Special Pairs with sum as Prime Number

问题

A number N is given in the range 1 <= N <= 10^50. A function F(x) is defined as the sum of all digits of a number x. We have to find the count of number of special pairs (x, y) such that:
1. 0 <= x, y <= N
2. F(x) + F(y) is prime in nature
We have to count (x, y) and (y, x) only once. Print the output modulo 1000000000 + 7

My approach:
Since the maximum value of sum of digits in given range can be 450 (If all the characters are 9 in a number of length 50, which gives 9*50 = 450). So, we can create a 2-D array of size 451*451 and for all pair we can store whether it is prime or not.
Now, the issue I am facing is to find all the pairs (x, y) for given number N in linear time (Obviously, we cannot loop through 10^50 to find all the pairs). Can someone suggest any approach, or any formula (if exists), to get all the pairs in linear time.

回答1:

You can create a 2-D array of size 451*451 and for all pair we can store whether it is prime or not. At the same time if you know how many numbers less than n who have F(x)=i and how many have F(x)=j, then after checking (i+j) is prime or not you can easily find a result with the state (i,j) of 2-D array of size 451*451.

So what you need is finding the total numbers who have F(x) =i.

You can easily do it using digit dp:

Digit DP for finding how many numbers who have F(x)=i:

string given=convertIntToString(given string);
int DP[51][2][452]= {-1};
Initially all index hpolds -1;
int digitDp(int pos,int small,int sum)
{
    if(pos==given.size())
    {
        if(sum==i) return 1;
        else return 0;
    }
    if(DP[pos][small][sum]!=-1)return DP[pos][small][sum];
    int res=0;
    if(small)
    {
        for(int j=0; j<=9; j++)res=(res+digitDp(pos+1,small,sum+j))%1000000007;
    }
    else
    {
        int hi=given[pos]-'0';
        for(int j=0; j<=hi; j++)
        {
            if(j==hi)res=(res+digitDp(pos+1,small,sum+j))%1000000007;
            else res=(res+digitDp(pos+1,1,sum+j))%1000000007;
        }
    }
    return DP[pos][small][sum]=res;
}

This function will return the total numbers less than n who have F(x)=i.

So we can call this function for every i from 0 to 451 and can store the result in a temporary variable.

int res[452];
for(i=0;i<=451;i++){
  memset(DP,-1,sizeof DP);
  res[i]=digitDp(0,0,0);
}

Now test for each pair (i,j) :

int answer=0;
for(k=0;k<=451;k++){
   for(int j=0;j<=451;j++){
       if(isprime(k+j)){
         answer=((log long)answer+(long long)res[k]*(long long)res[j])%1000000007;
      }
   }
}

finally result will be answer/2 as (i,j) and (j,i) will be calculated once.

Although there is a case for i=1 and j=1 , Hope you will be able to  handle it.

来源：https://stackoverflow.com/questions/56737445/special-pairs-with-sum-as-prime-number