兩個大小分別爲p*q和q*r的矩陣相乘時的運算次數計爲p*q*r。
第二行包含n+1個數,表示給定的矩陣。
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#include<cstring>
#include<iostream>
#include<algorithm>
#include<ctime>
#define MAX 1005
#define INF 0x3f3f3f3f
using namespace std;
long long mat[MAX], Min[MAX][MAX];
int count1, count2;
void dp(int n)
{
memset(Min, 0x3f, sizeof(Min));
int i, j, k;
long long t;
for (j = 2;j <= n;j++)
for (i = j-1, Min[j][j] =Min[i][i]= 0;i > 0;i--)
{
for (k = i;k < j;k++)
t=Min[i][k] + Min[k + 1][j] + mat[i - 1] * mat[j] * mat[k],count1++,
Min[i][j] = min(t, Min[i][j]);
}
}/*
void dp(int n)
{
long long i,j,k,t;
memset(Min, 0x3f, sizeof(Min));
for (i = 1; i <= n; i++)
Min[i][i] = 0;
for (int r = 2; r <= n; r++)
for (i = 1; i <= n - r + 1; i++)
{
j = i + r - 1;
Min[i][j] = Min[i + 1][j] + mat[i - 1] * mat[i] * mat[j];
for (k = i + 1; k < j; k++)
t = Min[i][k] + Min[k + 1][j] + mat[i - 1] * mat[k] * mat[j],count2++,
Min[i][j] =min(Min[i][j], t);
}
}*/
int main()
{
clock_t start, finish;
int n, v;
cin >> n;
for (int i = 0;i <= n;i++)
{
cin >> v;
mat[i] = (long long)v;
}
dp(n);
cout << Min[1][n] << endl;
return 0;
}