本文介绍了检查数字是否是bash中的质数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试编写bash脚本以查找数字是否为质数,但我找不到我的脚本出了什么问题
I am trying to write a bash script to find if a number is prime, but i can't find what is wrong with my script
#!/bin/bash #set -x echo -n "enter a number " read isPrime count=2 x=0 while [ $count -lt $isPrime ]; do if [ `expr $isPrime % $count`-eq 0 ]; then echo "not prime" fi count=`expr $count + 1` done echo " it is prime" #set +x 推荐答案使用 factor 很容易.但是,如果您以某种方式需要脚本,则可以实现类似以下的操作.我不确定这是否是最好的算法,但是这比您的算法有效.
Using factor would be easy. But if you somehow need a script, I would implement something like following. I'm not sure whether this is the best algorithm, but this is way efficient than yours.
function is_prime(){ if [[ $1 -eq 2 ]] || [[ $1 -eq 3 ]]; then return 1 # prime fi if [[ $(($1 % 2)) -eq 0 ]] || [[ $(($1 % 3)) -eq 0 ]]; then return 0 # not a prime fi i=5; w=2 while [[ $((i * i)) -le $1 ]]; do if [[ $(($1 % i)) -eq 0 ]]; then return 0 # not a prime fi i=$((i + w)) w=$((6 - w)) done return 1 # prime } # sample usage is_prime 7 if [[ $? -eq 0 ]]; then echo "not a prime" else echo "it's a prime" fi您可以找到有关所使用算法的说明此处
You can find an explanation about the used algorithm here
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检查数字是否是bash中的质数
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