1.蒙特卡洛求圆周率 向区域内随即撒点 当点的数目足够多时,落在圆的点数目与在正方形点数目成正比 即圆的面积和正方形的面积成正比 可以得出计算圆周率的算法 DARTS=100000000 hits=0.0 clock() for i in range(1,DARTS+1): x,y=random(),random() dist=sqrt(x**2+y**2) if dist <=1.0: hits=hits+1 pi=4*(
import math import time scale=30 s,m,=1,2 total,s,n,t=0.0,1,1.0,1.0 print("执行开始".center(scale//2, "-")) start = time.perf_counter() for i in range(scale+1): total+=t n+=2 s=-s t=s/n k=total*4 a = '*' * i b = '.' * (scale - i) c = (i/sc
Crazy TankTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5033 Accepted Submission(s): 1017 Problem DescriptionCrazy Tank was a famous game about ten years ago. Every child liked it. Time fli
Problem G: 圆周率 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 155 Solved: 99[Submit][Status][Web Board] Description YT大学附小举办背诵圆率 PI 的比赛.谁背的正确的位数越多,谁为胜者.很多小学生背的位数很多,但是往往会有少数位置的数是错误的.为了快速加测出错误的圆周率,将圆周率 PI 小数点后的数字求模进行简单验证. 例如:某学生背的圆周率为 3.15,则1510 mod 9
Martyr2项目实现--Number部分的问题求解 (1) Find Pi to Nth Digit Find Pi to Nth Digit 问题描述: Find PI to the Nth Digit – Enter a number and have the program generate PI up to that many decimal places. Keep a limit to how far the program will go. 翻译: 给定一个整数N,让程序生成精确