[Projecte1_] projecteuler_037

// Problem 37 // 14 February 2003 // The number 3797 has an interesting property. being Prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797,797, 97, and 7. similarly we can work from right to left: 3797,379, 37, and 3. /// find the sum of the only eleven primes that are both truncatable from left to right and right to left. //// note: 2, 3, 5, and 7 are not considered to be truncatable primes. # include <iostream> # include <windows. h> # include <cmath> # include <ctime> using namespace STD; // determines whether a number is a prime number. bool isprimenum (INT num) {If (Num % 2 = 0 & num> 2) | num <= 1) {return false;} int sqrtnum = (INT) SQRT (double) num); For (INT I = 3; I <= sqrtnum; I + = 2) {If (Num % I = 0) {return false ;}} return true ;} // determine whether it is a truncatable prime boo L checktruncatablenum (const int num) {int currentnum = num; // remove the number from the right to the left. The original number has been determined here, And the while (currentnum! = 0) {If (! Isprimenum (currentnum) {return false;} currentnum/= 10;} // remove the number int tendigit = 10 from left to right; currentnum = num % tendigit; while (currentnum! = Num) {If (! Isprimenum (currentnum) {return false;} tendigit * = 10; currentnum = num % tendigit;} return true;} void F1 () {cout <"Void F1 () "<Endl; large_integer timestart, timeend, freq; queryperformancefrequency (& freq); queryperformancecounter (& timestart); const int min_num = 11; // starting from 11, because the question requires exclusion of, const int max_count = 11; // a total of 11 int sum = 0; // The total number of records int COUNT = 0; // total records for (INT I = min_num; count <max_count; I + = 2) {If (checktruncatablenum (I) {cout <I <Endl; count ++; sum + = I ;}} cout <"Total:" <sum <Endl; queryperformancecounter (& timeend ); cout <"Total milliseconds is" <(double) (timeend. quadpart-timestart. quadpart) * 1000/freq. quadpart <Endl; time_t currenttime = Time (null); char timestr [30]; ctime_s (timestr, 30, effecttime ); cout <Endl <"by godmoon" <Endl <timestr;} // main function int main () {F1 (); Return 0 ;} /* void F1 () 2337537331331737379731373797739397 total is 748317 total milliseconds is 453.591by godmoonsat Nov 05 14:09:20 2011 */