Problem:
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Matrix-chain Multiplication

The goal of the matrix-chain multiplication problem is to find the most efficient way to multiply given $n$ matrices $M_1, M_2, M_3,...,M_n$.

Write a program which reads dimensions of $M_i$, and finds the minimum number of scalar multiplications to compute the maxrix-chain multiplication $M_1M_2...M_n$.

Input

In the first line, an integer $n$ is given. In the following $n$ lines, the dimension of matrix $M_i$ ($i = 1...n$) is given by two integers $r$ and $c$ which respectively represents the number of rows and columns of $M_i$.

Output

Print the minimum number of scalar multiplication in a line.

Constraints

$1 \leq n \leq 100$

$1 \leq r, c \leq 100$

Sample Input 1

6
30 35
35 15
15 5
5 10
10 20
20 25

Sample Output 1

15125

Solution in Python3 (judge verdict: Runtime Error)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import sys
import re
import math
import collections
import itertools
import functools
import logging

logging.basicConfig(level=logging.DEBUG, format="DBG: %(message)s")

def dbg(msg, *args, **kwargs):
    logging.debug(msg, *args, **kwargs)

Mat = collections.namedtuple("Mat", "r c")

def mat_mul(ms):
    return Mat(ms[0].r, ms[-1].c)

def cost(m1, m2):
    return m1.r * m1.c * m2.c

def solve(n, ms):
    if n == 1: return 0

    dp = {}
    for i in range(n):
        dp[(i,i)] = 0

    for l in range(2, n+1):
        for i in range(n-l+1):
            cost_min = math.inf
            for j in range(l-1):
                dbg("l, i, j: %d %d %d", l, i, j)
                m_lhs = mat_mul(ms[i:i+j+1])
                m_rhs = mat_mul(ms[i+j+1:i+l])
                dbg("m_lhs, m_rhs: %r %r", m_lhs, m_rhs)
                cost_lhs = dp[(i,i+j)]
                cost_rhs = dp[(i+j+1,i+l-1)]
                dbg("cost_lhs, cost_rhs: %d %d", cost_lhs, cost_rhs)
                cost_mul = cost(m_lhs, m_rhs)
                dbg("cost_mul: %d", cost_mul)
                cost_now = cost_mul + cost_lhs + cost_rhs
                dbg("cost_now: %d", cost_now)
                if cost_min > cost_now:
                    cost_min = cost_now
            dp[(i,i+l-1)] = cost_min

    dbg(dp)
    return dp[(0,n-1)]

def main():
    n = int(input())

    ms = []
    for _ in range(n):
        r,c = map(int, input().split())
        ms.append(Mat(r,c))

    print(solve(n, ms))

if __name__ == "__main__": main()