Description
The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on
the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.
Input
The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.
Output
Output must contain a single integer - the minimal score.
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
题意：在一列数中取数，价值为取出的数乘上左右两个数，最左最右两个数不能取出，问根据取数方法不同，求出总价值和最小。
#include<stdio.h>
int min(int a,int b)
{
return a>b?b:a;
}
int main()
{
int n,a[105],dp[105][105];
while(scanf("%d",&n)>0)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
for(int j=1;j<=n;j++)
dp[i][j]=0;
}
for(int r=2;r<n;r++)
for(int i=1;i<=n-r;i++)
{
int j=i+r;
dp[i][j]=dp[i+1][j]+a[i]*a[i+1]*a[j];
for(int k=i+2;k<j;k++)
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
}
printf("%d\n",dp[1][n]);
}
}