int main(int argc, char *argv[])
{
/* n记录斐波那契额数的下标 */
long f1, f2, fn, n;double t;
f1 = 1; f2 = 1; n = 3;
while(1) {
/* 以下的一些运算说明见上文 */
fn = f1 + f2;
f1 = f2 % get9digits;
f2 = fn % get9digits;
/* log(phi) log(sqrt(5)/5) */
t = n * 0.20898764024997873 + (-0.3494850021680094);
/* 这里已经令f2 = fn了 */
if (isPandigital(f2)
&& /* 取t的小数部分 */
isPandigital( (int)( pow(10.0, t - (int)t + 8) ) )
)
break;
n++;
}
printf("%d\n", n);
return 0;
}
int32 problem104() {
BigInteger f1(1);
BigInteger f2(1);
int32 k = 2;
string front = "123456789";
string back = "123456789";
while (true) {
k++;
BigInteger temp = f1 + f2;
f1 = f2;
f2 = temp;
if (temp.numberOfDigits() < 18) {
continue;
}
for (int32 i = 0; i < 9; i++) {
back[i] = temp.getNthDigitFromRight(i) + '0';
}
if (isPandigital(back)) {
for (int32 i = 0; i < 9; i++) {
front[i] = temp.getNthDigitFromLeft(i) + '0';
}
if (isPandigital(front)) {
return k;
}
}
}
}
int main()
{
int mask[10];
memset(mask, 0, sizeof(mask));
mask[0] = 1; // no zeroes allowed.
int i, k, j, l, m;
int temp = 0;
long int sum = 0;
for (i=1;i < 10; i++)
{
mask[i] = 1;
for (k=1; k < 10; k++)
{
if (mask[k]) continue;
mask[k] = 1;
for (j=1; j<10; j++)
{
if (mask[j]) continue;
mask[j] = 1;
for (l=1; l<10; l++)
{
if (mask[l]) continue;
mask[l] = 1;
for (m=1; m<10; m++)
{
if (mask[m]) continue;
mask[m] = 1;
temp = (i*10+k)*(j*100+l*10+m);
if (isPandigital(temp,mask)) {sum += temp; printf("%d * %d = %d\n",(i*10+k),(j*100+l*10+m),temp);}
temp = (i)*(k*1000 + j*100+l*10+m);
if (isPandigital(temp,mask)) {sum += temp; printf("%d * %d = %d\n",(i),(k*1000 + j*100+l*10+m),temp);}
mask[m] = 0;
}
mask[l] = 0;
}
mask[j] = 0;
}
mask[k] = 0;
}
mask[i] = 0;
}
printf("Sum: %ld\n",sum);
}
int64_t Euler_32()
{
int a, b, c, alist[50], dex, sum;
int counter = 1;
for (c = 1000; c < 10000; c++) {
for (a = 2; a < 100; a++) {
if (c % a != 0) {
continue;
} else {
b = c / a;
}
if ((a < 10) && (b < 1000)) {
continue;
} else {
if (isPandigital(a, b, c) != 0) {
alist[counter] = c;
counter++;
}
}
}
}
sum = 0;
for (dex = 1; dex < counter; dex++) {
if (alist[dex] != alist[dex - 1]) {
sum += alist[dex];
}
}
return (int64_t) sum;
}
int main(){
uint r = 0;
for(uint i = 1; i < 10000; i++){
for(uint j = 2; j < 10; j++){
string s = "";
for(uint k = 1; k < j; k++){
s += intToString(i * k);
}
if(s.size() > 9)
break;
if(s.size() == 9 && isPandigital(stoi(s), 1, 9))
if((uint)stoi(s) > r){
r = stoi(s);
}
}
}
cout << r << endl;
return 0;
}
int main() {
int i;
for (i=7654321; i>=1234567; i-=2) {
if (isPandigital(i)) {
if (isPrime(i)) {
printf ("The largest pandigital prime is: %d\n", i);
break;
}
}
}
return 0;
}
int main(){
unsigned long i = 0;
for(i=FIRST_NBRE; i>=LAST_NBRE; i-=2){
if(isPandigital(i)){
if(isPrime(i)){
printf("%ld\n", i);
return 0;
}
}
}
return 0;
}
int main() {
int temp = 0, answer = 0;
for (int i = 9876; i > 9200; --i) {
temp = i * 100000 + i * 2;
if (isPandigital(temp)) {
answer = temp;
break;
}
}
std::cout << "The answer is " << answer << std::endl;
return 0;
}
void calc(int seed)
{
char bigBuffer[256] = "";
char buffer[20];
unsigned int multiple = 1;
while (strlen(bigBuffer) < 9)
{
sprintf(buffer, "%d", seed * multiple);
strcat(bigBuffer, buffer);
multiple++;
}
if (strlen(bigBuffer) == 9)
if (isPandigital(bigBuffer))
printf("%s : seed=%d n=%d\n", bigBuffer, seed - 1, multiple);
}
int main(int argc, char **argv)
{
const long MAX_NUM = boost::lexical_cast<long>(argv[1]);
//Error checking
if(argc > 2)
{
std::cerr << " >> ERROR: Too many arguments" << std::endl;
return 1;
}
if(MAX_NUM < 0)
{
std::cerr << " >> ERROR: Argument must be positive or zero" << std::endl;
return 1;
}
boost::dynamic_bitset<> numList(MAX_NUM + 1);
numList.flip();
long index = 1, step = 0;
//Sieve
while((index * index) <= MAX_NUM)
{
index++;
if(numList[index] == 0)
{
continue;
}
step = (index > 2) ? 2 * index : index;
for(long i = index * index; i <= MAX_NUM; i += step)
