Python实现素数求法：从判断到批量生成攻略

python求某个范围的素数

defis_prime(m):""判断m是否为素数"""foriinrange(2,int(m**(1/2))+1):ifm%i==0:returnFalseeelse:returnTruedefall_prime(m,n):""删除m到n之间的所有素数"""prime=[]foriinrange(m,n+1):ifis_prime(i):primes.append(i)returnprimesprint(all_prime(50,100))

缩进程序为如图所示

python怎么输出100以内素数

您可以使用Python的循环和if决策语句来实现此目的：foriinrange(2,101):flag=Trueforjinrange(2,i):ifi%j==0:flag=Falsebreakiffflag:print(i)扩展：除了使用for循环和if语句，还可以使用Python的itertools模块中的com函数来获取100以内的素数：importitertoolsfromitertoolsimportcompressprimes=[iforiinrange(2,101)ifnot0inmap(lambdax:i%x,range(2,i))]list(compress(range(2,101),primes))

求10-1000之间所有的素数Python

classPrime:res=[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,\383,389,397,401,409,419,421,\431,433,439,443,449,457,461,\463,467,479,487,491,499,503,\509,521,523,541,547,557,563,\569,571,577,587,593,599,601,\607,613,617,619,631,641,643,\647,653,659,661,673,677,683,\691,701,709,719,727,733,739,\743,751,757,761,769,773,787,\797,809,811,821,823,827,829,\839,853,857,859,863,877,881,\883,887,907,911,919,929,937,\941,947,953,967,971,977,983,\991,997]defgetList(self):returnself.res