#include<iostream>
#include<iomanip>
using namespace std;

fun(int l,int n,int m[]) {
	int i,j,k,r;
	int **a = new int*[n];
	for(i=0;i<n;i++) {
		a[i] = new int[n];
	}
	for(i=0;i<n;i++) {
		for(j=0;j<n;j++) {
			a[i][j] = 0;
		}
	}
	for(i=0;i<n;i++) {
		for(j=0,k=i;k!=n;j++,k++) {
			if(j==k) a[j][k] = 0;
			else {
				int min = 1000000;
				for(r=j;r<k;r++) {
					int temp = a[j][r] + a[r+1][k] + m[j] * m[r+1] * m[k+1];
					cout<<m[j] * m[r+1] * m[k]<<endl;
					if(temp < min) min = temp;
				}
				a[j][k] = min;
				cout<<j<<":"<<k<<endl;
			}
		}
	}
	return a[0][n-1];
}

int main() {
	int n[] = {50,10,40,30,5};
	cout<<fun(0,4,n)<<endl;
	return 0;
}

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#include<iostream>
using namespace std;

//运用动态规划的思想解决矩阵连乘问题
int matrixMul(int i, int j, int p[]) {
    int **array;//array[i][j]
    array = new int*[j + 1];
    for (int k1 = 0; k1 < j + 1; k1++) {
        array[k1] = new int[j + 1];

        for (int k2 = 0; k2 < j + 1; k2++)
            array[k1][k2] = 0;
    }

    for (int f = 1; f < j + 1; f++) {
        for (int k1 = i, k2 = f; k1 < j + 1 && k2 < j + 1; k1++, k2++) {
            int min = 100000000;
            if (k1 == k2)
                break;
            for (int k = k1; k < k2; k++) {
                int temp = array[k1][k] + array[k + 1][k2] + p[k1 - 1] * p[k] * p[k2];
                if (temp < min)
                    min = temp;
            }

            array[k1][k2] = min;
        }
    }
    
    return array[i][j];
}

int main() {
    int p[] = { 50, 10, 40, 30, 5 };
    int result = matrixMul(1, 4, p);
    cout << result << endl;

    return 0;
}