{
numList[i] = 0;
}
}
//Output results
for(long i = 2; i <= MAX_NUM; ++i)
{
if(numList[i] && isPandigital(i))
{
std::cout << i << '\n';
}
}
return 0;
}
int main()
{
std::set<int> products;
for (int a = 1; a <= 9876; a++)
{
for (int b = a; b <= 9876; b++)
{
int l = len(a) + len(b) + len(a * b);
if (l > 9)
break;
if (l == 9 && isPandigital(a, b, a * b))
products.insert(a * b);
}
}
std::cout << std::accumulate(products.begin(), products.end(), 0) << '\n';
}
int main(void)
{
// 31427 = ceil(sqrt(987654321)), max necessary prime divisor
Primes primes = precomputePrimes(5000, 31427);
// printf("Max prime: %d, count: %d", primes.p[primes.num-1], primes.num);
int i;
for (i = 1; i < 9876543; i++)
{
if (isPandigital(i) && isPrime(i, primes))
{
printf("%d ", i);
}
}
return 0;
}
int main()
{
std::set<int> products;
for (int i = 2; i < 9999; i++) {
for (int j = 2; j < 9999; j++) {
int k = i*j;
int digitsI[10];
int digitsJ[10];
int digitsK[10];
int numDigitsI = fillDigits(i, digitsI);
int numDigitsJ = fillDigits(j, digitsJ);
int numDigitsK = fillDigits(k, digitsK);
int nDigits = numDigitsI + numDigitsJ + numDigitsK;
if (nDigits < 9)
continue; // jump to next in the inner loop
if (nDigits > 9)
break; // jump to next in outer loop
// we have nine digits, see if the identity is pandigital
// stuff all in same array
int digits[10];
int numDigits = 0;
numDigits = stuffArray(numDigits, digits, numDigitsI, digitsI);
numDigits = stuffArray(numDigits, digits, numDigitsJ, digitsJ);
numDigits = stuffArray(numDigits, digits, numDigitsK, digitsK);
if (isPandigital(numDigits, digits)) {
std::cout << "i " << i << " j " << j << " k " << k << " numDigits " << numDigits << "\n";
products.insert(k);
}
}
}
std::cout << "unique products: " << "\n";
int sum = 0;
for (std::set<int>::const_iterator i = products.begin(); i != products.end(); ++i) {
sum += *i;
std::cout << *i << "\n";
}
std::cout << "sum: " << sum << "\n";
}
int32 problem38() {
// We are given that 9 x (1,2,3,4,5) = 918273645, a 1-9 pandigital number.
// So our number must begin with a 9 if it is larger. One digit numbers other
// than 9 won't create a concatenated product starting with a 9. A two digit
// number starting with 9 will have a concatenated product whose size is
// 2 -> 5 -> 8 -> 11 -> ... so it cannot work. Similarly for 3-digit numbers
// and 5-digit and larger numbers. So our concatenated product is formed by
// a 4-digit integer starting with 9 and (1,2) if it is larger than 918273645.
// Otherwise, 918273645 is the largest. So we check all 4-digit numbers
// beginning with 9.
int32 largest = 918273645;
for (int32 i = 9000; i < 10000; i++) {
int32 concatenatedProduct = i * 100000 + 2 * i;
if (concatenatedProduct > largest && isPandigital(concatenatedProduct)) {
largest = concatenatedProduct;
}
}
return largest;
}
uint64_t e032(){
std::set<uint64_t> products;
uint64_t sum = 0;
for(uint32_t i = 2; i < 99; ++i){
for(uint32_t j = 123; j < 9876 / 2; ++j){
if(i < 10 && j < 1000) continue;
uint64_t product = i * j;
if(product > 9876) break;
if(isPandigital(i, j, product))
products.insert(product);
}
}
for(auto it = products.begin(); it != products.end(); ++it)
sum += *it;
return sum;
}
std::string Problem038::get_answer() {
int highest = 0;
for (int i = 1; i < 10000; ++i) {
std::string result = "";
for (int j = 1; ; ++j) {
result += std::to_string(i * j);
if (result.size() > 9) {
break;
}
if (result.size() == 9) {
if (isPandigital(result)) {
if (atoi(result.c_str()) > highest) {
highest = atoi(result.c_str());
}
}
break;
}
}
}
return std::to_string(highest);
}
int main()
{
int max = 0;
for(int i = 1; i < 10000; ++i)
{
std::string tmp = "";
for(int j = 1; tmp.size() < 10; ++j)
{
std::stringstream ss;
ss << i * j;
tmp += ss.str();
if(isPandigital(tmp))
if( max < atoi(tmp.c_str()))
max = atoi(tmp.c_str());
}
}
std::cout << "Answer: " << max << std::endl;
return 0;
}
