• 关于 prime 的搜索结果

回答

更正您的代码, user_number = int(input("Enter a number: ")) a = list(range(2, user_number)) def prime(x): isPrime = True # Assumne that the number is prime for e in a: if user_number % e == 0: # If divisible, it's not a prime number isPrime = False break if isPrime: print(str(user_number) + " is a prime number.") # Not divisible by any number in range [2,user_number), i.e. a prime number else: print(str(user_number) + " is not a prime number.") prime(user_number)

几许相思几点泪 2019-12-29 18:52:37 0 浏览量 回答数 0

回答

上学期交的作业,已通过老师在运行时间上的测试 #include <stdio.h> #include <stdlib.h> unsigned long prime1,prime2,ee; unsigned long *kzojld(unsigned long p,unsigned long q) //扩展欧几里得算法求模逆 { unsigned long i=0,a=1,b=0,c=0,d=1,temp,mid,ni[2]; mid=p; while(mid!=1) { while(p>q) {p=p-q; mid=p;i++;} a=c*(-1)*i+a;b=d*(-1)*i+b; temp=a;a=c;c=temp; temp=b;b=d;d=temp; temp=p;p=q;q=temp; i=0; } ni[0]=c;ni[1]=d; return(ni); } unsigned long momi(unsigned long a,unsigned long b,unsigned long p) //模幂算法 { unsigned long c; c=1; if(a>p) a=a%p; if(b>p) b=b%(p-1); while(b!=0) { while(b%2==0) { b=b/2; a=(a*a)%p; } b=b-1; c=(a*c)%p; } return(c); } void RSAjiami() //RSA加密函数 { unsigned long c1,c2; unsigned long m,n,c; n=prime1*prime2; system("cls"); printf("Please input the message:\n"); scanf("%lu",&m);getchar(); c=momi(m,ee,n); printf("The cipher is:%lu",c); return; } void RSAjiemi() //RSA解密函数 { unsigned long m1,m2,e,d,*ni; unsigned long c,n,m,o; o=(prime1-1)*(prime2-1); n=prime1*prime2; system("cls"); printf("Please input the cipher:\n"); scanf("%lu",&c);getchar(); ni=kzojld(ee,o); d=ni[0]; m=momi(c,d,n); printf("The original message is:%lu",m); return; } void main() { unsigned long m; char cho; printf("Please input the two prime you want to use:\n"); printf("P=");scanf("%lu",&prime1);getchar(); printf("Q=");scanf("%lu",&prime2);getchar(); printf("E=");scanf("%lu",&ee);getchar(); if(prime1<prime2) {m=prime1;prime1=prime2;prime2=m;} while(1) { system("cls"); printf("\t*******RSA密码系统*******\n"); printf("Please select what do you want to do:\n"); printf("1.Encrpt.\n"); printf("2.Decrpt.\n"); printf("3.Exit.\n"); printf("Your choice:"); scanf("%c",&cho);getchar(); switch(cho) { case '1':RSAjiami();break; case '2':RSAjiemi();break; case '3':exit(0); default:printf("Error input.\n");break; } getchar(); } }

