51nod lYK and GCD

Source: Internet
Author: User
Tags gcd

1678 lYK and GCD base time limit: 2 seconds Space limit: 131072 KB score: 80 Difficulty: 5-level algorithm problem

This day, lYK again and gcd on the bar.
It has an n number of columns and it wants to do two things.

1: theai Change to B.
2: Given a number I, ask allgcd (i ,j ) =1    < Span id= "mathjax-span-19" class= "Mrow" > aj Sum of   .

The first line is two numbers n,q (1<=n,q<=100000). The next line of N represents AI (1<=ai<=10^4). Next, the Q line reads a number a (1<=a<=2) in each line first. If A=1, represents the first operation, followed by two numbers I and B. (1<=i<=n,1<=b<=10^4). If b=2, represents the second operation, followed by a number I. (1<=i<=n).
An answer is indicated for each query output line.
Input example
5 31 2 3 4 52 41 3 12 4
Output example

Consider auxiliary array f[i] to represent the sum of all a arrays of a multiple of the subscript I.

For example there are 5 numbers, then f[1]=a[1]+a[2]+a[3]+a[4]+a[5],f[2]=a[2]+a[4],f[3]=a[3],f[4]=a[4],f[5]=a[5].
For each modification, we only need to ask for all the factors of I, and then change the value in the F array that is labeled its factor. For all inquiry operations, find out all the factors of I p1,p2,p3 ... Then the answer is ΣU[PI]*F[PI]. where u is the Mobius function. Total complexity is the sum of the number of factors for I in all operations.

Using the----of the capacity repulsion theorem

Add each number to its approximate numbers first---

And then each time we use the tolerance theorem to find and---the number of the I-biotin

Sum--and the solution to be asked.

#include <cstdio>#include<cmath>#include<vector>#include<cstring>#include<algorithm>using namespacestd; #defineLL Long LongVector<int> sta[200100]; intshu[220000]; intou[ -],ll; intqu[200100],KKP; LL pp[200100]; voidInitintN) {intsu[200100],kp=0; BOOLfa[200100]; memset (FA,true,sizeof(FA));  for(intI=2; i<=n;i++)      {          if(Fa[i]) {SU[KP++]=i; if(i<=sqrt (n)) for(intj=i*i;j<=n;j+=i) fa[j]=false; }      }       for(intI=2; i<=n;i++)      {          intLl=0; intkk=i;  for(intj=0; su[j]*su[j]<=kk;j++)          {              if(kk%su[j]==0) Ou[ll++]=Su[j];  while(kk%su[j]==0) KK/=Su[j]; }          if(kk>1) Ou[ll++]=KK; KKP=0; QU[KKP++]=-1;  for(intj=0; j<ll;j++) {KK=KKP;  for(intk=0; k<kk;k++) QU[KKP++]=qu[k]*ou[j]*-1; }           for(intj=1; j<kkp;j++) Sta[i].push_back (Qu[j]); }  }  intMain () {intn,k; /*freopen ("In.txt", "R", stdin); Freopen ("Wo.txt", "w", stdout);*/scanf ("%d%d",&n,&k);      Init (n); LL s=0, ans; memset (PP,0,sizeof(PP));  for(intI=1; i<=n;i++) {scanf ("%d",&Shu[i]);  for(intj=0; J<sta[i].size (); j + +)          {              if(sta[i][j]>0) Pp[sta[i][j]]+=Shu[i]; Elsepp[-sta[i][j]]+=Shu[i]; } s+=Shu[i]; }      intA,b,c;  while(k--) {scanf ("%d",&c); if(c==1) {scanf ("%d%d",&a,&b);  for(intj=0; J<sta[a].size (); j + +)              {                  if(sta[a][j]>0) Pp[sta[a][j]]-=Shu[a]; Elsepp[-sta[a][j]]-=Shu[a]; } s-=Shu[a]; Shu[a]=b;  for(intj=0; J<sta[a].size (); j + +)              {                  if(sta[a][j]>0) Pp[sta[a][j]]+=Shu[a]; Elsepp[-sta[a][j]]+=Shu[a]; } s+=Shu[a]; }          Else{scanf ("%d",&a); if(a==1) {printf ("%lld\n", s); Continue; } ans=0;  for(intI=0; I<sta[a].size (); i++)              {                  if(sta[a][i]<0) ans-=pp[-Sta[a][i]]; Elseans+=Pp[sta[a][i]]; } ans=s-ans; printf ("%lld\n", ans); }      }      return 0; }  
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This problem is my copy of the http://blog.csdn.net/leibniz_zhang/article/details/52318715 this big guy's = =

51nod lYK and GCD

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