Prime Digit Sums
Problem Statement :
Chloe is fascinated by prime numbers. She came across the number 283002 on a sign and, though the number is not prime, found some primes hiding in it by using the following rules: Every three consecutive digits sum to a prime: 283002 283002 283002 283002 Every four consecutive digits sum to a prime: 283002 283002 283002 Every five consecutive digits sum to a prime: 283002 283002 You must answer q queries, where each query consists of an integer, n. For each n, find and print the number of positive n-digit numbers, modulo 10^9 + 7, that satisfy all three of Chloe's rules (i.e., every three, four, and five consecutive digits sum to a prime). Input Format The first line contains an integer, q, denoting the number of queries. Each of the q subsequent lines contains an integer denoting the value of n for a query. Constraints 1 <= q <= 2 * 10^4 1 <= n <= 4 * 10^5 Output Format For each query, print the number of n-digit numbers satisfying Chloe's rules, modulo 10^9 +7, on a new line.
Solution :
Solution in C :
In C++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
#define mod 1000000007;
typedef long long ll;
int func(int n){
int sum=0;
while(n){
sum+=n%10;
n/=10;
}
return sum;
}
int p[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43};
int main(){
bool num3[1000],num4[10000];
int num5[100000];
for(int i=0;i<1000;i++){
num3[i]=false;
}
for(int i=0;i<10000;i++){
num4[i]=false;
}
for(int i=0;i<100000;i++){
num5[i]=-1;
}
int a=0,b=0,d=0;
int count1=0;
for(int i=0;i<1000;i++){
for(int j=0;j<14;j++){
if(func(i)==p[j]){
num3[i]=true;
//cout<<i<<endl;
count1++;
if(i>=100)
a++;
}
}
}
////cout<<" count of 3 "<<count1<<endl;
int count2=0;
for(int i=0;i<10000;i++){
if(num3[i%1000]&&num3[i/10]){
for(int j=0;j<14;j++){
if(func(i)==p[j]){
num4[i]=true;
//cout<<i<<endl;
count2++;
if(i>=1000)
b++;
}
}
}
}
//cout<<"Count of 4 "<<count2<<endl;
int count=0;
vector<int> num;
for(int i=0;i<100000;i++){
if(num4[i%10000]&&num4[i/10]){
for(int j=0;j<14;j++){
if(func(i)==p[j]){
num5[i]=count;
//cout<<i<<endl;
count++;
num.push_back(i);
if(i>=10000)
d++;
}
}
}
}
//cout<<"count of 5 "<<count<<endl;
// cout<<count<<endl;
vector<int> gr[count];
int count3=0;
for(int i=0;i<count;i++){
int temp=num[i];
temp/=10;
for(int j=0;j<count;j++){
if(temp==num[j]%10000){
gr[i].push_back(j);
//cout<<i<<" "<<j<<" "<<num[i]<< " "<<num[j]<<endl;
count3++;
}
}
}
//cout<<count3<<endl;
ll dp[2][count];
for(int i=0;i<count;i++){
dp[0][i]=1;
if(num[i]<10000){
dp[0][i]=0;
}
}
ll res[400007];
res[1]=9;
res[2]=90;
res[3]=a;
res[4]=b;
res[5]=d;
int r=1,c=0;
for(int i=6;i<400001;i++){
res[i]=0;
for(int j=0;j<count;j++){
dp[r][j]=0;
for(int k=0;k<gr[j].size();k++){
dp[r][j]+=dp[c][gr[j][k]];
}
dp[r][j]%=mod;
res[i]+=dp[r][j];
}
res[i]%=mod;
r=1-r;
c=1-c;
}
int q;
scanf("%d",&q);
for(int a0 = 0; a0 < q; a0++){
int n;
scanf("%d",&n);
cout<<res[n]<<endl;
//cout<<count<<endl;
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static boolean isPrime[] = new boolean[46];
public static int results[] = new int[400001];
public static boolean isGood(int n, int val) {
int max = Math.min(n, 5);
int mod = 100;
for (int i = 3; i <= max; i++) {
mod *= 10;
int d = val;
for (int j = 0; j < (max - i + 1); j++) {
int d2 = d % mod;
int sum = 0;
while (d2 > 0) {
sum += d2 % 10;
d2 /= 10;
}
if (!isPrime[sum]) {
return false;
}
d = d / 10;
}
}
return true;
}
public static void initSimple() {
for(int i=100; i<1000; i++) {
if(isGood(3,i)) {
results[3]++;
}
}
for(int i=1000; i<10000; i++) {
if(isGood(4,i)) {
results[4]++;
}
}
results[1]=9;
results[2]=90;
}
public static void init() {