一键天涯 2019-12-02 01:26:32 0 浏览量 回答数 0

回答

上学期交的作业,已通过老师在运行时间上的测试 #include <stdio.h> #include <stdlib.h> unsigned long prime1,prime2,ee; unsigned long *kzojld(unsigned long p,unsigned long q) //扩展欧几里得算法求模逆 { unsigned long i=0,a=1,b=0,c=0,d=1,temp,mid,ni[2]; mid=p; while(mid!=1) { while(p>q) {p=p-q; mid=p;i++;} a=c*(-1)*i+a;b=d*(-1)*i+b; temp=a;a=c;c=temp; temp=b;b=d;d=temp; temp=p;p=q;q=temp; i=0; } ni[0]=c;ni[1]=d; return(ni); } unsigned long momi(unsigned long a,unsigned long b,unsigned long p) //模幂算法 { unsigned long c; c=1; if(a>p) a=a%p; if(b>p) b=b%(p-1); while(b!=0) { while(b%2==0) { b=b/2; a=(a*a)%p; } b=b-1; c=(a*c)%p; } return(c); } void RSAjiami() //RSA加密函数 { unsigned long c1,c2; unsigned long m,n,c; n=prime1*prime2; system("cls"); printf("Please input the message:\n"); scanf("%lu",&m);getchar(); c=momi(m,ee,n); printf("The cipher is:%lu",c); return; } void RSAjiemi() //RSA解密函数 { unsigned long m1,m2,e,d,*ni; unsigned long c,n,m,o; o=(prime1-1)*(prime2-1); n=prime1*prime2; system("cls"); printf("Please input the cipher:\n"); scanf("%lu",&c);getchar(); ni=kzojld(ee,o); d=ni[0]; m=momi(c,d,n); printf("The original message is:%lu",m); return; } void main() { unsigned long m; char cho; printf("Please input the two prime you want to use:\n"); printf("P=");scanf("%lu",&prime1);getchar(); printf("Q=");scanf("%lu",&prime2);getchar(); printf("E=");scanf("%lu",&ee);getchar(); if(prime1<prime2) {m=prime1;prime1=prime2;prime2=m;} while(1) { system("cls"); printf("\t*******RSA密码系统*******\n"); printf("Please select what do you want to do:\n"); printf("1.Encrpt.\n"); printf("2.Decrpt.\n"); printf("3.Exit.\n"); printf("Your choice:"); scanf("%c",&cho);getchar(); switch(cho) { case '1':RSAjiami();break; case '2':RSAjiemi();break; case '3':exit(0); default:printf("Error input.\n");break; } getchar(); } }

游客886 2019-12-02 01:26:26 0 浏览量 回答数 0

新用户福利专场,云服务器ECS低至96.9元/年

新用户福利专场,云服务器ECS低至96.9元/年

问题

用c++编写一个类输出100到200的素数

a123456678 2019-12-01 20:06:53 1738 浏览量 回答数 1

问题

分解素因数的问题,输出总是1,跪求各位大神

a123456678 2019-12-01 19:24:16 660 浏览量 回答数 1

问题

该代码是否正确以检查数字是否为质数?

几许相思几点泪 2019-12-29 18:52:22 1 浏览量 回答数 1

回答

java api里的Object.hashcode(): (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the Java programming language.) 看代码可以知道, hashcode()是一个native函数, 但实际JVM会不会直接用对象地址来做hashcode有待探讨, 因为现代的JVM 堆都是分代管理的, 一个Object很可能在一次 GC后改变其对象地址.而对一个对象来说, 其生命周期内的hashcode是不会变的. Effective Java里提到自定义hashcode()的recipe(实际上在Eclipse里, 右键-> Source -> Generate hashcode and Equals可以代劳).大概例子这样: private String s; private int a; private short b; private Date d; @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + a; result = prime * result + b; result = prime * result + ((d == null) ? 0 : d.hashCode()); result = prime * result + ((s == null) ? 0 : s.hashCode()); return result; }

蛮大人123 2019-12-02 01:57:33 0 浏览量 回答数 0

问题

PHP的Big-O列表

保持可爱mmm 2020-01-15 16:54:12 1 浏览量 回答数 1

问题

返回x的除数的最小质数

is大龙 2020-03-23 17:16:43 0 浏览量 回答数 1

回答

对 n 进行分解质因数,应先找到一个最小的质数 k,然后按下述步骤完成: (1)如果这个质数恰等于 n,则说明分解质因数的过程已经结束,打印出即可。 (2)如果 n<>k,但 n 能被 k 整除,则应打印出 k 的值,并用 n 除以 k 的商,作为新的正整数 n,重复执行第一 步。 (3)如果 n 不能被 k 整除,则用 k+1 作为 k 的值,重复执行第一步。 package cskaoyan; import org.junit.Test; public class cskaoyan4 { @Test public void zhiYinShu() { java.util.Scanner in = new java.util.Scanner(System.in); int number = in.nextInt(); int prime = 2; in.close(); if (number < 2) { return; } else { System.out.print(number + "="); while (number != prime) { if (number % prime == 0) { System.out.print(prime + "*"); number = number / prime; } else { prime = nextPrime(prime); } } System.out.print(number); } } public int nextPrime(int number) { number = number + 1; while (true) { if (isPrime(number)) { return number; } else { number = number + 1; } } } public boolean isPrime(int number) { boolean flag = true; if (number < 2) { flag = false; } else if (number == 2) { flag = true; } else { for (int i = 2; i <= Math.sqrt(number); i++) { if (number % i == 0) { flag = false; break; } } } return flag; } }

