n = 1
temp =
value = 3
while n <= 10000:
for i in temp:
if value%i == 0:
break
if i == temp[-1]:
temp.append(value)
n += 1
value += 1
print(temp)
#include <stdio.h> #include <math.h>int isPrime(int num);int main() { int i,num=3; for(i=1;i!=10001;num+=2) { if(isPrime(num))i++; } printf("%d",num-2); return 0; } int isPrime(int sum) { int i,choose=1; for(i=3;i<=sqrt(sum);i+=2) { if(sum%i==0) { choose=0; break; } } return choose; }
本帖最后由 776667 于 2016-10-8 11:12 编辑
def euler(x):
count = 0
number = 2
while count < x:
if number == 2:
number += 1
count += 1
continue
for i in range(2,number):
if not number%i:
break
else:
count += 1
number += 2
return number - 2
if __name__ == '__main__':
print(euler(10001))
i = 5
j = 2
list1 =
while True:
for k in list1:
if(i%k == 0):
break
if k == list1:
list1.append(i)
j += 1
if j == 10001:
break
i += 2
print(i)
我的20来秒时间，答案104743
import time
import math
start=time.clock()
def isPrime(n):
if n<2:
return False
elif n==2:
return True
else:
m=int(math.sqrt(n))
for i in range(2,m+1):
if n%i==0:
return False
return True
count=0
i=2
while count<10001:
if isPrime(i):
count+=1
i+=1
continue
else:
i+=1
print(i-1)
end=time.clock()
print("耗时"+str(end-start)+"s")
public class The10001Prime {
public static boolean isPrime(int n){
boolean flag = true;
for(int i = 2;i < n;i++){
if(n % i == 0){
flag = false;
break;
}
}
return flag;
}
public static void main(String[] args){
int number = 1;
int j = 0;
for(j = 3;number < 10001;j = j +2){
if(isPrime(j)){
number += 1;
}
}
System.out.println("第10001个质数是：" + (j - 2));
}
}
zxc123qwe 发表于 2016-4-26 13:04
这代码怎么弄上去的，还能粘贴复制显示行数。教教我吧
看这个帖子
http://bbs.fishc.com/forum.php?mod=viewthread&tid=1742&page=1#pid11009
本帖最后由 梦想绘制者 于 2016-11-12 14:11 编辑
# Python 3.5实现求第10001个质数
def isPrime(n):
if n <= 1:
return False
else:
half = int(n ** 0.5) # n // 2
for i in range(2,half + 1):
if n % i == 0:
return False
return True
def nthPrime(nth):
primeCount = 0
num = 1
while 1:
num += 1
if isPrime(num):
primeCount += 1
else:
continue
if primeCount == nth:
return num
nth = 10001
print('第%d个质数为%d'%(nth, nthPrime(nth)))
速度还蛮快的, 1s多就出结果。
>>>
第10001个质数为104743
t = time.clock()
def isprime(n):
if (n%2==0 and n!=2) or n==1: return False
for x in range(3,int(math.sqrt(n))+1,2):
if n%x==0 or n%5==0: return False
else: return True
def euler07(num=10001):
"""第10001个质数是多少？"""
count = 0
n = 0
while count<num:
n += 1
if isprime(n): count+=1
return {count:n}
print(euler07(10001),'time:',time.clock()-t)
{10001: 104743} time: 0.45091703770432073
# encoding:utf-8
from math import sqrt
from time import time
def findNPrimes(number):
if number == 1:
return
elif number == 2:
return
else:
# 初始化一个素数列表
l_pn =
n = 3
# 循环遍历，判断是否为素数
while len(l_pn) < number:
n += 2
# 标志位
ispm = True
sqr_n = int(sqrt(n))
for t in l_pn:
if n % t == 0:
ispm = False
break
if t > sqr_n:
break
if ispm:
l_pn.append(n)
return l_pn
number = int(input('请输入要查找的数值：'))
start = time()
print('第 %d 个素数是: %d' % (number, max(findNPrimes(number))))
print('cost %.3f sec' % (time() - start))
第 10001 个素数是: 104743
cost 0.135 sec
此代码使用matlab编程
Problem7所用时间为1.1085秒
Problem7的答案为104743
%题目7：找出第10001个质数
