(PAT乙级)数素数(Python)
2018-02-17 14:56
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令Pi表示第i个素数。现任给两个正整数M <= N <= 104,请输出PM到PN的所有素数。输入格式:输入在一行中给出M和N,其间以空格分隔。输出格式:输出从PM到PN的所有素数,每10个数字占1行,其间以空格分隔,但行末不得有多余空格。输入样例:
这题也跟前面一道素数题一样,换了几种算法也没能通过第四个测试点,最终也是用c++写的代码通过了,思路比较简单,而素数的算法依然是用了网上那位老哥的算法,原创者也不知道是哪位,如果原作者看到了,请联系我。
python:import sys
x = input().split()
n = int(x[0])
m = int(x[1])
def get(num):#素数算法
if num == 2 or num == 3:
return True
if num%6 != 1 and num%6 != 5:
return False
tps = int(num**0.5)
for i in range(5,tps+1,6):
if num%i == 0 or num%(i+2) == 0:
return False
return True
counter = 0
printer = 0
i = 1
while counter < m:#区间为2到m,不用算出(2,10000)的素数
i = i+1
if get(i):
counter = counter + 1
if counter >= n and counter <= m:
printer = printer + 1
sys.stdout.write(str(i))
else:
continue
if printer%10 == 0:
sys.stdout.write('\n')
elif counter != m:
sys.stdout.write(' ')
c++:#include<stdio.h>
#include<math.h>
#include <iostream>
using namespace std;
bool sushu(int number)
{
if (2 == number||3 == number) {return true;}
if (number %6 != 1&& number %6 != 5) {return false;}
int temp = sqrt(number);
for (int i =5; i<= temp; i += 6)
{
if (0 == number %i || 0 == number %(i+2))
{
return false;
}
}
return true;
}
int main()
{
int n,m;
cin>>n>>m;
int counter = 0,printer = 0;
for ( int i = 2 ; counter != m;++i)
{
if (sushu(i))
{
counter = counter +1;
if (n <= counter && counter <= m) { ++printer; printf("%d",i);}
else {continue;}
if (printer % 10 == 0) {printf("\n");}
else if (counter != m) {printf(" ");}
}
}
return 0;
}
5 27输出样例:
11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103
这题也跟前面一道素数题一样,换了几种算法也没能通过第四个测试点,最终也是用c++写的代码通过了,思路比较简单,而素数的算法依然是用了网上那位老哥的算法,原创者也不知道是哪位,如果原作者看到了,请联系我。
python:import sys
x = input().split()
n = int(x[0])
m = int(x[1])
def get(num):#素数算法
if num == 2 or num == 3:
return True
if num%6 != 1 and num%6 != 5:
return False
tps = int(num**0.5)
for i in range(5,tps+1,6):
if num%i == 0 or num%(i+2) == 0:
return False
return True
counter = 0
printer = 0
i = 1
while counter < m:#区间为2到m,不用算出(2,10000)的素数
i = i+1
if get(i):
counter = counter + 1
if counter >= n and counter <= m:
printer = printer + 1
sys.stdout.write(str(i))
else:
continue
if printer%10 == 0:
sys.stdout.write('\n')
elif counter != m:
sys.stdout.write(' ')
c++:#include<stdio.h>
#include<math.h>
#include <iostream>
using namespace std;
bool sushu(int number)
{
if (2 == number||3 == number) {return true;}
if (number %6 != 1&& number %6 != 5) {return false;}
int temp = sqrt(number);
for (int i =5; i<= temp; i += 6)
{
if (0 == number %i || 0 == number %(i+2))
{
return false;
}
}
return true;
}
int main()
{
int n,m;
cin>>n>>m;
int counter = 0,printer = 0;
for ( int i = 2 ; counter != m;++i)
{
if (sushu(i))
{
counter = counter +1;
if (n <= counter && counter <= m) { ++printer; printf("%d",i);}
else {continue;}
if (printer % 10 == 0) {printf("\n");}
else if (counter != m) {printf(" ");}
}
}
return 0;
}
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