珍宝珠 2020-02-13 18:18:20 0 浏览量 回答数 0

回答

在第二个代码段中,第二次输入后您什么也没做。也许您忘记了再次调用forloop?这对我有用,可以按照您的描述进行。以下内容对您有用吗?当然,使用函数对此更优雅。 x = int(input()) for i in range(2,x): if(x % i ==0): print("not Prime") break else : print("Prime") x = int(input()) for i in range(2,x): if(x % i ==0): print("not Prime") break else : print("Prime") 这是我的pycharm中的运行方式 回答来源:stackoverflow

is大龙 2020-03-24 23:33:24 0 浏览量 回答数 0

回答

gram rem (c) W.Buchanan rem Jan 2002 Function check_prime(ByVal val As Long) As Boolean Dim primes primes = Array(1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397) check_prime = False For i = 0 To 78 If (val = primes(i)) Then prime = True End If Next i check_prime = prime End Function Function decrypt(ByVal c, ByVal n, ByVal d As Long) Dim i, g, f As Long On Error GoTo errorhandler If (d Mod 2 = 0) Then g = 1 Else g = c End If For i = 1 To d / 2 f = c * c Mod n g = f * g Mod n Next i decrypt = g Exit Function errorhandler: Select Case Err.Number ' Evaluate error number. Case 6 status.Text = "Calculation overflow, please select smaller values" Case Else status.Text = "Calculation error" End Select End Function Function getD(ByVal e As Long, ByVal PHI As Long) As Long Dim u(3) As Long Dim v(3) As Long Dim q, temp1, temp2, temp3 As Long u(0) = 1 u(1) = 0 u(2) = PHI v(0) = 0 v(1) = 1 v(2) = e While (v(2) <> 0) q = Int(u(2) / v(2)) temp1 = u(0) - q * v(0) temp2 = u(1) - q * v(1) temp3 = u(2) - q * v(2) u(0) = v(0) u(1) = v(1) u(2) = v(2) v(0) = temp1 v(1) = temp2 v(2) = temp3 Wend If (u(1) < 0) Then getD = (u(1) + PHI) Else getD = u(1) End If End Function Function getE(ByVal PHI As Long) As Long Dim great, e As Long great = 0 e = 2 While (great <> 1) e = e + 1 great = get_common_denom(e, PHI) Wend getE = e End Function Function get_common_denom(ByVal e As Long, ByVal PHI As Long) Dim great, temp, a As Long If (e > PHI) Then While (e Mod PHI <> 0) temp = e Mod PHI e = PHI PHI = temp Wend great = PHI Else While (PHI Mod e <> 0) a = PHI Mod e PHI = e e = a Wend great = e End If get_common_denom = great End Function Private Sub show_primes() status.Text = "1" no_primes = 1 For i = 2 To 400 prime = True For j = 2 To (i / 2) If ((i Mod j) = 0) Then prime = False End If Next j If (prime = True) Then no_primes = no_primes + 1 status.Text = status.Text + ", " + Str(i) End If Next i status.Text = status.Text + vbCrLf + "Number of primes found:" + Str(no_primes) End Sub Private Sub Command1_Click() Dim p, q, n, e, PHI, d, m, c As Long p = Text1.Text q = Text2.Text If (check_prime(p) = False) Then status.Text = "p is not a prime or is too large, please re-enter" ElseIf (check_prime(q) = False) Then status.Text = "q is not a prime or is too large, please re-enter" Else n = p * q Text3.Text = n PHI = (p - 1) * (q - 1) e = getE((PHI)) d = getD((e), (PHI)) Text4.Text = PHI Text5.Text = d Text6.Text = e m = Text7.Text c = (m ^ e) Mod n Text8.Text = c m = decrypt(c, n, d) Text9.Text = m Label12.Caption = "Decrypt key =<" + Str(d) + "," + Str(n) + ">" Label13.Caption = "Encrypt key =<" + Str(e) + "," + Str(n) + ">" End If End Sub Private Sub Command2_Click() End End Sub Private Sub Command3_Click() frmBrowser.Show End Sub Private Sub Command4_Click() Call show_primes End Sub

