• Source 
    1. #include<stdio.h>
    2. #include<cstring>
    3. #include<iostream>
    4. using namespace std;
    5. // assuming maximum subscribers and books is 1000
    6. int dp[1001][1001],pages[1001],n,m;
    7. //dp[][] is memset to -1 from main
    8. // Assuming books are numbered 1 to n
    9. // change value of MAX based on your constraints
    10. int MAX = 1000000000;
    11. int rec(int position , int sub )
    12. {
    13.    // These two are the base cases
    14.    if(position > n)
    15.    {
    16.       if(sub == 0)return 0;
    17.       return MAX;
    18.    }
    19.    if(sub == 0)
    20.    {
    21.       if(position > n)return 0;
    22.       return MAX;
    23.    }
    24.  
    25.    // If answer is already computed for this state return it
    26.    if(dp[position][sub] != -1)return dp[position][sub];
    27.  
    28.    int ans = MAX,i,sum = 0;
    29.    for(i = position; i <= n;i++)
    30.    {
    31.       sum += pages[i];
    32.       // taking the best of all possible solutions
    33.       ans = min(ans,max(sum,rec(i+1,sub-1)));
    34.    }
    35.    dp[position][sub]=ans;
    36.    return ans;
    37. }
    38.  
    39. int main()
    40. {
    41.    int test,i,start,ans,pos,sum;
    42.    cin >> test;
    43.    while(test -- )
    44.    {
    45.       cin >> n >> m;
    46.       memset(dp,-1,sizeof(dp));
    47.       for(i = 1;i <=n ;i++) cin >> pages[i];
    48.  
    49.       // This is the minimized maximum number of
    50.       // pages any subscriber gets
    51.       cout << rec(1,m) <<endl;
    52.  
    53.  
    54.       // Now printing the assignment,
    55.       // basically the part you were looking for
    56.       // Once dp array is filled its easy to find an optimal assignment
    57.       start = 1;
    58.       while( start <= n )
    59.       {
    60. 	 ans = MAX,pos=-1,sum=0;
    61. 	 for(i = start; i <= n; i++)
    62. 	 {
    63. 	    sum += pages[i];
    64. 	    if(ans > max(sum ,rec(i+1,m-1)))
    65. 	    {
    66. 	       ans = max(sum,rec(i+1,m-1));
    67. 	       pos = i;
    68. 	    }
    69. 	 }
    70. 	 m--;
    71. 	 for(i = start;i <= pos;i++)cout << pages[i] << " ";
    72. 	 cout << " / ";
    73. 	 start = pos+1;
    74.       }
    75.       cout << endl;
    76.    }
    77.    return 0;
    78. }