Go: package main

/*
連鎖行列積
*/

import (
	"bufio"
	"bytes"
	"fmt"
	"log"
	"math"
	"os"
	"strconv"
	"strings"
)

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

type mem *[][]int

func main() {
	scanner := makeScanner(1024)
	n := eGetInt(scanner)
	p := make([]int, n+1)
	for i := 0; i < n; i++ {
		rowcol := eGetInts(scanner)
		p[i] = rowcol[0]
		p[i+1] = rowcol[1]
	}
	m := make([][]int, n)
	for i := 0; i < n; i++ {
		m[i] = make([]int, n)
	}
	// 初期値の格納
	for i := 0; i < n; i++ {
		for j := i; j < n; j++ {
			if i == j {
				m[i][j] = 0 //  同じ行列へのコストは当然0
			} else {
				m[i][j] = math.MaxInt64
			}
		}
	}

	// num : 行列の数. num == 1の場合は初期値0を格納ずみ
	for num := 2; num <= n; num++ {
		// s : スタート行列番号
		for s := 0; s+num <= n; s++ {
			// t : 終了行列番号
			t := s + num - 1
			// k : 仕切
			for k := s; k < t; k++ {
				m[s][t] = min(m[s][t], m[s][k]+m[k+1][t]+p[s]*p[k+1]*p[t+1])
			}
		}
	}
	fmt.Println(m[0][n-1])
}

////////////////////////////////////////////////////////////////////////////////
// ライブラリ
////////////////////////////////////////////////////////////////////////////////
func makeScanner(maxByte int) *bufio.Scanner {
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Buffer(make([]uint8, 0, 8192), maxByte)
	return scanner
}
func makeWordScanner() *bufio.Scanner {
	scanner := makeScanner(8192)
	scanner.Split(bufio.ScanWords)
	return scanner
}
func eGetLine(r *bufio.Scanner) string {
	if r.Scan() {
		return r.Text()
	}
	err := r.Err()
	if err != nil {
		log.Fatal(err)
	}
	// EOF
	return ""
}
func eGetInt(r *bufio.Scanner) int {
	line := eGetLine(r)
	return eAtoi(line)
}
func eGetFields(r *bufio.Scanner) []string {
	line := eGetLine(r)
	return strings.Fields(line)
}
func eGetInts(r *bufio.Scanner) []int {
	fields := eGetFields(r)
	ints := make([]int, len(fields))
	for i := 0; i < len(ints); i++ {
		ints[i] = eAtoi(fields[i])
	}
	return ints
}
func eGetChars(r *bufio.Scanner) []string {
	line := eGetLine(r)
	return strings.Split(line, "")
}
func eAtoi(s string) int {
	n, err := strconv.Atoi(s)
	if err != nil {
		log.Fatal(err)
	}
	return n
}

type ints []int

func (s ints) String() string {
	var buf bytes.Buffer
	buf.WriteString("[")
	sep := ""
	for _, v := range s {
		buf.WriteString(fmt.Sprintf("%s%d", sep, v))
		sep = ", "
	}
	buf.WriteString("]")
	return buf.String()
}

func reverse(a ints) ints {
	r := make(ints, len(a))
	for i := 0; i < len(a); i++ {
		r[i] = a[len(a)-1-i]
	}
	return r
}
func max(a ints) (int, bool) {
	if len(a) == 0 {
		return 0, false
	}
	m := a[0]
	for _, e := range a {
		if e > m {
			m = e
		}
	}
	return m, true
}

// func min(a []int) int {
// 	m := a[0]
// 	for _, v := range a {
// 		if v < m {
// 			m = v
// 		}
// 	}
// 	return m
// }

func sum(a ints) int {
	if len(a) == 0 {
		return 0
	}
	sum := 0
	for _, e := range a {
		sum += e
	}
	return sum
}
func find(n int, a ints) int {
	for i, v := range a {
		if n == v {
			return i
		}
	}
	return -1
}

C++: #include <bits/stdc++.h>

using namespace std;

#define FOR(i,s,e) for((i)=(s);(i)<(int)(e);(i)++)
#define REP(i,e) FOR(i,0,e)