美人迟暮 2019-12-02 01:27:04 0 浏览量 回答数 0

回答

什么是质数:质数(Prime number),又称素数,指在大于1的自然数中,除了1和该数自身外,无法被其他自然数整除的数(也可定义为只有1与该数本身两个因数的数)。——via维基百科简单来说就是,只能除以1和自身的数(需要大于1)就是质数。举个栗子,5这个数,从2开始一直到4,都不能被它整除,只有1和它本身(5)才能被5整除,所以5就是一个典型的质数。那么想计算出一个随机数是不是质数用Python应该怎么写呢?首先第一句话肯定是接受用户输入的数字:n = int(input("please enter the number:"))接着要计算该数是不是质数,那么就要从2开始一直除到该数之前的那个自然数,很明显是一个数字范围:for i in range(2, n):在循环体里面,每次循环当然就是要判断当次除法是否是整除,这里可以使用求模运算,也就是取余,当余数为0时,该数就不是质数:if n % i == 0: print("%d is not a prime number!" % n) break这个break意思就是当该数不是质数时,就跳出整个循环,该数就不是我们要的数字了。那么,所有循环迭代都完成后还没有找出能整除的情况的话,那么可以判断该数就是一个质数,所以:else: print("%d is a prime number!" % n)那么此时,所有代码就写好了,不过为了看起来简单,没有罩一层是否大于1的判断,用户输入的数字默认需要大于1:n = int(input("please enter the number:"))for i in range(2, n): if n % i == 0: print("%d is not a prime number!" % n) breakelse: print("%d is a prime number!" % n)这里要细细品味这段代码,else其实不是和if是一对,而是和for并排的,我们常见的是if…else…或者if…elif…else诸如此类,但其实for也可以和else搭配出现,在这段代码里,当某一次遍历结果余数为0后,break生效,那循环就结束了,那与之成对出现的else代码也就不执行了;当所有遍历结束后没有一次余数为0,那该循环就转到else开始执行,打印输出“该数为质数”。最后我们来随便输2个数字看看功能有没有实现:please enter the number:1111 is a prime number! please enter the number:2121 is not a prime number!

世事皆空 2019-12-02 01:07:38 0 浏览量 回答数 0

问题

如何在python中接受输入和接收输出后再次请求输入

is大龙 2020-03-24 23:33:15 0 浏览量 回答数 1

回答

package cskaoyan; public class cskaoyan45 { public static void main(String[] args) { // TODO Auto-generated method stub java.util.Scanner in = new java.util.Scanner(System.in); long prime = 0; long count = 1; long ret = 0; long i = 9; while (!isPrime(prime)) { System.out.println("请输入一个素数:"); prime = in.nextLong(); } while (i % prime != 0) { i = i * 10 + 9; count++; } ret = i; System.out.println("素数" + prime + "能整除" + count + "个9组成的数" + ret); in.close(); } private static boolean isPrime(long number) { boolean flag = true; if (number < 2) { flag = false; } else if (number < 4) { flag = true; } else if (number % 2 == 0) { flag = false; } else { for (long i = 3; i < Math.sqrt(number) + 1; i++) { if (number % i == 0) { flag = false; break; } } } return flag; } }

珍宝珠 2020-02-13 17:49:24 0 浏览量 回答数 0

回答

我在我的机器上,执行到s = PrimeND()这块是没有错误的。我用的是Python2.7.2, 但你的错误提示,应该是类型转换的问题,查查这块吧。######而且,你写的程序,好像有问题,如果你是想判断你输入的数是个什么 类型的数的话。 ###### IDLE 2.6.6 ==== No Subprocess ==== >>> def isprime(n): flag = 1 for i in xrange(2,int(n**0.5)+1): if n%i==0: flag = 0 break return flag==1 and 'prime' or 'Not prime' >>> isprime(3) 'prime' >>> isprime(45) 'Not prime' >>> isprime(50) 'Not prime' >>> isprime(2) 'prime' >>> isprime(17) 'prime' >>> ######PrimeND方法连个返回值都没有。 所以看不懂:s = PrimeND()  ######没有进行类型转换,需要 n = int(n)###### 引用来自“oschina的ID”的答案 没有进行类型转换,需要 n = int(n) +1 ###### n = input('请任意输入一个自然数') n = int(n) def PrimeND(): for i in range(2,n): if (n % i == 0): print ('您输入的是合数') else: print ('您输入的是素数') if n == 0 and 1 : print ('0和1既不是素数也不是合数') else: PrimeND() 语法对了 但求素数的算法不对吧 ######算法不对的

