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
Queries with Fixed Length
Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon
View Solution →QHEAP1
This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element
View Solution →Jesse and Cookies
Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t
View Solution →Find the Running Median
The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.
View Solution →Minimum Average Waiting Time
Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h
View Solution →Merging Communities
People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w
View Solution →