Integer division - the interval dp (gravel merger)

This is not a number to a to divide, but to an integer bits divided

How a positive integer N (N length <20) divided into M (M> 1) parts, so that the product of the maximum M portions. N, M input from the keyboard, and outputs the maximum one division manner.

The first line of a positive integer T (T <= 10000), expressed T set of data.

The next two lines each line T positive integers N, M.

For each test

The first line of maximum output.

The second line of the output division scheme, the sequence is divided into M number of N outputs, the number of opening between the two elements by spaces.

Sample input

1
199 2

Sample Output

171
19 9

这是递归思想，动态规划是正向的，而判断后是逆向的，输出时运用回溯，达到正向输出的目的
以下是代码

#include <the iostream> 
#include <cstdio> 
#include <CString>
 the using  namespace STD; 
unsigned Long  Long T, n-[ 21 is ], N2, N3 [ 21 is ] [ 21 is ], X, Son [ 1000 ] [ 1000 ], F [ 21 is ] [ 21 is ], m; // for the maximum, with an unsigned long integer 
String N1;
 int printf1 ( int A, int B) // output functions, backtracking 
{
     IF (B == 0 ) return  0 ; 
    printf1 (son [a] [b ], b-1);
    for(int i=son[a][b]+1;i<=a;i++)
        cout<<n[i];
    cout<<" ";
}
int main()
{
    cin>>t;
    for(int l=1;l<=t;l++)
    {
        memset(n,0,sizeof(n));
        memset(son,0,sizeof(son));
        cin>>n1>>m;
        n2=n1.length();
        for(int i=0;i<=n2;i++)
            for(int j=0;j<=n2;j++)
            {
                f[i][j]=0;
                //n3[i][j]=1;
            }    
        f[0][0]=1;
        for(int i=1;i<=n2;i++)
        {
            n[i]=n1[i-1]-'0';
            //cout<<n[i];
        }
        for(int i=1;i<=n2;i++)
        {
            x=n[i];
            for(int j=i;j<=n2;j++)
            {
                n3[i][j]=x;
                x*=10;
                x+=n[j+1];
                //cout<<n3[i][j]<<" "<<i<<" "<<j<<endl;
            }
        }
        for(int i=1;i<=n2;i++)
        {
            for(int j=1;j<=m&&j<=i;j++)
            {
                for(int k=1;k<=i;k++)
                {
                    if(f[i][j]<f[k-1][j-1]*n3[k][i])
                    {
                        f[i][j]=f[k-1][j-1]*n3[k][i];
                        //cout<<f[i][j];
                        son[i][j]=k-1;//记录分割点
                    }
                    
                }
            }
        }
        cout<<f[n2][m]<<endl;
        if(m==n2)//特判，防止输出紊乱
            for(int i=1;i<=n2;i++)
                cout<<n[i]<<" ";
        else printf1(n2,m);
        cout<<endl;
    }
}

 石子合并

Playground around a circular stones piled placed N, Stone is to have order to merge into a pile. Each time only a predetermined selected adjacent stack 2 into a new pile and the new pile of stones number, remember that the combined score.

Test design an algorithm to calculate the N stones piled into a stack the maximum combined score.

A first data line test positive integer N, 1≤N≤2000, N stones piled expressed. The second row has the number N, respectively, represent the number of each pile of stones.

The output common line 1, the maximum score

Sample input

4
4 4 5 9

Sample Output

54
最终一堆一定是前一次合并后，剩下的两堆相加的最优解。

State transition equation
provided t [i, j] denotes the i-th stack from the stack to the sum of the j-number of stones.
Fmax (i, j) represents the i-heap of stones from the merger to the maximum score of the j-heap of stones
Fmin (i, j) represents the i-heap of stones from the combined minimum score of j-heap of stones (see the meaning of the questions not required)

附上代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m[4001],m1[4001][4001],f[4001][4001],x,ma;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>m[i];
    }
    for(int i=1;i<=n;i++)
    {
        m[i+n]=m[i];
    }
    for(int i=1;i<=2*n-1;i++)
    {
        x=m[i];
        for(int j=i+1;j<=2*n-1;j++)
        {
            x+=m[j];
            m1[i][j]=x;
        }
    }
    for(int i=2*n-1;i>=1;i--)
    {
        for(int j=i;j<=2*n-1;j++)
        {
            f[i][j]=max(f[i+1][j],f[i][j-1])+m1[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(ma<f[i][i+n-1])ma=f[i][i+n-1];
    }
    cout<<ma;
}