素数表
来源:互联网 发布:公司网络 编辑:程序博客网 时间:2024/05/17 04:47
Description
The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his company. The cryptographic keys are created from the product of two primes, and are believed to be secure because there is no known method for factoring such a product effectively.
What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key.
What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key.
Input
The input consists of no more than 20 test cases. Each test case is a line with the integers 4 <= K <= 10 100 and 2 <= L <= 10 6. K is the key itself, a product of two primes. L is the wanted minimum size of the factors in the key. The input set is terminated by a case where K = 0 and L = 0.
Output
For each number K, if one of its factors are strictly less than the required L, your program should output "BAD p", where p is the smallest factor in K. Otherwise, it should output "GOOD". Cases should be separated by a line-break.
Sample Input
143 10143 20667 20667 302573 302573 400 0
Sample Output
GOODBAD 11GOODBAD 23GOODBAD 31
#include<iostream>#include<cstdio>#include<string>#include<cmath>using namespace std;int kw[1001];bool zhengchu(int n,int lkw)//是否能整除?{int sum=0;for(int i=lkw-1;i>=0;i--)sum=(sum*1000+kw[i])%n; if(sum%n==0) return true; return false;}bool sushu[1000001]={0};//0为素数void ss(){sushu[0]=1;sushu[1]=1;for(int i=2;i*i<1000005;i++){if(sushu[i]==0){for(int j=2*i;j<1000005;j+=i)sushu[j]=1;}}}int main(){string s;int n;cin>>s;ss();while(scanf("%d",&n)!=EOF){if(n==0)break; memset(kw,0,sizeof(kw)); int sum=0; int i=0; for(i=0;i<s.length();i++) { int ii=(s.length()+2-i)/3-1; kw[ii]=s[i]-'0'+10*kw[ii]; }for(i=2;i<n;i++){if(!sushu[i]&&zhengchu(i,(s.length()+2)/3)){cout<<"BAD"<<' '<<i<<endl;break;}}if(i==n)cout<<"GOOD"<<endl;cin>>s;}}
从这道题,可以学到
判断素数的简便方法
bool sushu[1000001] = { 0 };//0为素数
void ss()
{
sushu[0] = 1;
sushu[1] = 1;
for (int i = 2; i*i<1000005; i++)
{
if (sushu[i] == 0)
{
for (int j = 2 * i; j<1000005; j += i)
sushu[j] = 1;
}
}
}
还有大数据的进制转换,将大数转化为千进制数据,加快处理,
Int kw[10000];
对s输入的字符串进行进制转换,
memset(kw,0,sizeof(kw));
int i=0;
for(i=0;i<s.length();i++)
{
int ii=(s.length()+2-i)/3-1;
kw[ii]=s[i]-'0'+10*kw[ii];
}
注意高位在kw的后面,首先处理,然后把处理剩下的数据*1000+前一位。
int kw[1001];
bool zhengchu(int n,int lkw)
{
//是否能整除?
int sum=0;
for(int i=lkw-1;i>=0;i--)
sum=(sum*1000+kw[i])%n;
if(sum%n==0)
return true;
return false;
}
0 0
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数表
- 素数 表
- 素数表
- 素数表
- 素数表
- 素数表
- disruptor:CAS实现高效(伪)无锁阻塞队列实践
- actor Java 实现
- c语言:双向链表的实现
- Android 常见分辨率(mdpi、hdpi 、xhdpi、xxhdpi )及屏幕适配注意事项
- Java并发编程之——Amino框架
- 素数表
- 面试必备之:MFC socket编程(浅出+深度:服务端和客户端端口问题)
- 多项式的输出
- 应用程序的通信成本
- 大数据架构:flume-ng+Kafka+Storm+HDFS 实时系统组合
- 【消息队列MQ】各类MQ比较
- Apache Kafka:下一代分布式消息系统
- JAVA远程操作Zookeeper示例
- 互联网产品经理常用工具有哪些