POJ 2739 素數篩選

POJ 2739

   關鍵是找10000內質數的方法

#include <iostream>using namespace std;void create_primes(int * prime,int &size){//找出10000以內所有的質數，用因子排除法int notPrime[10001] = {false};int i ;for (i = 2; i < 10001 ;i++){int x = i*2;while(x<10001){notPrime[x] = true;x += i;}}for (i = 2; i < 10001;i++){if (!notPrime[i]){prime[size++]=i;}}}int main(){int prime[10000];int sizeofprime = 0;create_primes(prime,sizeofprime);int num,result=0;while (cin >> num,num != 0){for(int i = 0;i < sizeofprime;i++){if (prime[i] > num){break;}int n = num;int index = i;        while(n > 0){n -= prime[index];//反覆試探用num減去從某個比之小的質數開始的連續質數看是否可以減為0index++;}if (n == 0)result++;}}cout<<result<<endl;result = 0;}return 0;}