Square root digital expansion
It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.
The square root of two is 1.41421356237309504880…, and the digital sum of the first one hundred decimal digits is 475.
For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.
众所周知，如果一个自然数的平方根不是整数，那么就一定是无理数。这样的平方根的小数部分是无限不循环的。
2的平方根为1.41421356237309504880…，它的小数点后一百位数字的和是475。
对于前一百个自然数，求所有无理数平方根小数点后一百位数字的总和。
解题
问题：如何求无理数的一百位小数？这真是无理取闹
在上面给的博客中给了一个很好的方法
对于 数 n 我们需要去根号n，如下很有意思的规律
def Suqareroot(n,digits): limit = 10**(digits+ 1) a = 5*n b = 5 while b < limit: if a>= b: a -= b b +=10 else: a *= 100 b = int(b/10) * 100 + 5 return int(b/100)
说明下：
1.题目让求的是小数点前100位的值，包括整数位
2.上面算法只有最后b/100 是根号n的近似解，这里是去小数点的，只有为什么不是b表示不理解
JAVA
package Level3; import java.math.BigInteger; import java.util.ArrayList; public class PE080{ void run(){ int j = 1; int res = 0; for(int i=1;i<=100;i++){ if(j*j==i){ j++; continue; } res += Int_Sum(Squareroot(i,100)); } System.out.println(res); } private Integer Int_Sum(BigInteger b){ int res = 0; String str = b.toString(); for(int i=0;i<str.length() ;i++){ res += str.charAt(i) - ‘0‘; } return res; } private BigInteger Squareroot(int n,int digits){ // 定义上界 BigInteger limit = new BigInteger("10").pow(digits+1); BigInteger five = new BigInteger("5"); BigInteger ten = new BigInteger("10"); BigInteger hunderd = new BigInteger("100"); BigInteger a = new BigInteger(n+"").multiply(five); BigInteger b = five; while( b.compareTo(limit) < 0){ if(a.compareTo(b) >=0){ a = a.subtract(b); b = b.add(ten); }else{ a = a.multiply(hunderd); b = b.divide(ten).multiply(hunderd).add(five); } } return b.divide(hunderd); } public static void main(String[] args){ long t0 = System.currentTimeMillis(); new PE080().run(); long t1 = System.currentTimeMillis(); long t = t1 - t0; System.out.println("running time="+t/1000+"s"+t%1000+"ms"); } }
40886
running time=0s34ms
Python
import time ; def Suqareroot(n,digits): limit = 10**(digits+1) a = 5*n b = 5 while b < limit: if a>= b: a -= b b +=10 else: a *= 100 b = int(b/10) * 100 + 5 return int(b/100) def Int_Sum(n): res = sum(map(lambda x:int(x),unicode(n))) return res if __name__==‘__main__‘: t0 = time.time() limit = 1000000 result = 0 j = 1 for i in range(1,101): if j*j == i: j+=1 continue result += Int_Sum(Suqareroot(i,100)) print result t1 = time.time() print "running time=",(t1-t0),"s" # 40886 # running time= 0.039999961853 s
这样的 程序好无节操
from decimal import Decimal,getcontext getcontext().prec=102 N = set(range(2,100)) - set([4,9,16,25,36,49,64,81]) s = 0 for n in N: d = Decimal(n).sqrt() s += sum([int(i) for i in str(d).replace(".","")[:100]]) print(s)
Project Euler 80：Square root digital expansion 平方根数字展开
原文：http://www.cnblogs.com/theskulls/p/4986652.html