Multiplication Puzzle POJ

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

1.和石子归并差不多的问题。

正确的代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;

int N;
int dp[105][105],a[105];

int main()
{   scanf("%d",&N);
    for(int i=1;i<=N;i++) scanf("%d",&a[i]);
    memset(dp,0,sizeof(dp));
    for(int i=1;i<N-1;i++) dp[i][i+2]=a[i]*a[i+1]*a[i+2];
    for(int len=3;len<N;len++){
        for(int i=1;i<=N&&i+len<=N;i++){
            int j=len+i;
            for(int k=i+1;k<j;k++){
                if(dp[i][j]==0) dp[i][j]=dp[i][k]+dp[k][j]+a[i]*a[k]*a[j];
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
            }
        }
    }
    printf("%d
",dp[1][N]);
}

 错误的代码：调整k的时候遗漏某些情况，如1 2 3 4 5，取2之后再取4，这种情况会被漏掉！

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;

const int INF=1000000000;

int N;
int dp[105][105],a[105];

int main()
{   scanf("%d",&N);
    for(int i=1;i<=N;i++) scanf("%d",&a[i]);
    memset(dp,0,sizeof(dp));
    for(int i=1;i<N-1;i++) dp[i][i+2]=a[i]*a[i+1]*a[i+2];
    for(int len=3;len<N;len++){
        for(int i=1;i<=N&&i+len<=N;i++){
            int j=len+i;
            dp[i][j]=INF;
            for(int k=i;k<j;k+=len-1){
                int tem;
                if(k==i) tem=dp[i][k]+dp[k+1][j]+a[i]*a[k+1]*a[j];
                else tem=dp[i][k]+dp[k+1][j]+a[i]*a[k]*a[j];
                if(dp[i][j]>tem) dp[i][j]=tem; 
            }
        }
    }
    printf("%d
",dp[1][N]);
}