kun坤 2020-05-29 15:52:31 0 浏览量 回答数 0

回答

#include <stdio.h> #include <string.h> #include <time.h> #include <stdlib.h> #include <math.h> int str2num(char *str) //字符转数字 { int i=0,num=0; for(i=0;i<(int)strlen(str);i++) num=num*10+str[i]-'0'; return num; } float CarmSqrt(float x) //求平方根 系统的太慢,用了别人的 { union { int intPart; float floatPart; }convertor; union { int intPart; float floatPart; }convertor2; convertor.floatPart = x; convertor2.floatPart = x; convertor.intPart = 0x1FBCF800 + (convertor.intPart >> 1); convertor2.intPart = 0x5f3759df - (convertor2.intPart >> 1); return 0.5f*(convertor.floatPart + (x * convertor2.floatPart)); }//可以不用,用sqrt()也可以 int isPrime(int n) //判断是否为素数 { int i=0,k=2; k=(int)CarmSqrt(n); for(i=2;i<=k;i++) { if(n%i==0) break; } if(i>k) return 1; else return 0; } int rnd(int max) //生成随机数 2~max 用来生成e, { //取系统时间做随机数种子 int range,n; int min=2,flag=0; time_t t; double j; range=max-min; t=time(NULL); srand((unsigned)t); n=rand(); j=((double)n/(double)RAND_MAX); n=(int)(j*(double)range); n+=min; return n; } int co_prime(int a ,int b) // 求互质 { int c; do { if(b==1) return 1; c=a%b; a=b; b=c; }while(c!=0); return 0; } void get_d_e(int p,int q) { int r,t,e,d,i=2,k=0; if(isPrime(p)!=1||isPrime(q)!=1) { printf("Invaild Parameters\nshould be PRIME!\n"); printf("Usage:RSA Prime1 Prime2\n"); exit(0); } r=p*q; t=(p-1)*(q-1); e=rnd(t)/10; while(co_prime(t,e)!=1) { e=e+1; } for(d=2;d<t;d++) { if((e*d)%t==1) break; } printf("d=%d e=%d r=%d\n",d,e,r); } void en(int n,int e,int r) { //加密 } void de(int c,int d,int r) { //解密 } void main(int argc,char* argv[]) { int n1,n2; if(argc!=3&&argc!=5) { printf("Invaild Parameters!\n"); printf("Usage: \nRSA -e Express e r\n"); printf("RSA -d Ciphertext d r\n"); printf("RSA Prime1 Prime2\n");//错误输出 exit(0); } else if(argv[1][0]!='-') { n1=str2num(argv[1]); n2=str2num(argv[2]); get_d_e(n1,n2); exit(0); } else if(argv[1][1]=='e') //加密 { n1=str2num(argv[3]); n2=str2num(argv[4]); en(str2num(argv[2]),n1,n2); exit(0); } if(argv[1][1]=='d') //解密 { n1=str2num(argv[3]); n2=str2num(argv[4]); en(str2num(argv[2]),n1,n2); exit(0); } else{ printf("Invaild Parameters!\n"); printf("Usage: \nRSA -e Express e r\n"); printf("RSA -d Ciphertext d r\n"); printf("RSA Prime1 Prime2\n"); exit(0); } }

琴瑟 2019-12-02 01:26:32 0 浏览量 回答数 0

回答

def is_prime2(num): for i in range(2, num): if not (num % i): return False return True def factorization(num): if num < 2: print('This number should be a positive integer greater than one!') return elif is_prime2(num): print("This number must be composite!") return else: lst = [x for x in range(2, num) if is_prime2(x)] str1 = '' while int(num) != 1: for i in lst: if not num % i: str1 += str(i) + '*' num /= i return '*'.join(sorted(str1[:-1].split('*'))) if __name__ == '__main__': print(factorization(90))

