題目鏈接
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
10150 + 50205 + 10505 = 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
15020 + 1205 + 1015 = 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
題意：給你n個數字，要求不能刪除兩端點的數字，然後刪除其他數字的代價是該數字和左右相鄰數字的乘積，問把數字（除端點）刪完後的最小總代價。
思路：dp【i】【j】代表區間【i，j-1】的刪除完的最大價值
#include <iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=105;
int n,a[maxn],dp[maxn][maxn];
int main()
{
scanf("%d",&n);
memset(dp,0x3f3f3f3f,sizeof(dp));
for(int i=1;i<=n;++i) scanf("%d",&a[i]),dp[i][i]=0;
for(int len=1;len<=n;++len)
{
for(int l=2;l+len-1<=n;++l)
{
int r=l+len-1;
for(int k=l;k<r;++k)
dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]+a[k]*a[l-1]*a[r]);
}
}
printf("%d\n",dp[2][n]);
}