Arrays.fill(isPrime, 0, isPrime.length, true);
isPrime[0] = false;
isPrime[1] = false;
for (int i = 2; i < 45; i++) {
if (isPrime[i]) {
for (int j = 2; j < i; j++) {
if (i % j == 0) {
isPrime[i] = false;
}
}
}
}
ArrayList<Integer> good = new ArrayList<Integer>();
for (int i = 0; i < 100000; i++) {
if (isGood(5, i)) {
good.add(i);
}
}
List<Integer> childs[] = new List[good.size()];
for (int i = 0; i < good.size(); i++) {
int val = (good.get(i) % 10000) * 10;
childs[i] = new ArrayList<Integer>();
for (int j = 0; j < 10; j++) {
if (good.contains(val + j)) {
childs[i].add(good.indexOf(val + j));
}
}
}
int buf1[] = new int[good.size()];
int buf2[] = new int[good.size()];
int temp[];
int count = 0;
for (int i = 0; i < buf1.length; i++) {
if (good.get(i) >= 10000) {
buf1[i] = 1;
count++;
}
}
results[5] = count;
int idx = 0;
for (int i = 6; i < results.length; i++) {
Arrays.fill(buf2, 0, buf2.length, 0);
count = 0;
for (int j = 0; j < buf1.length; j++) {
if (buf1[j] > 0) {
List<Integer> next = childs[j];
for (int k = 0; k < next.size(); k++) {
idx = next.get(k);
buf2[idx] += buf1[j];
buf2[idx] %= 1000000007;
count += buf1[j];
count %= 1000000007;
}
}
}
results[i]=count;
temp=buf1;
buf1=buf2;
buf2=temp;
}
}
public static void main(String[] args) {
init();
initSimple();
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int n = in.nextInt();
System.out.println(results[n]);
}
}
}
In C :
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef unsigned long long ull;
typedef unsigned int ui;
#define swap_(x, y)
{
ui *z = x;
x = y;
y = z;
}
#define ac 20
#define ec (18 * 18)
#define mod 1000000007
#define brutec 13
ui brutes[] = {0, 9, 90, 303, 280, 218, 95,
101, 295, 513, 737, 668, 578}; // see haskell file
ui as[ec * ac];
ui startv[] = {17, 6, 6, 2, 0, 0, 9, 3, 9, 3, 0, 0, 6, 4, 13, 15, 13, 15};
ui endv[] = {5, 11, 11, 26, 20, 20, 7, 7, 7, 5, 24, 24, 5, 5, 2, 2, 2, 1};
// 18x18 times 18x18
void multmm(ui *a, ui *b, ui *c) {
for (int i = 0; i < 18; ++i) {
for (int j = 0; j < 18; ++j) {
ull x = 0;
for (int k = 0; k < 18; ++k) {
x += (ull)a[i * 18 + k] * (ull)b[k * 18 + j];
}
c[i * 18 + j] = (ui)(x % mod);
}
}
}
// 18x18 times 18x1
void multmv(ui *a, ui *b, ui *c) {
for (int i = 0; i < 18; ++i) {
ull x = 0;
for (int k = 0; k < 18; ++k) {
x += (ull)a[i * 18 + k] * (ull)b[k];
}
c[i] = (ui)(x % mod);
}
}
// 1x18 times 18x1
ui inprod(ui *a, ui *b) {
ull x = 0;
for (int k = 0; k < 18; ++k) {
x += (ull)a[k] * (ull)b[k];
}
return (ui)(x % mod);
}
void copy(ui *a, ui *b, int n) {
for (int i = 0; i < n; ++i) {
a[i] = b[i];
}
}
void ini() {
ui _a[] = {
0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
ui _b[ec];
ui *a = _a, *b = _b;
for (int i = 0; i < ac; ++i) {
copy(&as[i * ec], a, ec);
multmm(a, a, b);
swap_(a, b);
}
}
ui solve(int n) {
if (n < brutec) {
return brutes[n];
}
n -= 12;
ui _v[18], _u[18];
ui *v = _v, *u = _u;
copy(v, startv, 18);
ui *a = as;
while (n) {
if (n % 2) {
multmv(a, v, u);
swap_(v, u);
}
n /= 2;
a += ec;
}
return inprod(v, endv);
}
int main() {
ini();
int q;
scanf("%d", &q);
for (int i = 0; i < q; ++i) {
int n;
scanf("%d", &n);
ui x = solve(n);
printf("%u\n", x);
}
return 0;
}
In Python3 :
mod = 10**9 + 7
A = [0,0,0,0,0,2,3,15,2,0]
B = [0,0,0,0,0,4,9,13,17,2]
results = [1,9,90,303,280,218,95,101,295]
for _ in range(int(input())):
n = int(input())
for i in range(len(A),n):
A.append((2*(A[i-5] + B[i-5])) % mod)
B.append((A[i-1] + A[i-5] + 2*B[i-5]) % mod)
print(results[n] if n < 9 else
(5*A[n-1] + 9*A[n-2] + 19*A[n-3] + 6*A[n-4] + 3*A[n-5]
+ 2*B[n-1] + 5*B[n-2] + 20*B[n-3] + 5*B[n-4] + 4*B[n-5]) % mod)
View More Similar Problems
Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →