SGU111 大数开方 Evolution

Problem: You are given natural number X. Find such maximum integer number that it square is not greater than X.(X.length<1000)

解法：想用java水过但是跪了。上网查到了笔算开方，不过另一段短小的程序却更神奇。完全不会= 。=  于是收录在这里了。

Solution: A small piece of code can solve it but I don't know why.  _(:3」∠)_

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int l;
int work(int o,char *O,int I) {
char c, *D=O ;
if (o>0) {
for (l=0;D[l];D[l++]-=10) {
D[l++] -= 120;
D[l] -= 110;
while (!work(0,O,l)) D[l] += 20;
putchar((D[l]+1032)/20);
}
putchar(10);
}
else {
c = o+(D[I]+82)%10-(I>l/2)*(D[I-l+I]+72)/10-9;
D[I] += I<0?0:!(o=work(c/10,O,I-1))*((c+999)%10-(D[I]+92)%10);
}
return o;
}
int main() {
char s[1200];
s[0]='0';
scanf("%s",s+1);
if (strlen(s)%2 == 1) work(2,s+1,0);
else work(2,s,0);
return 0;
}