JAVA BigInteger class use Case 1: To find the integer portion of the square root of a large integer (SGU) __java

Very Simple problem

Time limit:0.25 sec, Memory limit:1 MB

You are given natural number X. Find such maximum integer number that it square
is not greater than X.

Input

Input file contains number X (1≤x≤10^1000).

Output

Write answer in output file.

Sample Input

16

Sample Output

4

It is simple to ask for an integral part of the square root of a positive integer X that may be up to 1000 digits.

Beginners Java, Java BigInteger class to achieve Newton's iterative square root (many library functions do not know, on their own to write slightly).

Import Java.math. * ;
Import java.io. * ;

public class Solution ... {
Convert a string to a positive integer
static int str2int (String str)//0 <= the integer <= 2 ^ 31-1
... {
int ret = 0;
for (int i = 0, j = str.length (); i < J; i + +)
... {
RET = RET * + str.charat (i)-' 0 ';
}
return ret;
}
Convert a positive integer to a string
static String int2str (int i)//0 <= i <= 2 ^ 31-1
... {
StringBuffer sb = new StringBuffer ("0000000000");
int digit = 0;
if (i = = 0)
... {
return "0";
}
while (i > 0)
... {
Digit + +;
int low = i% 10;
I/= 10;
Sb.setcharat (10-digit, (char) (Sb.charat (10-digit) + low);
}
Return sb.substring (10-digit);
}
The integer portion of the square root of the positive integer (this is copied from the Hacker ' s Delight book)
static int isqrt (int x)//0 <= x <= 2 ^ 31-1
... {
int B, m = 0x40000000, y = 0;
while (M!= 0)
... {
b = y | M
Y >>= 1;
if (x >=b)
... {
x = b;
Y |= m;
}
M >>= 2;
}
return y;
}

public static void Main (string[] args) throws IOException
... {
BufferedReader br = new BufferedReader (
New InputStreamReader (system.in)
);
String str = br.readline ();
BigInteger Bi []