#define all(o) (o).begin(), (o).end()
#define psb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))

typedef long long ll;
typedef pair<int, int> PII;

const double EPS = 1e-10;
const int INF = 1e+9;

const int N = 100;
int n;
int s[N+1];       //M_i : s[i-1] x s[i] (i=1,...,n)
int dp[N+1][N+1]; //dp[i][j] = (M_i * ... * M_j)????????????????¨?????????°???????°????(1<=i<=j<=n)

int main() {
  int i, j, k, l;
  scanf("%d ", &n);
  for (i=0; i<n; i++) scanf("%d%d ", s+i, s+i+1);
  for (i=1; i<=n; i++) dp[i][i] = 0;  //1????????????????¨?????????°???0
  for (l=2; l<=n; l++)  //M_i ... M_i????????°???????????§????????????
    for (i=1; i<=n-l+1; i++) {
      j = i + l - 1; 
      dp[i][j] = INF;
      for (k=i; k<j; k++)
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + s[i-1]*s[k]*s[j]);
    }

  printf("%d\n", dp[1][n]);

  return 0;
}

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

#define N_MAX 105
#define INF (1 << 21)

long P[N_MAX];
long dp[N_MAX][N_MAX];

int main() {
    long n;
    cin >> n;
    for (long i = 0; i < n; i++) {
        cin >> P[i] >> P[i + 1];
    }

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

    for (long l = 1; l <= n - 1; l++) {
        for (long i = 1; i + l <= n; i++) {
            long j = i + l;
            dp[i][j] = INF;
            for (long k = i; k <= j; k++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + P[i - 1] * P[j] * P[k]);
            }
        }
    }

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

    return 0;
}

C++: #include <bits/stdc++.h>
using namespace std;
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef long long ll;
const long long INF = 1LL << 60;
typedef pair<int, int> P;
struct RC {
    int r, c;
};

int dp[101][101];
int main()
{
    int N;
    cin >> N;
    vector<RC> rc(N);
    rep(i, N) {
        int r, c;
        cin >> r >> c;
        rc[i] = {r, c};
    }

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

    repd(w, 1, N) {
        rep(i, N - w) {
            int j = i + w;
            repd(k, i, j) {
                chmin(dp[i][j], dp[i][k] + dp[k + 1][j] + rc[i].r * rc[k].c * rc[j].c);
            }
        }
    }

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

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

#define N 101

vector<vector<int> > dp(N, vector<int>(N));
int 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 <stdio.h>

int N;
int R[200];
int dp[200][200];

int min(int x, int y) { return x < y ? x : y; }

int main(void) {
	scanf("%d", &N);
	for (int i = 0; i < N; i ++) {
		scanf("%d %d", &R[i], &R[i+1]);
	}
	for (int i = 0; i < N; i ++) {
		for (int j = 0; j < N; j ++) {
			dp[i][j] = 1 << 20;
		}
	}
	for (int i = 0; i < N; i ++) {
		dp[i][0] = 0;
	}
	for (int j = 0; j < N; j ++) {
		for (int i = 0; i + j < N; i ++) {
			// M_j ... M_{j+i}のi+1個の積について考える
			for (int k = 0; k < j; k ++) {
				// (M_j ... M_{j+k}) (M_{j+k+1} ... M_{j+i})
				//   (k+1個の積)            (i-k個の積)
				dp[i][j] = min(dp[i][j], dp[i][k] + dp[i+k+1][j-k-1] + R[i] * R[i+k+1] * R[i+j+1]);
			}
		}
	}
	printf("%d\n", dp[0][N-1]);
	return 0;
}

C++: #include <iostream>
#include <limits.h>

using namespace std;

int dp[100][100];
int p[101];
int n;

void solve() {
	for (int m = n; m > 0; m--) {
		for (int t = 0, i = 0, j = n - m + 1; t < n && j < n; t++) {
			dp[i][j] = INT_MAX;
			for (int k = i; k < j; k++) {
				int temp = dp[i][k] + dp[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
				if (temp < dp[i][j]) {
					dp[i][j] = temp;
				}
			}
			i++;
			j++;
		}
	}
}

int main() {
	int a, b;
	cin >> n >> a;

	p[0] = a;
	for (int i = 1; i < n; i++) {
		cin >> a >> b;
		p[i] = a;
	}
	cin >> a;
	p[n] = a;

	solve();
	cout << dp[0][n - 1] << endl;

	return 0;
}