Lucky Numbers
Problem Statement :
A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between a and b inclusive, are lucky? For example, a=20 and b=25. Each number is tested below: digit digit squares value sum squares sum 20 2 4,0 4 21 3 4,1 5 22 4 4,4 8 23 5 4,9 13 24 6 4,16 20 25 7 4,25 29 We see that two numbers, 21, 23 and 25 are lucky. Note: These lucky numbers are not to be confused with Lucky Numbers Function Description Complete the luckyNumbers function in the editor below. It should return an integer that represents the number of lucky numbers in the given range. luckyNumbers has the following parameter(s): a: an integer, the lower range bound b: an integer, the higher range bound Input Format The first line contains the number of test cases T. Each of the next T lines contains two space-separated integers, a and b. Constraints 1 <= T <= 10^4 1 <= a <= b <= 10^18 Output Format Output T lines, one for each test case in the order given.
Solution :
Solution in C :
In C++ :
#include <iostream>
#include <cstring>
using namespace std;
typedef long long s64;
const int D = 18;
const int MD = 9;
const s64 MAX = 1000000000000000000ull;
s64 dp[D][D*MD+1][D*MD*MD+1];
int hi[D];
bool isPrime[D*MD*MD+1];
s64 go(int idx, int sd, int ssd, bool limited) {
if (idx == D) return isPrime[sd] && isPrime[ssd] ? 1 : 0;
s64& ret = dp[idx][sd][ssd];
if (!limited && ret >= 0) return ret;
int upper = (limited ? hi[idx] : 9);
s64 r = 0;
for (int i = 0; i <= upper; i++)
r += go(idx+1, sd+i, ssd+i*i, (limited && i == upper));
if (!limited) ret = r;
return r;
}
s64 go(s64 n) {
if (n == MAX) return go(MAX-1);
if (n <= 1) return 0;
for (int i = D-1; i >= 0; i--) hi[i] = (int)(n % 10), n /= 10;
return go(0, 0, 0, true);
}
int main() {
for (int i = 2; i <= D*MD*MD; i++) isPrime[i] = true;
for (int i = 2; i <= D*MD*MD; i++)
if (isPrime[i]) for (int j = i+i; j <= D*MD*MD; j += i) isPrime[j] = false;
memset(dp, 0xff, sizeof(dp));
int t; cin >> t;
while (t--) {
s64 a, b; cin >> a >> b;
cout << (go(b)-go(a-1)) << endl;
}
return 0;
}
In Java :
import java.io.*;
import java.util.Arrays;
/**
* @author Sergey Vorobyev (svorobyev@spb.com)
*/
public class Solution {
MyTokenizer in;
PrintWriter out;
public static void main(String[] args) throws Exception {
Solution instance = new Solution();
instance.go();
}
static class MyTokenizer {
private BufferedReader in;
private String[] buffer = new String[]{};
private int uk;
MyTokenizer() {
in = new BufferedReader(new InputStreamReader(System.in));
}
private String nextToken() throws IOException {
while (true) {
if (uk < buffer.length && !buffer[uk].isEmpty()) {
return buffer[uk++];
}
if (uk >= buffer.length) {
uk = 0;
buffer = in.readLine().split(" ");
} else if (buffer[uk].isEmpty()) {
uk++;
}
}
}
int ni() throws IOException {
return Integer.parseInt(nextToken());
}
long nl() throws IOException {
return Long.parseLong(nextToken());
}
}
private int ni() throws IOException {
return in.ni();
}
private long nl() throws IOException {
return in.nl();
}
boolean[] isPrime;
long[] fact;
long[][][] d;
long[][][] d2;
long[][][][] bufferedD;
int[] primes;
int[] nextPrime;
int primesCount;
static final int MAX_DIGITS = 18;
static final int MAX_SQ_SUM = 81 * MAX_DIGITS + 1;
static final int MAX_PRIME = 3000;
static final int MAX_SUM = 9 * MAX_DIGITS + 1;
void fillPrime() {
primes = new int[MAX_PRIME];
nextPrime = new int[MAX_PRIME];
isPrime = new boolean[MAX_PRIME];
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;
int predPrime = -1;
for (int i = 2; i < MAX_PRIME; i++) {
if (isPrime[i]) {
for (int ii = predPrime + 1; ii <= i; ii++) {
nextPrime[ii] = primesCount;
}
