close
標題:
C++ Program pf seiving numbers
發問:
Hey everyone, i need a program that uses the seive of Eratosthenes(don't know how to speel, sorry!) to find all the prime numbers from 1-300. Can it be as simple as possible and pls be quick! 10000x thx! =) (pls don't include iostream.h, thx!)
最佳解答:
Here's the program. As usual, if you need details, see http://mathmate.brinkster.net/ and look under programming. #include < stdio.h> #include < stdlib.h> #include < math.h> int main(int argc, char *argv[]) { int a[301],i,j; for(i=0;i<=300;i++)a[i]=1; // initialize, 1=prime, 0=composite for(i=2;i<=sqrt(300);i++) { // skip over composites if(a[i]!=0)for(j=2*i;j<=300;j+=i) a[j]=0; // mark multiples of primes except first } // print primes for(i=1;i<=300;i++)if(a[i]!=0)printf("%d ",i); } 2008-01-19 21:56:27 補充： The above code is based on making an array of numbers from 1 to 300, all initialized to 1. Starting from 2, multiples of 2 (4,6,8,...)are marked zero (composite). Then it proceeds to the next number if it is not a composite. 2008-01-19 21:56:59 補充： For 3, it marks zero for 6,9,12,15.... Four (4) has previously been marked composite, so it will not be necessary to mark multiples. This will continue until the square root of 300 (17), which is the maximum integer whose square is below 300.
C++ Program pf seiving numbers
發問:
Hey everyone, i need a program that uses the seive of Eratosthenes(don't know how to speel, sorry!) to find all the prime numbers from 1-300. Can it be as simple as possible and pls be quick! 10000x thx! =) (pls don't include iostream.h, thx!)
最佳解答:
Here's the program. As usual, if you need details, see http://mathmate.brinkster.net/ and look under programming. #include < stdio.h> #include < stdlib.h> #include < math.h> int main(int argc, char *argv[]) { int a[301],i,j; for(i=0;i<=300;i++)a[i]=1; // initialize, 1=prime, 0=composite for(i=2;i<=sqrt(300);i++) { // skip over composites if(a[i]!=0)for(j=2*i;j<=300;j+=i) a[j]=0; // mark multiples of primes except first } // print primes for(i=1;i<=300;i++)if(a[i]!=0)printf("%d ",i); } 2008-01-19 21:56:27 補充： The above code is based on making an array of numbers from 1 to 300, all initialized to 1. Starting from 2, multiples of 2 (4,6,8,...)are marked zero (composite). Then it proceeds to the next number if it is not a composite. 2008-01-19 21:56:59 補充： For 3, it marks zero for 6,9,12,15.... Four (4) has previously been marked composite, so it will not be necessary to mark multiples. This will continue until the square root of 300 (17), which is the maximum integer whose square is below 300.
- 3-6月時事題(20分)
- New Sir - @2866 一問@1@
- 201+202+203+...+401
- 係hk賣淫犯法--
- 兩用住宅的條例一問 work-home policy
- 怎樣做網站的登入系統？
- 淺水灣點去-
- Initial and Final States ( Chemistry Problem)
- 番禺...
- 中交建設可嗎？
此文章來自奇摩知識+如有不便請留言告知
其他解答:7638E7481407D16B
文章標籤
全站熱搜
留言列表