june-fu 2020-02-02 13:35:09 0 浏览量 回答数 0

回答

您陷入无限循环,因为如果不满足返回条件,则不更改n的值,因为您可以看到返回条件仅在您的数字x是2的倍数时得到满足改变 : if x % n == 0: x += 1 与: while n <= x: if x % n == 0: return x n += 1 为了优化代码,您可以搜索一个以质数n来除以int(math.sqrt(x)+ 1)的x: import math def smallest_prime_factor(x): """Returns the smallest prime number that is a divisor of x""" # Start checking with 2, then move up one by one n = 2 max_n = int(math.sqrt(x) + 1) while n < max_n: if x % n == 0: return n n += 1 return x 甚至更好的是,您可以使用Eratosthenes筛子快速生成质数并针对x进行测试: # Sieve of Eratosthenes # Code by David Eppstein, UC Irvine, 28 Feb 2002 # http://code.activestate.com/recipes/117119/ def gen_primes(y): """ Generate an infinite sequence of prime numbers. """ # Maps composites to primes witnessing their compositeness. # This is memory efficient, as the sieve is not "run forward" # indefinitely, but only as long as required by the current # number being tested. # D = {} # The running integer that's checked for primeness q = 2 while q < y: if q not in D: # q is a new prime. # Yield it and mark its first multiple that isn't # already marked in previous iterations # yield q D[q * q] = [q] else: # q is composite. D[q] is the list of primes that # divide it. Since we've reached q, we no longer # need it in the map, but we'll mark the next # multiples of its witnesses to prepare for larger # numbers # for p in D[q]: D.setdefault(p + q, []).append(p) del D[q] def smallest_prime_factor(x): """Returns the smallest prime number that is a divisor of x""" # Start checking with 2, then move up one by one return next((i for i in gen_primes(int(math.sqrt(x) + 1)) if x % i == 0), x) 回答来源:stackoverflow

is大龙 2020-03-23 17:16:53 0 浏览量 回答数 0

回答

def is_prime(num): for i in range(2, num): if num % i: continue else: return True return False if __name__ == '__main__': total = 0 for j in range(101, 201): if not is_prime(j): print(j) total += 1 print('The total is %d' % total)

june-fu 2020-02-01 12:38:58 0 浏览量 回答数 0

回答

<script type="text/javascript"> var prime = function(len){ var i,j; var arr = []; for(i = 1; i < len; i++){ for(j=2; j < i; j++){ if(i%j === 0) { break; } } if(i <= j && i !=1){ arr.push(i); } } return arr; }; document.write(prime(100)); </script>

小旋风柴进 2019-12-02 02:15:33 0 浏览量 回答数 0

问题

如何使这个质数递归函数更有效

kun坤 2019-12-27 17:45:51 0 浏览量 回答数 1

回答

我实际上通过增加一个条件使它更有效率: def has_divisors(n, i=2): """ Check if a number is prime or not :param n: Number to check :param i: Increasing value that tries to divide :return: True if prime, False if not """ if n <= 1: return False if i == n: return True if n <= 2 and n > 0: return True if i * i > n: return True if n % i == 0: return False return has_divisors(n, i + 1) 感谢所有试图帮助我的人。