predPrime = i;
primes[primesCount++] = i;
for (int j = i * i; j < MAX_PRIME; j += i) {
isPrime[j] = false;
}
}
}
}
void fillD2() {
d2 = new long[MAX_DIGITS + 1][MAX_SUM][MAX_SQ_SUM];
for (int dig = 0; dig <= MAX_DIGITS; dig++) {
for (int sum = 0; sum + dig * 9 < MAX_SUM; sum++) {
for (int isum = 0; isum <= Math.min(9*dig, MAX_SUM - sum); isum++) {
if (isPrime[sum + isum]) {
for (int sqSum = 0; sqSum <= dig*81; sqSum++) {
d2[dig][sum][sqSum] += d[dig][isum][sqSum];
}
}
}
}
}
}
private void go() throws Exception {
long time = System.currentTimeMillis();
in = new MyTokenizer();
out = new PrintWriter(System.out);
fillPrime();
int FAC_MAX = MAX_DIGITS + 1;
fact = new long[FAC_MAX];
fact[0] = 1;
for (int i = 1; i < FAC_MAX; i++) {
fact[i] = fact[i - 1] * i;
}
d = new long[MAX_DIGITS + 1][MAX_SUM][MAX_SQ_SUM];
int[] counts = new int[10];
rec(0 /*pos*/, MAX_DIGITS, 0, 0, 1, true);
fillD2();
int T = ni();
for (int cas = 1; cas <= T; cas++) {
long A = nl();
long B = nl();
out.println(getLucky(B) - getLucky(A - 1));
}
out.close();
}
private long getLucky(long x) {
if (x == 1000000000000000000L) {
return getLucky(x - 1);
}
int n = log10(x);
int[] digits = toDigits(x, n);
return calc(digits);
}
private void rec(int pos, int remind, int sum, int sqSum, long div, boolean update) {
int used = MAX_DIGITS - remind;
if (update) {
d[used][sum][sqSum] += fact[used] / div;
}
if (pos == 10) {
return;
}
int pos2 = pos * pos;
for (int i = 0; i <= remind; i++) {
rec(pos + 1, remind - i, sum, sqSum, div * fact[i], i != 0);
sum += pos;
sqSum += pos2;
}
}
private int[] toDigits(long x, int n) {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[n - i - 1] = (int) (x % 10);
x /= 10;
}
return res;
}
private long calc(int[] digits) {
int n = digits.length;
long res = 0;
int sum = 0;
int sumSq = 0;
for (int i = 0; i < n; i++) {
int nmi = n - 1 - i;
int maxprime = 81 * n;
if (digits[i] < 5) {
for (int j = 0; j < digits[i]; j++) {
sum += j;
sumSq += j * j;
for (int p2i = nextPrime[sumSq]; primes[p2i] < maxprime; p2i++) {
int isumSq = primes[p2i] - sumSq;
res += d2[nmi][sum][isumSq];
}
sum -= j;
sumSq -= j * j;
}
} else {
for (int p2i = nextPrime[sumSq]; primes[p2i] < maxprime; p2i++) {
int isumSq = primes[p2i] - sumSq;
res += d2[n-i][sum][isumSq];
}
for (int j = digits[i]; j < 10; j++) {
sum += j;
sumSq += j * j;
for (int p2i = nextPrime[sumSq]; primes[p2i] < maxprime; p2i++) {
int isumSq = primes[p2i] - sumSq;
res -= d2[nmi][sum][isumSq];
}
sum -= j;
sumSq -= j * j;
}
}
sum += digits[i];
sumSq += digits[i] * digits[i];
}
// inclusive
res += isPrime[sum] && isPrime[sumSq] ? 1 : 0;
return res;
}
private int log10(long x) {
int res = 0;
while (x != 0) {
res++;
x /= 10;
}
return res;
}
}
In C :
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <malloc.h>
#include <assert.h>
#define NUMDIGITS 19
#define NUMSIEVE ((NUMDIGITS-1)*9*9)
// Sieve for generating primes
int p[1+NUMSIEVE];
// Lookup table for factorials
unsigned long long factorial[NUMDIGITS+1];
// To store binned integers
int bin[10];
// Stuff for number lists
typedef struct {
short s;
short q;
unsigned long long mult;
} Element;
static Element *lists[NUMDIGITS+1];
unsigned int listlen[NUMDIGITS+1];
unsigned int listidx[NUMDIGITS+1];
unsigned long long table[1+(NUMDIGITS-2)*9][1+(NUMDIGITS-2)*9*9];
static unsigned long long lookup[NUMDIGITS-1][1+(NUMDIGITS-2)*9][1+(NUMDIGITS-2)*9*9];
void make_list_1(int digit, int n, unsigned long long num, short s, short q, int N)
{
int i;
Element *e = lists[n] + listidx[n]++;
e->s = s;
e->q = q;
e->mult = factorial[n]/num;
if (n==N) return;
for (i=digit; i<=9; i++) {
bin[i]++;
make_list_1(i, n+1, num*bin[i], s+i, q+i*i, N);
