UVa 10023 square root: High precision and open square formula

10023-square Root

Time limit:3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=964

The Problem

You are are to determinate X by given Y, from expression

The Input

The the number of test cases, followed by a blank line.

Each test case is the input contains a positive integer Y (1<=y<=101000), with no blanks or leading zeroes in it.

It is guaranteed, this for given Y, X would be always an integer.

Each test case is separated by a.

The Output

For each test case, your program should print X in the same format as Y is given in input.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

1

7206604678144

Sample Output

2684512

Using the open square formula can be broken, Newton's method incredibly will tle. (Java almost tle)

Complete code:

/*0.562s*/#include <cstdio> #include <cstring> const int MAXN = 1010;  
    
Char S[MAXN], RE[MAXN];  
    int big (char s1[], char s2[]) {int len1, len2, I, q;  
    q = 0;  
    while (s1[q] = = ' 0 ') q++;  
    strcpy (S1, S1 + q);  
        if (strlen (s1) = = 0) {s1[0] = ' 0 ';  
    S1[1] = 0;  
    } q = 0;  
    while (s2[q] = = ' 0 ') q++;  
    strcpy (s2, s2 + q);  
        if (strlen (s2) = = 0) {s2[0] = ' 0 ';  
    S2[1] = 0;  
    } len1 = strlen (S1);  
    Len2 = strlen (s2);  
    if (Len1 > Len2) return 1;  
    else if (Len1 < len2) return 0; else {for (i = 0; i < len1 i++) {if (s1[i) > S2[i]) retur  
            n 1;  
        else if (S1[i] < s2[i]) return 0;  
} return 0;  
    } void Mul (char s[], int t, char re[])//Multiply {int left, I, J, K, Len;  
    char c;left = 0;  
    j = 0;  
        for (i = strlen (s)-1; I >= 0; i--) {k = T * (s[i]-"0") + left;  
        Re[j++] = (k% 10) + ' 0 ';  
    left = K/10;  
        while (Left > 0) {re[j++] = (left% 10) + ' 0 ';  
    Left/= 10;  
    } Re[j] = 0;  
    Len = strlen (re);  
        for (i = 0; i < LEN/2 i++) {c = re[i];  
        Re[i] = re[len-1-i];  
    Re[len-1-I] = C;  
} return;  
    } void Sub (char a[], char b[])//minus {int left, len1, Len2, temp, J;  
    Len1 = strlen (a)-1;  
    Len2 = strlen (b)-1;  
    left = 0;  
        while (len2 >= 0) {temp = A[len1]-b[len2] + left;  
            if (Temp < 0) {temp = 10;  
        left =-1;  
        else left = 0;  
        A[LEN1] = temp + ' 0 ';  
        len1--;  
    len2--; while (len1 >= 0) {temp = a[len1]-' 0' + left;  
            if (Temp < 0) {temp = 10;  
        left =-1;  
        else left = 0;  
        A[LEN1] = temp + ' 0 ';  
    len1--;  
    } j = 0;  
    while (a[j] = = ' 0 ') j + +;  
    strcpy (A, A + j);  
        if (strlen (a) = = 0) {a[0] = ' 0 ';  
    A[1] = 0;  
} return;  
    } void Sqr (char s[], char re[])//root {char TEMP[MAXN];  
    Char LEFT[MAXN];  
    Char P[MAXN];  
    int I, J, Len1, Len2, q;  
    Len1 = strlen (s);  
        if (len1% 2 = 0) {Left[0] = s[0];  
        LEFT[1] = s[1];  
        LEFT[2] = 0;  
    j = 2;  
        else {left[0] = s[0];  
        LEFT[1] = 0;  
    j = 1;  
    } re[0] = ' 0 ';  
    RE[1] = 0;  
    q = 0;  
        while (J <= len1) {mul (Re, temp);  
        Len2 = strlen (temp); for (i = 9; I >= 0; i--) {temp[len2-1] = i + ' 0';  
            Mul (temp, I, p);  
        if (!big (p, left));  
        } re[q++] = i + ' 0 ';  
        RE[Q] = 0;  
        Sub (left, p);  
        Len2 = strlen (left);  
        LEFT[LEN2] = S[j];  
        Left[len2 + 1] = s[j + 1];  
        Left[len2 + 2] = 0;  
    J + + 2;  
    int main (void) {int t;  
    scanf ("%d", &t);  
        while (t--) {scanf ("%s", s);  
        SQR (S, re);  
        int i = 0;  
        while (re[i] = = ' 0 ') i++;  
        strcpy (Re, re + i);  
        printf ("%s\n", re);  
    if (t) putchar (' \ n ');  
return 0; }

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

/*2.818s*/
    
import java.io.*;  
Import java.util.*;  
Import java.math.*;  
    
public class Main {  
    static final BigInteger two = new BigInteger ("2");  
    static Scanner cin = new Scanner (new Bufferedinputstream (system.in));  
    
    public static void Main (string[] args) {  
        int n = cin.nextint ();  
        while (n--> 0) {  
            BigInteger y = Cin.nextbiginteger ();  
            BigInteger C = y, temp;  
            do {  
                temp = y;  
                y = Temp.add (c.divide (y)). Divide (two);  
            }  
            while (Y.compareto (temp) = = 1);  
            System.out.println (y);  
            if (n > 0)  
                System.out.println ();  
        }  
    }  
}