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 rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(ri,n) for(int ri = (int)(n-1); ri >= 0; ri--)
#define rep2(i,x,n) for(int i = (int)(x); i < (int)(n); i++)
#define repit(itr,x) for(auto itr = x.begin(); itr != x.end(); itr++)
#define rrepit(ritr,x) for(auto ritr = x.rbegin(); ritr != x.rend(); ritr++)
#define ALL(n) begin(n), end(n)
using ll = long long;
using namespace std;

int main(){
  int n;
  cin >> n;
  vector<int> p(n+1);
  rep(i, n){
    int a, b;
    cin >> a >> b;
    if(i != 0) p.at(i+1) = b;
    else{
      p.at(0) = a;
      p.at(1) = b;
    }
  }
  vector<vector<ll>> dp(n, vector<ll>(n, 0));
  rep2(l, 1, n){
    rep(i, n - l){
      int j = i + l;
      dp.at(i).at(j) = LLONG_MAX;
      rep2(k, i, j){
        ll add = dp.at(i).at(k) + dp.at(k+1).at(j) + p.at(i) * p.at(k+1) * p.at(j+1);
        dp.at(i).at(j) = min(dp.at(i).at(j), add);
      }
    }
  }
  // rep(i, n){
  //   rep(j, n){
  //     printf("%9d ", (int)dp.at(i).at(j));
  //   }
  //   cout << endl;
  // }
  cout << dp.at(0).at(n-1) << endl;
  return 0;
}

C++: #include<iostream>
#include<algorithm>
using namespace std;
static const int N=100;

int main(){
  int n,i,l,k,p[N+1],m[N+1][N+1];
  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++){
      int j=i+l-1;
      m[i][j]=(1<<21);
      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++: #include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int INF = 1e9;

int main()
{
  int n = 0;
  fscanf(stdin, "%d", &n);
  vector<ll> rows(n + 5), cols(n + 5);
  for (int i = 0; i < n; ++i) {
    fscanf(stdin, "%lld %lld", &rows[i + 1], &cols[i + 1]);
  }
  vector<vector<ll> > dp(n + 5, vector<ll>(n + 5, 0));
  for (int l = 2; l <= n; ++l) {
    for (int i = 1; i <= n - l + 1; ++i) {
      int j = l + i - 1;
      dp[i][j] = std::numeric_limits<ll>::max();
      for (int k = i; k <= j - 1; ++k) {
        dp[i][j] = min(dp[i][j],
                       dp[i][k] + dp[k + 1][j] + rows[i] * cols[k] * cols[j]);
      }
    }
  }
  fprintf(stdout, "%lld\n", dp[1][n]);
  return 0;
}

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

int MCM(int n){
int i,j,k,l,m[101][101],p[101],x;

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]=10000000;
        for(k=i;k<=j;k++){
            x=m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j];
            if(m[i][j]>x)m[i][j]=x;
            }
        }
    }
return m[i][j];
}

int main(){

int i,n,m[101];

scanf("%d",&n);
for(i=0;i<n;i++){
    scanf("%d %d",&m[i],&m[i+1]);
   }
printf("%d\n",MCM(n));
return 0;
}

C++: #include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>

#define mp       make_pair
#define pb       push_back
#define all(x)   (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define repi(i,a,b) for(int i = (a); i < (b); i++)

using namespace std;

typedef    long long          ll;
typedef    unsigned long long ull;
typedef    vector<bool>       vb;
typedef    vector<int>        vi;
typedef    vector<vb>         vvb;
typedef    vector<vi>         vvi;
typedef    pair<int,int>      pii;

const int INF=1<<29;
const double EPS=1e-9;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};
pii A[200];//firstは行secondを列とする
int dp[200][200];//dp[i][j]:=Ai...Ajの最小スカラー積とする
int n;
int f[200][200];//最適な括弧を挿入する位置
void solve(){
	for(int i =0;i < n;i++){
		for(int j =0;j < n;j++){
			if(i==j)dp[i][j]=0;
			else dp[i][j]=INF;
		}
	}
	
	for(int k =1;k <n;k++){//鎖の数
		for(int i =0;i <n-k;i++){//どこから始めるか
			for(int j =i;j <i+k;j++){//どこに括弧を挟むか
				int tmp =A[i].first*A[j].second*A[i+k].second;
				if(dp[i][i+k] >dp[i][j]+dp[j+1][i+k]+tmp){
					dp[i][i+k]=dp[i][j]+dp[j+1][i+k]+tmp;
					f[i][i+k] =j;
				}
			}
		}
	}
}

void print(int i,int j){//解を表示する
	if(i==j){
		cout <<"A"<<i;
	}else{
		cout <<"(";
		print(i,f[i][j]);
		print(f[i][j]+1,j);
		cout <<")";
	}
}


int main(){
	cin>>n;
	for(int i =0;i < n;i++){
		cin>>A[i].first>>A[i].second;
	}
	solve();
	cout <<dp[0][n-1]<<endl;
	//print(0,n-1);
	return 0;
}

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

static const int NMAX = 100;
static const int RMAX = 100;
static const int CMAX = 100;
static const int INF = (1 << 29);

int main() {
    int n; cin >> n;
    int M[NMAX+1][NMAX+1];
    int r[RMAX+1];
    int c[CMAX+1];
    int min_cost;

    for (int i = 1; i < n+1; ++i) cin >> r[i] >> c[i];

    for (int i = 0; i < n; ++i) {
        for (int j = 1; j < n+1-i; ++j) {
            for (int k = 1+i; k < n+1; ++k) {
                if (j==k) {
                    M[j][k] = 0;
                    continue;
                }

                min_cost = INF;
                for (int l = j; l < k; ++l) {
                    min_cost = min(min_cost, M[j][l] + M[l+1][k] + (r[j]*c[l]*c[k]));
                }
                M[j][k] = min_cost;
            }
        }
    } 

    // for (int i = 1; i < n+1; ++i) {
    //     for(int j = 1; j < n+1; ++j) {
    //         cout << M[i][j] << " ";
    //     }
    //     cout << endl;
    // }

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


    return 0;
}

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

typedef long long ll;
#define inf 1e9+7

int n;
ll dp[102][102];
ll p[102];

void init(){
    for(int i=0;i<=n;i++){
        for(int j=0;j<=n;j++){
            dp[i][j]=inf;
        }
        dp[i][i]=0;
    }
}

void cal(int l, int r){
    if(l>=r)return;
    for(int k=l;k<r;k++){
        cal(l,k);
        cal(k+1,r);
        dp[l][r] = min(dp[l][r], dp[l][k]+dp[k+1][r]+p[l]*p[k+1]*p[r+1]);
    }
}

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

    cin >> n;
    init();
    cin >> p[0] >> p[1];
    ll a,b;
    for(int i=1;i<n;i++){
        cin >> a >> b;
        p[i+1]=b;
    }
    cal(0,n-1);
    cout << dp[0][n-1] << endl;
    return 0;
}