kun坤 2019-12-27 17:45:56 0 浏览量 回答数 0

回答

c++prime 奥本海默,或外文

沉默术士 2019-12-02 01:23:33 0 浏览量 回答数 0

回答

C prime plus 十分经典 从入门到精通 算法导论 在语法基本掌握的基础下可以尝试学习算法

玄学酱 2019-12-02 01:20:41 0 浏览量 回答数 0

回答

@hantsy请求帮助~~这是大香蕉的大号?大号一般指上厕所大便,小号指上厕所小便 恩,这个就是传说中的@小香蕉 他爹 @小香蕉 两种方法: 1.settings.xml添加jdk-1.4profile到 <activeProfiles>中。 2.项目中的pluginRepositories是用来查找mavenPlugin的,不是查找dependencies. 将 <pluginRepositories><pluginRepository><id>prime-repo</id><name>PrimeFacesMavenRepository</name><url>http://repository.primefaces.org</url><releases><enabled>false</enabled></releases></pluginRepository></pluginRepositories> 改成 <repositories><repository><id>prime-repo</id><name>PrimeFacesMavenRepository</name><url>http://repository.primefaces.org</url></repository></repositories> 刚试了一下,还是这样pluginRepositories是我后来见不起作用,然后就改了下尝试的。忘改回去了,我加一下activeProfiles试试 引用来自“hantsy”的答案 两种方法: 1.settings.xml添加jdk-1.4profile到 <activeProfiles>中。 2.项目中的pluginRepositories是用来查找mavenPlugin的,不是查找dependencies. 将 <pluginRepositories><pluginRepository><id>prime-repo</id><name>PrimeFacesMavenRepository</name><url>http://repository.primefaces.org</url><releases><enabled>false</enabled></releases></pluginRepository></pluginRepositories> 改成 <repositories><repository><id>prime-repo</id><name>PrimeFacesMavenRepository</name><url>http://repository.primefaces.org</url></repository></repositories> 加参数-U 我把你的项目down下来用mvncompile跑了一下,可以成功,不知道你要执行哪个命令?--因为我上传的代码中注释掉了all-theme那个jar的依赖运行的命令是mvncleanpackage你还是先看看Sonatype官方网站那几本免费的Maven书吧。 测试一下 是因为 <mirrorOf>*</mirrorOf>吧,这个设置会把所有的请求都镜像到你配置的oschina上面去吧

爱吃鱼的程序员 2020-06-22 13:22:58 0 浏览量 回答数 0

回答

常见SSL算法:RSA、哈希签名算法:SHA256、加密位数:2048 如果在独立服务器配置SSL或者有条件的使用: 推荐SSL算法:ECC、哈希签名算法:SHA256、加密位数:prime256v1

行者武松 2019-12-02 01:26:44 0 浏览量 回答数 0

回答

#include<stdio.h> int main(void) { int a, b; //takes the two input numbers printf("Enter the two numbers\n"); scanf("%d %d", &a, &b); //checks if the entered number is invalid if((a < 0) || (b < 0)) { printf("Invalid numbers"); return -1; } //for each number between a and b, checks either it is prime or not for(int i = a; i <= b; i++) { int count = 0; //counts the number of divisors of the number i with remainder 0 for(int j = 1; j <= i; j++) { if(i % j == 0) { count++; } } //if number of divisors of the i is less than 3, which means the //number is prime, it prints the number if(count < 3) { printf("%d ", i); } } return 0; }

