Go: package main

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

type Matrix struct {
	r int // num of row
	c int // num of column
}

type Range struct {
	start int // index, inclusive
	end   int // index, inclusive
}

func (r Range) length() int {
	return r.end - r.start + 1
}

func calcMultiplication(x, y Matrix) (Matrix, int) {
	if x.c != y.r {
		panic("cannnot calculate")
	}
	return Matrix{x.r, y.c}, x.r * y.c * x.c
}

func calcMinMatrixChainMultiplication(matrices []Matrix, r Range, memo map[Range]int) (Matrix, int) {
	if v, ok := memo[r]; ok {
		return Matrix{matrices[r.start].r, matrices[r.end].c}, v
	}
	if r.length() == 0 {
		panic("must not be empty")
	} else if r.length() == 1 {
		return matrices[r.start], 0
	}

	var minCount int = math.MaxInt32

	for i := 0; i < r.length()-1; i++ {
		formerRange := Range{r.start, r.start + i}
		latterRange := Range{r.start + i + 1, r.end}
		matF, countF := calcMinMatrixChainMultiplication(matrices, formerRange, memo)
		matL, countL := calcMinMatrixChainMultiplication(matrices, latterRange, memo)
		_, countFL := calcMultiplication(matF, matL)
		count := countF + countL + countFL
		if count < minCount {
			minCount = count
		}
	}
	memo[r] = minCount
	return Matrix{matrices[r.start].r, matrices[r.end].c}, minCount
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	n, _ := nextInt(sc)
	matrices := make([]Matrix, n)
	for i := 0; i < n; i++ {
		r, _ := nextInt(sc)
		c, _ := nextInt(sc)
		matrices[i] = Matrix{r, c}
	}
	memo := make(map[Range]int)
	_, resultCount := calcMinMatrixChainMultiplication(matrices, Range{0, len(matrices) - 1}, memo)
	fmt.Println(resultCount)
}

func nextInt(sc *bufio.Scanner) (int, error) {
	sc.Scan()
	return strconv.Atoi(sc.Text())
}

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;
}

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

#define REP(i, n) for(int i = 0; i < (n); i++)

typedef pair<int, int> pii;

const int N_SIZE = 100;

int dp[N_SIZE + 1][N_SIZE + 1];

void solve(int n, pii rc[])
{
  REP(i, n)
    dp[i][i] = rc[i].first * rc[i].second;
  REP(i, n - 1)
    dp[i][i + 1] = dp[i][i] * rc[i + 1].second;
  for(int i = 2; i < n; i++){
    for(int j = 0, k = i + j, m = 100000000; k < n; j++, k++){
      for(int l = j, x; l < k; l++){
	x = ((j == l) ? dp[j][l] * rc[k].second : dp[j][l]) + ((l + 1 == k) ? dp[l + 1][k] * rc[j].first : dp[l + 1][k]);
	if(j != l && l + 1 != k)
	  x += rc[j].first * rc[l].second * rc[k].second;
	m = min(m, x);
      }
      dp[j][k] = m;
      m = 100000000;
    }
  }
}
 
int main()
{
  int n;
  cin >> n;
  pii rc[n];
  REP(i, n)
    cin >> rc[i].first >> rc[i].second;
  solve(n, rc);
  for(int i = 0; i < n; i++){
    for(int j = 0; j < n; j++){
      cout << " " << dp[i][j];
    }
    cout << endl;
  }
  cout << dp[0][n - 1] << endl;
}

C++: #include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <tuple>
#include <climits>

#define DEBUG

#ifdef DEBUG

#define printd printf

#else

#define printd(...)

