
求助代碼查錯(cuò)批處理實(shí)現(xiàn)Leibniz公式計(jì)算圓周率最后由 pcl_test 于 -5-22 20:29根據(jù)公式π=4*[1+1/3-1/5+1/7……+1/(2*2n+1)-1/(2*2n+3)]計(jì)算π
新手請(qǐng)教錯(cuò)在哪兒。nclick="copycode($('code0'));">復(fù)制代碼
- @echo off
- set /a k=0
- for /l %%i in (0,2,10000) do (
- set /a aa=2*%%i+1
- set /a cc=2*%%i+3
- set /a kk=kk+1/aa-1/cc
- set /a pi=4*kk
- echo %pi%
- )
- pause

