long long 范围内的开根
int Sqrt(int x) { if (x == 0) return 0; double last = 0; double res = 1; while (res != last) { last = res; res = (res + x / res) / 2; } return int(res); }
浮点数的开根
double fun(double x) { if (x == 0) return 0; double last = 0.0; double res = 1.0; while (res != last) { last = res; res = (res + x / res) / 2; } return res; }
java 大数套牛顿迭代
import java.math.BigInteger; import java.math.*; import java.math.BigInteger; import java.util.Scanner; import java.util.*; public class Main { public static void bigSqrt(){ Scanner cin=new Scanner(System.in); String s=cin.next(); BigInteger remain=BigInteger.ZERO; BigInteger odd=BigInteger.ZERO; BigInteger ans=BigInteger.ZERO; // remain=BigInteger.ZERO; // odd=BigInteger.ZERO; // ans=BigInteger.ZERO; int group=0,k=0; if(s.length()%2==1) { group=s.charAt(0)-‘0‘; k=-1; } else { group=(s.charAt(0)-‘0‘)*10+s.charAt(1)-‘0‘; k=0; } for(int j=0;j<(s.length()+1)/2;j++) { if(j!=0) group=(s.charAt(j*2+k)-‘0‘)*10+s.charAt(j*2+k+1)-‘0‘; odd=BigInteger.valueOf(20).multiply(ans).add(BigInteger.ONE); remain=BigInteger.valueOf(100).multiply(remain).add(BigInteger.valueOf(group)); int count=0; while(remain.compareTo(odd)>=0) { count++; remain=remain.subtract(odd); odd=odd.add(BigInteger.valueOf(2)); } ans=ans.multiply(BigInteger.TEN).add(BigInteger.valueOf(count)); } System.out.println(ans); cin.close(); return; } public static void main(String[] args) { Scanner cin=new Scanner(System.in); //int t=cin.nextInt(); bigSqrt(); cin.close(); } }
原文地址:https://www.cnblogs.com/ccut-ry/p/9651982.html
时间: 2024-11-09 11:15:13