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  • UVa10023手动开大数平方算法

    题目链接:UVa 10023

     1 import java.math.BigInteger;
     2 import java.util.Scanner;
     3 public class Main {
     4     public static void sqrt(BigInteger bi){
     5         String str;
     6         str=bi.toString();
     7         int m=str.length();
     8         if(m%2!=0)
     9         str="0"+str;
    10         BigInteger a,b,c,d,ans;
    11         b=BigInteger.valueOf(0);
    12         c=BigInteger.valueOf(0);
    13         ans=BigInteger.valueOf(0);
    14         try{
    15             for(int i=0;i<m;i+=2){
    16                 a=b.multiply(new BigInteger("100")).add(new BigInteger(str.substring(i,i+2)));
    17                 for(int j=0;j<10;j++){
    18                     d=c.multiply(new BigInteger("20")).add(BigInteger.valueOf(j+1)).multiply(BigInteger.valueOf(j+1));
    19                     if(d.compareTo(a)>0){
    20                         c=c.multiply(new BigInteger("20")).add(BigInteger.valueOf(j)).multiply(BigInteger.valueOf(j));
    21                         b=a.subtract(c);
    22                         ans=ans.multiply(new BigInteger("10")).add(BigInteger.valueOf(j));
    23                         c=ans;
    24                         
    25                         break;
    26                     }
    27                 }
    28             }
    29         }catch(Exception e){
    30             e.getStackTrace();
    31         }
    32         System.out.println(ans);
    33     }
    34     public static void main(String args[]){
    35         Scanner cin=new Scanner(System.in);
    36         int n;
    37         n=cin.nextInt();
    38         for(int k=0;k<n;k++){
    39             if(k!=0)
    40                 System.out.println();
    41             sqrt(cin.nextBigInteger());
    42         }
    43     }
    44 }

     

     

     

     

     

     

     

     

     

     

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  • 原文地址:https://www.cnblogs.com/mypsq/p/4723855.html
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