Poj1651multiplication Puzzle (take number, interval DP)

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 The card taken and the numbers on the cards in the left and on the it. It is not a allowed to take out the first and the last card in the row. After the final move, only the cards was left in the row. 

The goal was to take cards in such order as to MI Nimize The total number of scored points. 

For example, if cards in the row contain numbers 1 5, p Layer might take a card with 1, then + and, 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. and 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, 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

Test instructions: In a column number, the value is taken out of the number of times on the left and right two numbers, the leftmost two number can not be taken out, asked according to the method of taking the number of different, to find the total value and minimum.

#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]);
    }
}