#endif

int main() {
    int n;
    int shapes[100][2];

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

    constexpr uint64_t INF = UINT64_MAX;
    uint64_t dp[100][100];
    for (int i = 0; i < 100; i++) {
        dp[i][i] = 0;
    }
    for (int width = 2; width <= n; width++) {
        for (int l = 0; l <= n - width; l++) {
            auto min = INF;
            for (int split = l + 1; split < l + width; split++) {
                auto times = dp[l][split - 1] + dp[split][l + width - 1] + shapes[l][0] * shapes[split][0] * shapes[l + width - 1][1];
                if (times < min) {
                    min = times;
                }
            }
            dp[l][l + width - 1] = min;
        }
    }
    std::cout << dp[0][n - 1] << std::endl;

    return 0;
}

C++: #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;

vector<ll> r(101);
vector<ll> c(101);
vector<vector<ll>> DP(101,vector<ll> (101,-1));

ll inf=1e18;



ll dp(ll l,ll h) {
  if(DP[l][h]>-1) {
    return DP[l][h];
  }
  ll res=inf;
  if(l+1==h) {
    res=0;
  }
  else {
    for(ll i=l+1;i<h;i++) {
      res=min(res,dp(l,i)+dp(i,h)+r[i]*r[l]*c[h-1]);
    }
  }
  return DP[l][h]=res;
}

int main() {
  ll n;
  cin >> n;
  for(ll i=0;i<n;i++) {
    cin >> r[i] >> c[i];
  }
  cout << dp(0,n) << endl;
}

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

using namespace std;

static const int N = 100;
static int m[N + 1][N + 1] = {}, p[N + 1] = {};

static inline void matrixChainMultiplication(int n) {
  for (int l = 2; l <= n; l++) {
    for (int i = 1; i <= n - l + 1; i++) {
      int j = i + l - 1;
      m[i][j] = INT_MAX;
      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]); 
      }
    }
  }
}

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

  matrixChainMultiplication(n);
  
  cout << m[1][n] << endl;

  return 0;
}

C++: #include<iostream>
using namespace std;
int min(int a,int b){return a<b?a:b;}
int p[101],m[101][101];
int main()
{	int n;
	cin>>n;
	for(int i=1;i<=n;i++)//总共有n+1个有效数据
		cin>>p[i-1]>>p[i];//保证只有n个有效的数据重复的不管
	for(int i=1;i<=n;i++)
		m[i][i]=0;
	for(int l=2;l<=n;l++)//这个l状态积累量
	{
		for(int i=1;i<=n-l+1;i++)
		{	int j=i+l-1;
			m[i][j]=(1<<21);//极大取极小,这个地方要有括号
			for(int k=i;k<=j-1;k++)//从i开始,对l开始累加
			{
				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;

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 <stdio.h>

#define MAX_N 100
#define INF 1e8

int main(){
  int N,i,j,k,l,q,r[MAX_N],c[MAX_N],p[MAX_N+1];
  int dp[MAX_N+1][MAX_N+1];
  scanf("%d",&N);
  for(i = 0 ; i < N ; i++){
    scanf("%d%d",&r[i],&c[i]);
  }
  p[0] = r[0];
  for(i = 1 ; i <= N ; i++){
    p[i] = c[i-1];
  }
  N++;
  for(i = 0 ; i < N ; i++){
    dp[i][i] = 0;
  }
  for(l = 2 ; l < N ; l++){
    for(i = 1 ; i < N-l+1 ; i++){
      j = i + l - 1;
      dp[i][j] = INF;
      for(k = i ; k <= j-1 ; k++){
        q = dp[i][k] + dp[k+1][j] + p[i-1]*p[k]*p[j];
        if(q < dp[i][j]){
          dp[i][j] = q;
        }
      }
    }
  }
  printf("%d\n",dp[1][N-1]);
  return 0;
}

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

int n, p = 0, min = 101, mini;
int v[101];

int main()
{
	cin >> n;

	for (int i = 0; i < n; i++) {
		cin >> v[i] >> v[i + 1];
		if (min > v[i]) {
			min = v[i];
			mini = i;
		}
	}

	for (int j = 0; j < n; j++)
		p += v[j] * v[j + 1];
	p += v[0] * v[n];

	if (mini == 0)
		p -= v[0] * v[1];
	else if (mini == n)
		p -= v[n - 1] * v[n];
	else
		p -= v[mini] * (v[mini + 1] + v[mini - 1]);

	p *= v[mini];

	cout << p << endl;

	return 0;
}