kun坤 2019-12-02 03:23:50 0 浏览量 回答数 0

回答

importjava.util.Iterator; importjava.util.List; classEmp{//emp表映射类 privateintempno; privateStringename; privateStringjob; privatedoublesal; privatedoublecomm; privateEmpmgr;//雇员领导 privateDeptdept;//雇员所属部门 publicEmp(){//无参构造 } publicEmp(intempno,Stringename,Stringjob,doublesal,doublecomm){ this.empno=empno; this.ename=ename; this.job=job; this.sal=sal; this.comm=comm; } //部分setter、getter略 publicvoidsetDept(Deptdept){//设置雇员所在部门 this.dept=dept; } publicDeptgetDept(){//取得雇员所在部门 returnthis.dept; } publicvoidsetMgr(Empmgr){//设置雇员领导 this.mgr=mgr; } publicEmpgetMgr(){//取得雇员领导 returnthis.mgr; } @Override publicinthashCode(){ finalintprime=31; intresult=1; longtemp; temp=Double.doubleToLongBits(comm); result=prime*result+(int)(temp^(temp>>>32)); result=prime*result+((dept==null)?0:dept.hashCode()); result=prime*result+empno; result=prime*result+((ename==null)?0:ename.hashCode()); result=prime*result+((job==null)?0:job.hashCode()); result=prime*result+((mgr==null)?0:mgr.hashCode()); temp=Double.doubleToLongBits(sal); result=prime*result+(int)(temp^(temp>>>32)); returnresult; } @Override publicbooleanequals(Objectobj){ if(this==obj) returntrue; if(obj==null) returnfalse; if(getClass()!=obj.getClass()) returnfalse; Empother=(Emp)obj; if(Double.doubleToLongBits(comm)!=Double .doubleToLongBits(other.comm)) returnfalse; if(dept==null){ if(other.dept!=null) returnfalse; }elseif(!dept.equals(other.dept)) returnfalse; if(empno!=other.empno) returnfalse; if(ename==null){ if(other.ename!=null) returnfalse; }elseif(!ename.equals(other.ename)) returnfalse; if(job==null){ if(other.job!=null) returnfalse; }elseif(!job.equals(other.job)) returnfalse; if(mgr==null){ if(other.mgr!=null) returnfalse; }elseif(!mgr.equals(other.mgr)) returnfalse; if(Double.doubleToLongBits(sal)!=Double.doubleToLongBits(other.sal)) returnfalse; returntrue; } @Override publicStringtoString(){//取得雇员信息 return"雇员编号:"+this.empno+",姓名:"+this.ename+",职位:"+this.job +",工资:"+this.sal+",佣金:"+this.comm; } } classDept{//dept表映射类 privateintdeptno; privateStringdname; privateStringloc; privateList<Emp>emps;//多个雇员 publicDept(){//无参构造 this.emps=newArrayList<Emp>(); } publicDept(intdeptno,Stringdname,Stringloc){ this();//调用无参构造 this.deptno=deptno; this.dname=dname; this.loc=loc; } //部分setter、getter略 publicvoidsetEmps(List<Emp>emps){ this.emps=emps; } publicList<Emp>getEmps(){ returnemps; } @Override publicStringtoString(){//取得部门信息 return"部门编号:"+this.deptno+",部门名称:"+this.dname+",位置:" +this.loc; } } publicclassTestDemo{ publicstaticvoidmain(Stringargs[]){ //1、第一层配置关系 Deptdept=newDept(10,"ACCOUNTING","NewYrok"); Empempa=newEmp(7369,"SMITH","CLERK",800.0,0.0); Empempb=newEmp(7566,"ALLEN","MANAGER",2450.0,0.0); Empempc=newEmp(7839,"KING","PRESIDENT",5000.0,0.0); empa.setMgr(empb);//设置雇员和领导的关系 empb.setMgr(empc);//设置雇员和领导的关系 empa.setDept(dept);//每个雇员属于一个部门 empb.setDept(dept);//每个雇员属于一个部门 empc.setDept(dept);//每个雇员属于一个部门 //每一个部门有多个雇员,通过对象数组表示多个雇员 dept.getEmps().add(empa); dept.getEmps().add(empb); dept.getEmps().add(empc); //2、第二层取得关系 System.out.println(dept); Iterator<Emp>iter=dept.getEmps().iterator(); while(iter.hasNext()){ Empemp=iter.next(); System.out.println(emp); if(emp.getMgr()!=null){//有领导 System.out.println("\t"+emp.getMgr()); } System.out.println("---------------------------------------------------"); } } }

游客pklijor6gytpx 2019-12-02 03:19:50 0 浏览量 回答数 0

回答

  公钥为17。   #include <stdio.h>   #include <math.h>   #include <stdlib.h>   //判断公钥e是否为素数,1成立,0不成立   int prime(int e);   //判断公钥e与(p-1)*(q-1)的最大公约数是否为1,1成立,0不成立   int gcd(int e,int pq);   //判断e*d余(p-1)*(q-1)是否为1,1成立,0不成立   int mod(int e,int d,int pq);   #define SIZE 1024   void main()   {   int p;   int q;   int e;//公钥   int d;//密钥   int pq;//(p-1)*(q-1)   int* eArray=(int*)malloc(sizeof(int)*SIZE);   int i=0;   int size;   printf("请输入素数p和q\n");   scanf("%d%d",&p,&q);   printf("请输入密钥d\n");   scanf("%d",&d);   pq=(p-1)*(q-1);   printf("p=%d\nq=%d\nd=%d(p-1)*(q-1)=%d\n",p,q,d,pq);   for(e=1;e<pq;e=e+2)   {   if(mod(e,d,pq)&&gcd(e,pq))   {   eArray[i]=e;   i++;   }   }   size=i;   printf("在0-%d可能的公钥数量为 %d \n",pq,size);   printf("公钥为:");   for(i=0;i<size;i++)   {   if(i%5==0)   printf("\n");   printf("%d ",eArray[i]);   }   printf("\n");   free(eArray);   }   int mod(int e,int d,int pq)   {   int ed=e*d;   if(ed%pq==1)   return 1;   return 0;   }   int gcd(int e,int pq)   {   if(prime(e)&&pq%e!=0)   return 1;   return 0;   }   int prime(int e)   {   for(int i=2;i<=int(sqrt(e)+1);i++)   {   if(e%i==0&&e!=2)   return 0;   }   return 1;   }

管理贝贝 2019-12-02 01:26:57 0 浏览量 回答数 0
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