摘要: Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 2 Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem DescriptionEddy's interest is very extensive, recently he is interested in prime number. Eddy discover th 阅读全文
posted @ 2012-04-14 11:02 剑不飞 阅读(130) 评论(0) 推荐(0) 编辑
摘要: Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4 Accepted Submission(s) : 3Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description哥德巴赫猜想大家都知道一点吧.我们现在不是想证明这个结论,而是想在程序语言内部能够表示的数集中,任意取出一个偶数,来寻找两个素数,使得其和等于该偶数.做好了这件实事,就能说明这个猜想是成立的.由于 阅读全文
posted @ 2012-04-14 10:30 剑不飞 阅读(180) 评论(0) 推荐(0) 编辑
摘要: 素数判定 Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 8 Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39 #include #define N 4004 using namespace std; bool i 阅读全文
posted @ 2012-04-14 10:08 剑不飞 阅读(169) 评论(0) 推荐(0) 编辑
摘要: 计算大素数因子#include #include #define N 20004 using namespace std; int isp[N]; int main() { int m,n,max,ans; memset(isp,0,sizeof(isp)); isp[1]=1; for(int i=2;i>m) { max=0; for(int i=0;i>n; if(isp[n]>max) { max=isp[n]; ans=n; } } cout<<ans<<endl; }... 阅读全文
posted @ 2012-04-14 09:08 剑不飞 阅读(135) 评论(0) 推荐(0) 编辑