bin[i]--;
}
}
void pack_list(int n)
{
int i, j, k, idx;
int imax=n*9, jmax=n*9*9, kmax=listlen[n];
Element *e;
for (i=0; i<=imax; i++)
for (j=0; j<=jmax; j++)
table[i][j] = 0;
for (k=0; k<kmax; k++) {
e = lists[n]+k;
table[e->s][e->q] += e->mult;
}
idx = 0;
for (i=0; i<=imax; i++) {
for (j=0; j<=jmax; j++) {
if (table[i][j]) {
e = lists[n]+idx++;
e->s = i;
e->q = j;
e->mult = table[i][j];
}
}
}
listlen[n] = idx;
}
void make_list(int N)
{
int i;
make_list_1(0, 0, 1, 0, 0, N);
for (i=1; i<=N; i++) {
pack_list(i);
}
}
unsigned long long num_lucky(int n, int ss, int sq)
{
int i, imax=listlen[n];
unsigned long long ires = 0;
if (lookup[n][ss][sq]) return lookup[n][ss][sq];
for (i=0; i<imax; i++) {
if (p[sq+lists[n][i].q]&&p[ss+lists[n][i].s]) {
ires += lists[n][i].mult;
}
}
if (ires)
lookup[n][ss][sq] = ires;
return ires;
}
int main()
{
int T;
int i, j;
int nsieve, sqrtnsieve;
int len, alen, blen;
char A[NUMDIGITS+1], B[NUMDIGITS+1];
int a[NUMDIGITS+1], b[NUMDIGITS+1];
unsigned long long ires;
int ssl, ssu, sql, squ;
// Lengths of lists of candidates
for (i=0; i<=NUMDIGITS; i++) {
if (i>0)
listlen[i] = ((i+9)*listlen[i-1])/i;
else
listlen[i] = 1;
lists[i] = (Element *)malloc(listlen[i]*sizeof(Element));
listidx[i] = 0;
}
// Make primes
nsieve = NUMSIEVE;
sqrtnsieve = sqrt(nsieve);
p[1] = 0;
for (i=2; i<=nsieve; i++)
p[i] = 1;
for (i=2; i<=sqrtnsieve; i++)
if (p[i]==1)
for (j=i+i; j<=nsieve; j+=i)
p[j] = 0;
// Make factorials
factorial[0] = 1;
for (i=1; i<=NUMDIGITS; i++)
factorial[i] = i*factorial[i-1];
// Create the master lists
make_list(17);
scanf("%d", &T);
while (T--) {
scanf("%s %s", A, B);
alen = strlen(A);
blen = strlen(B);
if (alen==19) {
alen = 18;
for (i=0; i<alen; i++)
A[i] = '9';
}
if (blen==19) {
blen = 18;
for (i=0; i<blen; i++)
B[i] = '9';
}
len = blen;
for (i=0; i<len; i++)
a[i] = (i<blen-alen)?0:(A[i-(blen-alen)]-'0');
for (i=0; i<len; i++)
b[i] = B[i]-'0';
ires = 0;
for(i=a[0]; i<=b[0]; i++)
ires += num_lucky(len-1, i, i*i);
ssl = sql = ssu = squ = 0;
for (j=1; j<len; j++) {
ssl += a[j-1];
sql += a[j-1]*a[j-1];
for (i=0;i<a[j];i++) {
ires -= num_lucky(len-j-1, ssl+i, sql+i*i);
}
ssu += b[j-1];
squ += b[j-1]*b[j-1];
for (i=9;i>b[j];i--) {
ires -= num_lucky(len-j-1, ssu+i, squ+i*i);
}
}
printf("%llu\n", ires);
}
for (i=0; i<=NUMDIGITS; i++)
if (lists[i]) free(lists[i]);
return 0;
}
In Python3 :
def lucky(N):
p = [int(x) for x in str(N)]
numDigits = len(p)
total = 0
curSum = 0
curSumSq = 0
while len(p):
for a in range(p[0]):
total += resolve(numDigits - 1, curSum + a, curSumSq + a**2)
numDigits -= 1
curSum += p[0]
curSumSq += p[0]**2
p.pop(0)
return total
def resolve(n, s, sq):
if (n,s,sq) in memo:
return memo[(n,s,sq)]
if n == 0:
if s in primes and sq in primes:
memo[(n,s,sq)] = 1
return 1
memo[(n,s,sq)] = 0
return 0
c = 0
for b in range(10):
c += resolve(n-1, s + b, sq + b**2)
memo[(n,s,sq)] = c
return c
memo = {}
primes = set([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999])
for test in range(int(input())):
[A,B] = [int(x) for x in input().split(' ')]
print(lucky(B+1)-lucky(A))
View More Similar Problems
Counting On a Tree
Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n
View Solution →Polynomial Division
Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie
View Solution →Costly Intervals
Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the
View Solution →The Strange Function
One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting
View Solution →Self-Driving Bus
Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever
View Solution →Unique Colors
You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti
View Solution →