function Output=Problem7(Input)
if nargin==0
Input=10001;
end
tic
Sum=0;
Start=1;
while (Sum<Input)
Start=Start+1;
Judge=Prime_Judge(Start);
if Judge==1
Sum=Sum+1;
end
end
Output=Start;
toc
disp('此代码使用matlab编程')
disp(['Problem7所用时间为',num2str(toc),'秒'])
disp(['Problem7的答案为',num2str(Output)])
end
%% 子程序
%验证其是否为素数
%若输入为素数则Judge为1，否则为0
function Judge=Prime_Judge(Input)
if Input==2
Judge=1;
else
node=floor(sqrt(Input))+1;
Judge=1;
for ii=2:node
ifmod(Input,ii)==0
Judge=0;
break
else
end
end
end
end
import time
def is_prime(number):
'判断是否为质数'
flag = True
for i in range(2, int(number ** 0.5) + 1):
if number % i == 0:
flag = False
return flag
def find_prime(number=1):
'寻找第number个质数'
prime = 1
i = 1
while i < number:
prime += 2
if is_prime(prime):
i += 1
return prime
start = time.clock()
print('第10001个质数为%d' %find_prime(10001))
end = time.clock()
print('程序执行了%fs。' %(end - start))
执行结果：
第10001个质数为104743
程序执行了1.150654s。
def ifprime(n):
k = 0
if n == 1 or n == 0:
k = 1
elif n % 2 == 0 and n != 2:
k = 1
else:
from math import sqrt
maxi = int(sqrt(n))
for i in range(3,maxi+1,2):
if n % i == 0:
k = 1
break
if k == 0:
return True
if k == 1:
return False
i = 1
j = 1
while True:
j +=2
if ifprime(j):
i +=1
if i== 10001:
print(j)
break
104743 平均运行时间0.2s
/*题目：求第10001个质数*/
#include <stdio.h>
#include <math.h>
int IsPrime(int x)
{
int i, flag;
flag = 1;
for (i=2; i<=(int)sqrt(x); i++)
{
if (x%i == 0)
{
flag = 0;
break;
}
}
return flag;
}
int main()
{
int i = 1;//存放结果
int n = 0;//数目
while (n<10002)
{
i++;
n = IsPrime(i) ? n+1 : n;
}
printf("第10001个质数是： %d \n", i);
return 0;
}
第10001个质数是： 104759
Process returned 0 (0x0) execution time : 0.177 s
这题要是算法不好不知道跑多久，我这效率应该还是不错的
zs=
i=5
def zzz(i):
for x in zs:
if i%x==0:
return 0
return 1
while True:
if zzz(i):
zs.append(i)
i+=2
if len(zs)>10000:
print(zs)
break
== RESTART: C:\Users\ASUS\AppData\Local\Programs\Python\Python35-32\test.py ==
104743
>>>
m=3
n=1
while True:
for i in range(2,m):
if m%i==0:
break
else:
n+=1
if n==10001:
print(m)
break
m+=1
import time
start = time.clock()
list_prime=
num=5
a=2
while 1:
ret=1
for i in range(2,int(num**0.5)+1):
if num%i ==0:
ret=0
break
if ret:
list_prime.append(num)
a+=1
if a==10001:
print(num)
break
num+=2
end = time.clock()
print("read:%f s" %(end - start))
104743
read:0.803580 s
99592938 发表于 2017-3-14 16:07
104743
read:0.803580 s
list_prime其实不用加进来，但发现几乎没有降低运行效率，就写上了，毕竟万一让输出1万个质数呢
本帖最后由 JonTargaryen 于 2017-3-28 14:21 编辑
The 10001st prime is: 104743
#include <stdio.h>
#include <math.h>
int is_prime(long);
int main(void)
{
long num = 3;
int count = 2;
while(count < 10001)
{
num += 2;
if(is_prime(num))
{
count++;
}
}
printf("The %dst prime is: %ld\n", count, num);
return 0;
}
int is_prime(long num)
{
long i;
for(i = 2; i <= sqrt(num); i++)
{
if(!(num % i))
{
return 0;
}
}
return 1;
}
JonTargaryen 发表于 2017-3-28 14:17
The 10001st prime is: 104743
import math
def is_prime(num):
for i in range(2, int(math.sqrt(num)) + 1 ):
if not (num % i):
return 0
return 1
def main():
num = 3
count = 2
while count < 10001:
num += 2
if is_prime(num):
count += 1
print("The 10001st prime is:", num)
if __name__ == "__main__":
main()