + 1;
            if (r <= N) {
                FOR(h, l + 1, r) { chmin(dp[l][r], dp[l][h] + dp[h][r] + a[l] * b[h - 1] * b[r - 1]); }
            }
        }
    }

    cout << dp[0][N] << endl;

    return 0;
}

C++: #include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <queue>
#include <list>
#include <stack>
#include <string>
#include <tuple>
#include <functional>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <map>
#include <random>
#include <cstdlib>
#define INT_MAX 2147483647
#define INT_MIN -2147483646
#define Loop(i, n) for(int i = 0; i < (int)n; i++)
#define Loop1(i, n) for(int i = 1; i <= (int)n; i++)
#define Loopr(i, n) for(int i = (int)n - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)n; i >= 1; i--)
using namespace std;
typedef long long int ll;

int memo[105][105];
int r[105], c[105];

int func(int a, int b) {
	if (memo[a][b] != INT_MIN) return memo[a][b];
	if (b - a == 0) { 
		memo[a][b] = 0;
	}
	else {
		int m = INT_MAX;
		for (int k = a;k < b;k++) {
			m = min(m, func(a, k) + func(k + 1, b) + r[a] * c[k] * c[b]);
		}
		memo[a][b] = m;
	}
	return memo[a][b];
}

int main() {
	int n;
	cin >> n;
	Loop(i, n) {
		Loop(j, n) {
			memo[i][j] = INT_MIN;
		}
	}
	Loop(i, n) {
		cin >> r[i] >> c[i];
	}
	cout << func(0, n - 1) << endl;
	return 0;
}

C++: #include<iostream>
#include<stdio.h>
#include<stdlib.h>

using namespace std;

int minana(int a,int b){
  if(a<b){
    return a;
  }
  else if(a>b){
    return b;
  }
  else return a;
}


int main(){
  int n;
  cin>>n;
  int r[100];
  int m[101][101];
  int j;
  int q=0;
  //int c[n];
  for(int i=1;i<=n;i++){
    cin>>r[i-1]>>r[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++){
      j=i+l-1;
      m[i][j]=(1<<21);
      for(int k=i;k<j;k++){
	q=m[i][k]+m[k+1][j]+r[i-1]*r[k]*r[j];
	m[i][j]=minana(m[i][j],q);
      }
    }
  }
  
  cout<<m[1][n]<<endl;
  return 0;
}

C++: #include "bits/stdc++.h"
using namespace std;
typedef long long ll;

template <typename TYPE>
void print_vec(const vector<TYPE>& v){
	for(int i=0; i<v.size(); i++){
		cout << v[i] << " ";
	}
	cout << endl;
}

template <typename TYPE>
void print_vec2(const vector<vector<TYPE>>& v){
  cout << endl; cout << "       ";
  for(int i=0; i<v[0].size(); i++) cout << i << "   "; 
  cout << endl;
	for(int i=0; i<v.size(); i++){
		cout << "i=" << i << ":   ";
		for(int j=0; j<v[i].size(); j++){
			if(v[i][j] == 0) cout << "\x1B[0m" << v[i][j] << "   ";
      else cout << "\x1B[31m" << v[i][j] << "   ";//https://stackoverrun.com/ja/q/12618775
		}
		cout << "\x1B[0m" << endl;
	}
}
 
ll chmin(ll &a, ll b) {return a = min(a, b);}

int main(){
	int n;
	cin >> n;
	vector<ll> row(n), col(n);// 計算に使うんでllにしときます
	for(int i=0; i<n; i++) cin >> row[i] >> col[i];
	const ll INF = 100100100100100;
	// dp[l][r]: 区間[l,r]の行列の積を計算するのに必要なスカラー乗算の最小回数 
	// ll ans = dp[0][n-1] となるよ
	vector<vector<ll>> dp(n+1, vector<ll>(n+1, INF));
	
	//初期化
	for(int l=0; l<n; l++){
		dp[l][l] = 0;//行列1つだけじゃあ計算回数は0
	}

	for(int w=1; w<n; w++){
		for(int l=0; l+w<n; l++){
			int r = l+w;
			for(int mid=l; mid<r; mid++){// 結合する位置をずらしていく
				ll num = row[l] * col[mid] * col[r];// a×b と b×c の計算回数は a * b * c
				num = row[l] * row[mid+1] * col[r];//こっちでもよい
				chmin(dp[l][r], dp[l][mid] + num + dp[mid+1][r]);// 例  a×b と b×c　と c×d　の計算回数は a*b*c + a*c*d 　と  b*c*d + a*b*d  の小さいほう 
			}
		}
	}

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

C++: #include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>

#define INF 100000001

using namespace std;

int DP[101][101];
int *p;

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

int main(){
    int N;
    cin >> N;
    for (int i = 1; i <= N; i++) DP[i][i] = 0;
    
    p = (int *)malloc(sizeof(int)* (N+1));
    
    cin >> p[0];
    cin >> p[1];
    int bb;
    for (int i = 2; i <= N; i++) {
        cin >> bb;
        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 = 3+j; i <= N; i++) {//????????°???????????????
            DP[i-2-j][i] = rec(i-2-j,i);
        }
    }
    
    cout << DP[1][N] << endl;
    
    return 0;
}

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

void solve(){
  int n; cin >> n;
  vector<int> p(n + 1);
  for(int i = 0; i < n; i++){
    cin >> p[i] >> p[i + 1];
  }
  vector<vector<int> > m(n + 1, vector<int>(n + 1));
  for(int i = 1; i <= n; i++) m[i][i] = 0;
  for(int i = 2; i <= n; i++){
    for(int j = 1; j <= n - i + 1; j++){
      int t = 1000000000;
      for(int k = j; k <= j + i - 2; k++){
        t = min(m[j][k] + m[k + 1][j + i - 1] + p[j - 1] * p[k] * p[j + i - 1], t);
      }
      m[j][j + i - 1] = t;
    }
  }
  cout << m[1][n] << endl;
  return;
}

int main(){
  solve();
  return 0;
}