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 <iostream>
using namespace std;

void MatrixChain(int, int[], int **, int **);
void print(int, int, int **);

int main()
{
    int n;
    cin >> n;
    int *tempP = new int[2 * n]; //p[0]为A1的行数；p[i]为Ai的列数
    
    int *p = new int[n + 1];    //p[0]为A1的行数；p[i]为Ai的列数
    int **m = new int *[n + 1]; //m[i][j]为Ai连乘到Aj的最少乘次数
    int **s = new int *[n + 1]; //s[i][j]为 Ai连乘到Aj的最优解的第一层断点
    for (int i = 1; i <= n; i++)
    {
        m[i] = new int[n + 1];
        s[i] = new int[n + 1];
    }
    
    for (int i = 0; i < 2 * n; i++)
    {
        cin >> tempP[i];
    }
    
    for (int i = 0; i <= n; i++)
    {
        p[i] = i == 0 ? tempP[0] : tempP[i*2-1];
    }
    
    // for (int i = 0; i <= n; i++)
    //     cin >> p[i];
    MatrixChain(n, p, m, s);
    cout    << m[1][n] << endl; //最优值存储在m[1][n]
        //print(1, n, s);
    
//    for (int idx = 1; idx <= n; idx++)
//    {
//        for (int jdx = 1; jdx <= n; jdx++)
//        {
//            cout << m[idx][jdx] << " ";
//        }
//        cout << endl;
//    }
    return 0;
}

void MatrixChain(int n, int p[], int **m, int **s)
{
    int i, j, k, temp;
    //以下双重循环对m[][]的上三角进行填表
    //填表顺序为自底向上，自左向右
    for (i = n; i >= 1; i--) //i表示行
    {
        for (j = i; j <= n; j++) //j表示列，因为只填上三角，所以j的初值为i
        {
            //以下按照矩阵连乘问题的递归公式来求每一个m[i][j]
            if (i == j)
            {
                m[i][j] = 0;
            }
            else
            {
                m[i][j] = m[i][i] + m[i + 1][j] + p[i - 1] * p[i] * p[j]; //若i与j不相等，m[i][j]的初值为断点为i时的最优值
                
                s[i][j] = i;                //记录当前断点位置为i
                for (k = i + 1; k < j; k++) //k用于尝试不同的断点
                {
                    temp = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
                    if (temp < m[i][j]) //当前k值作为Ai连乘到Aj的断点能获得更少计算量
                    {
                        m[i][j] = temp; //刷新最优值
                        s[i][j] = k;    //刷新相应断点
                    }
                }
            }
        }
    }
}

void print(int i, int j, int **s)
{
    if (i == j)
        cout << "A" << i;
    else
    {
        cout << "(";
        print(i, s[i][j], s);
        print(s[i][j] + 1, j, s);
        cout << ")";
    }
}

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

int main(){
    int n, p[101], c[101][101], k; 
    cin >> n;
    for (int i=1; i<=n; i++) {
        cin >> p[i-1] >> p[i];
    }
    for (int i=1; i<=n; i++) {
        c[i][i] = 0;
    }
    for (int i=2; i<=n; i++) {
        for (int j=1; j<=n-i+1; j++) {
            k = j + i -1;
            c[j][k] = 2147483647;
            for (int l=j; l <=k-1; l++) {
            c[j][k] = min(c[j][k], c[j][l] + c[l+1][k] + p[j-1]*p[l]*p[k]);
            }
        }
    }
    cout << c[1][n] << endl;
}

C++: #include <iostream>
#include <vector>
using namespace std;
const int INF = 1001001001;

int main() {
    int n;
    cin  >> n;
    vector<int> p(n+1);
    int temp;
    for(int i=0; i<n-1; i++) {
        cin >> p[i] >> temp;
    }
    cin >> p[n-1] >> p[n];
    
    vector<int> a(n);
    vector< vector<int> > A(n,a);
    for(int i=0; i<n; i++) A[i][i] = 0;

    for(int l=2; l<=n; l++) {
        for(int i=0; i<n-l+1; i++) {
            int j = i+l-1;
            A[i][j] = INF;
            for(int k=i; k<j; k++) {
                int q = A[i][k] + A[k+1][j] + p[i]*p[k+1]*p[j+1];
                if(q<A[i][j]) A[i][j] = q;
            }
        }
    }

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