数位小孩()数位dp

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

const int N = 100;
#define int long long
int f[N][10][2];


bool ck(int x, int y) {
    if (x < 0) return 1;
    if (x + y< 2) return false;
    x += y ;
    for (int i = 2; i <= x / i; i ++) 
        if (x % i == 0) return false;
    return true;
}
void init () {
     f[1][1][1] = 1;
    f[1][0][0] = 1;
     for (int i = 2; i <= 9; i ++) 
            f[1][i][0] = 1;
    
    for (int i = 2; i < N; i ++) 
        for (int j = 0; j <= 9; j ++)
            for (int k = 0; k <= 9; k ++) 
               if (ck(j, k)) {
                    if (j == 1) {
                        f[i][j][1] += f[i - 1][k][0] + f[i - 1][k][1];
                        f[i][j][0] = 0;
                    }
                    else {
                        f[i][j][1] += f[i - 1][k][1];
                        f[i][j][0] += f[i - 1][k][0]; 
                    }
               }
}

int dp (int n) {
    if (!n ) return 0;
    
    vector<int> nums;
    int res = 0, last = -1;//第一次一定满足条件
    while (n) nums.push_back (n % 10), n /= 10;
    bool ok = 0;
    for (int i = nums.size() - 1; i >= 0; i --) {
        int x = nums[i];
       
        for (int j  = (i == nums.size() - 1) ; j < x; j ++) {
            if (ck(last, j) ){
                 res += f[i + 1][j][1];
                 if (ok) res += f[i + 1][j][0];
            }
        } 
        if (x == 1) ok = 1;
        if (ck(last, x)) last = x;
       else break;
        if (!i&& ok) res ++;
    }
    
    for (int i = 1; i < nums.size(); i ++)
        for (int j = 1; j <= 9; j ++)
            res += f[i][j][1];
    return res;
}
signed main () {
    init();
    int l, r;

    cin >> l >> r;
    cout << dp(r) - dp(l - 1) << endl;
    //cout << dp (r) - dp (l - 1) << endl;
   
}