"Tyvj P1014" multiplication game

Time: 1000ms/Space: 131072kib/java class name: Main background Taiyuan Cheng-Cheng Middle School 2nd time simulation Fourth Road description multiplication game is performed on a line of cards. Each card consists of a positive integer. In each move, the player takes out a card, the score is multiplied by its number to the left and the right, so it is not allowed to take the 1th Zhang and last 1 cards. After the last move, there are only two cards left.
Your goal is to make the score and the smallest.
For example, if the number is 10 1 50 20 5, take 1, 20, 50, the total score is 10*1*50+50*20*5+10*50*5=8000
And take 50, 20, 1, the total score is 1*50*20+1*20*5+10*1*5=1150. Input format the first line of the input file includes the number of cards (3<=n<=100), the second line contains n 1-100 integers, separated by a space. Output format output file has only one number: Minimum score test Sample 1 input

6
1 5

Output

3650

TYVJ evaluation machine is not.　　I have to get the test data. This problem is obviously a dynamic programming problem, and it is the interval motion regulation.　　So how to establish a State transfer.　　First, it is easy to think that the large interval is from the community between the introduction, then there will be an optimal sub-structure. Then, no matter how the community between the big interval can not affect the answer.　　Then the state transfer equation is in the open.　　Set F[1][n-2] is the answer. F[I][J] = Min{f[i][k-1] + f[k + 1][j] + a[i-1] * a[k] * a[j + 1] | I < K < J} But we have to preprocess F[i][i] and F[i][i + 1] Case and put f[i][j] | The value of J < i is set to 0.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <cstdlib>6 using namespacestd;7 8 Long Longf[ the][ the];9 inta[ the];Ten intN; One  A intMain () - { -scanf"%d", &n); the      for(inti =0; I < n; ++i) -scanf"%d", &a[i]); -Memset (F,0x7f,sizeof(f)); -f[0][0] = F[n-1][n-1] =0; +      for(inti =1; I < n-1; ++i) F[i][i] = a[i-1] * A[i] * a[i +1]; -      for(inti =1; I < n-2; ++i) +F[i][i +1] = min (F[i][i] + f[i +1][i +1]/a[i] * a[i-1], F[i +1][i +1] + f[i][i]/a[i +1] * a[i +2]); A      for(inti =1; I < n; ++i) at          for(intj =0; J < I; ++j) -F[I][J] =0; -      for(inti = n-2; i >0; --i) -          for(intj = i +1; J < N-1; ++j) -              for(intK = i; K <= J; ++k) -F[i][j] = min (f[i][j], f[i][k-1] + f[k +1][J] + a[i-1] * A[k] * a[j +1]); incout << f[1][n-2] <<Endl; -     //System ("pause"); to     return 0; +}

"Tyvj P1014" multiplication game