Problem 50 auf projecteuler.net
Hier meine Brute Force Lösung zu Problem 50.
Befehl zum Compilieren und Ausführen sowie die Ausgabe mit Zeitmessung:
gcc -O3 -march=native -mtune=native -std=c11 -ffast-math -fopenmp Prob50.c -lm -o Prob50 && time ./Prob50
Result: 997651 with 543 terms
real 0m2.809s
user 0m16.577s
sys 0m0.012s
Und der Code:
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#include <stdio.h> #define ARRLENGTH(x) (sizeof(x) / sizeof((x)[0])) static unsigned int prmlst[78498]; inline static unsigned int isPrime(unsigned int n) { if(n == 1) return 0; else if(n == 2) return 1; else { int i, r = (int) __builtin_sqrt(n); for(i = 2; i <= r && n % i != 0; ++i); return (r == i - 1) ? 1 : 0; } } inline static unsigned int getConPrmCnt(unsigned int n) { unsigned int sum = 2, prm = 1, cnt = 1, sumcnt = 0; while(sum != n) { sum = prmlst[sumcnt++]; prm = 1; cnt = sumcnt; while(sum < n) { sum += prmlst[cnt++]; prm++; } } return (sum == n) ? prm : 0; } int main(void) { unsigned int idx = 0, res = 0; for(unsigned int i = 0; i < 1000000; i++) if(isPrime(i)) prmlst[idx++] = i; #pragma omp parallel for for(unsigned int i = 0; i < ARRLENGTH(prmlst); i++) { unsigned int tmp = getConPrmCnt(prmlst[i]); if(tmp > res) { res = tmp; idx = prmlst[i]; } } printf("Result: %u with %u terms\n", idx, res); } |
Veröffentlicht am 21. Mai 2016 von admin in C, Programmierung, projecteuler