[code3119] High-precision practice of large integer Open root

Question Description

Given a positive integer n, the value of the integer part after the opening of the square root of N. The number of bits in n is no more than 1000 bits.

Input

Read in a positive integer n not exceeding 1000 bits.

Output

Output the answers you ask

Input example

17

Output example

4

High-precision open root: Need to use the hand to open the square root of the method, I actually this method will not, is temporarily to the online learning

Online said the method is quite detailed, I am not detailed here, the following direct code:

High-precision templates require high-reduction, high-multiplication, and high-and-low.

1#include <iostream>2#include <algorithm>3#include <cmath>4#include <cstring>5#include <queue>6#include <cstdio>7 using namespacestd;8typedefLong LongLL;9InlineintRead ()Ten { One     intx=0, f=1;CharC=GetChar (); A      while(!isdigit (c)) {if(c=='-') f=-1; c=GetChar ();} -      while(IsDigit (c)) {x=x*Ten+c-'0'; c=GetChar ();} -     returnx*F; the } - Const intmaxn=1010; - structData - { +     intL,V[MAXN]; -Data () {l=1; memset (V,0,sizeof(v));} +Dataoperator= (Constdata&s) A     { atL=s.l;memset (V,0,sizeof(v)); -          for(intI=1; i<=l;i++) v[i]=S.v[i]; -         return* This; -     } -Dataoperator= (Const int&s) -     { inL=1; v[1]=s; -          while(v[l]>9) v[l+1]+=v[l]/Ten, v[l]%=Ten, l++; to         return* This; +     } -Dataoperator- (Constdata&s) the     { *Data c;c.l=l; $          for(intI=1; i<=c.l;i++) c.v[i]=V[i];Panax Notoginseng          for(intI=1; i<=c.l;i++) c.v[i]-=S.v[i]; -          for(intI=1; i<c.l;i++)if(c.v[i]<0) c.v[i]+=Ten, c.v[i+1]--; the          while(!C.V[C.L]) c.l--; +         returnC; A     } theDataoperator* (Const int&s) +     { -Data c;c.l=l; $          for(intI=1; i<=c.l;i++) c.v[i]=v[i]*s; $          for(intI=1; i<c.l;i++)if(c.v[i]>9) c.v[i+1]+= (C.v[i])/Ten, c.v[i]%=Ten; -          while(c.v[c.l]>9) c.v[c.l+1]+= (C.V[C.L])/Ten, c.v[c.l]%=Ten, c.l++; -         returnC; the     } -Dataoperator+ (Const int&s)Wuyi     { theData c=* This; -c.v[1]+=s;intI=1; Wu          while(c.v[i]>9) c.v[i+1]+=c.v[i]/Ten, c.v[i]%=Ten, i++; -C.l=Max (c.l,i); About         returnC; $     } -     BOOL operator<= (Constdata& t)Const  -     { -         if(L!=T.L)returnl<T.L; A          for(inti=l;i;i--)if(V[i]!=t.v[i])returnv[i]<T.v[i]; +         return 1; the     } - }a; $ intnum[510]; the voidScan (Data &s) the { the     CharCH[MAXN]; thescanf"%s", ch+1); S.l=strlen (ch+1); -      for(intI=1; i<=s.l;i++) s.v[i]=ch[s.l-i+1]-'0';  in     return; the }  the voidPrint (data s) { for(inti=s.l;i;i--) printf ("%d", S.v[i]);cout<<Endl;} About voidInit (data s)//set this number from one digit to the left every two bits (for example, 65536 into 6,55,36) the { the     if(s.l%2) the     { +num[0]=S.V[S.L]; -          for(intI=2; i<=s.l;i+=2) num[i/2]=s.v[s.l-i+1]*Ten+s.v[s.l-i];  the     }Bayi     Else  for(intI=1; i<=s.l;i+=2) num[i/2]=s.v[s.l-i+1]*Ten+s.v[s.l-i]; the     return; the }  - data Sqrt (data s) - { the data ans,t,q,p; theans=0; t=0; the     intlen= s.l%2? s.l/2+1: s.l/2;//dividing the number of blocks the      for(intI=0; i<len;i++) -     { theT= (t* -)+Num[i]; thep=ans* -; ans=ans*Ten;//the ANS will be multiplied by 10 in advance, and ans will not be updated if the following loop does not go in again . the          for(intj=9; j>=0; j--)94         { theq= (P+J) *J; the             if(q<=t) the             {98ans=ans+J; Aboutt=t-Q; -                  Break;101             }102         }103     }    104     returnans; the }106 intMain ()107 {108 Scan (a);109 Init (a); the Print (Sqrt (a));111     return 0; the}

[code3119] High-precision practice of large integer Open root