HDU5780 gcd Euler function

http://acm.hdu.edu.cn/showproblem.php?pid=5780

BC #85 1005

Ideas:

First primitive Simplification:x^gCD(a,b)?? −1

That is, in N, (the number of conventions is the logarithm of i) *x^i-1, where I is 22 greatest common divisor in N. Then the problem can be transformed into pre-processing I, and then summing, note o (n*300) = 1, normal card constants. must also be optimized

Because Ans=∑s[D]∗(x^? D?? −1), s[d]= greatest common divisor is the logarithm of D

< Span class= "Mopen" > < Span class= "Mclose" > we notice s[d] or (the number of conventions is the logarithm of i), that is, the logarithm of Inma within n/i, apparently using Euler to do

< Span class= "Mopen" > < Span class= "Mclose" >s[d]=2* (phi[1]+phi[2]+...+phi[n/d])-1

< Span class= "Mopen" > < Span class= "Mclose" > Note: D is different, but n/d, that is, s[d] may have many of the same, such as 10/6 10/7 10/8 10/9 10/10, so for S[d] the same items, we can use equal formula summation (fast Power + inverse new knowledge)

So we just have to find each paragraph s[d] can be j=n/(n/i), J for the last of the same s[d] subscript

1 //#pragma COMMENT (linker, "/stack:102c000000,102c000000")2#include <iostream>3#include <cstdio>4#include <cstring>5#include <sstream>6#include <string>7#include <algorithm>8#include <list>9#include <map>Ten#include <vector> One#include <queue> A#include <stack> -#include <cmath> -#include <cstdlib> the //#include <conio.h> - using namespacestd; - #definePi ACOs (-1.0) - Const intN = 1e6+Ten; + Const intMOD = 1e9+7; - #defineINF 0x7fffffff +typedefLong LongLL; A  at voidFre () {freopen ("In.txt","R", stdin);} -InlineintRead () {intx=0, f=1;CharCh=getchar (); while(ch>'9'|| ch<'0') {if(ch=='-') f=-1; Ch=getchar ();} while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; Ch=getchar ();}returnx*F;} -  - ll Pow_m (ll x,ll N) - { -LL res=1; in      while(n>0) -     { to         if(N &1) +res= (res*x)%MOD; -x= (x*x)%MOD; theN >>=1; *     } $     returnRes;Panax Notoginseng } -  the intPrime[n],sphi[n]; + intInv[n]; A voidE_fun () { thesphi[1]=1; +      for(intI=2; i<=n;i++){ -         if(!Sphi[i]) { $prime[++prime[0]]=i; $sphi[i]=i-1; -         } -          for(intj=1; j<=prime[0]&&i*prime[j]<=n;j++){ the             if(I%prime[j]) sphi[i*prime[j]]=sphi[i]* (prime[j]-1); -             Elsesphi[i*prime[j]]=sphi[i]*Prime[j];Wuyi         } the     } -      for(intI=1; i<=n;i++) sphi[i]= (sphi[i-1]+sphi[i])%MOD; Wu      -     //make a table to find the inverse element About     //inv[1] = inv[0] = 1; $     //for (int i = 2;i < n;i++) -     //Inv[i] = inv[mod%i]* (mod-mod/i)%mod; - } -  A  + voidEX_GCD (LL a,ll b,ll &d,ll &x,ll &y) { the     if(!b) { -D =A; $x =1; they =0; the     } the     Else { theEX_GCD (b, a%B, D, y, x); -Y-= x* (A/b); in     } the     //return x; the } About  the ll Sn (ll q,ll N) { the     if(q==1)returnN; the LL x,y,d; +EX_GCD (q1, mod,d,x,y); -     return(Pow_m (Q,n)-1) * ((x+mod)%mod)%MOD; the }Bayi  the intMain () { the E_fun (); -     intT; -t=read (); the      while(t--){ the         intX,n; theX=read (), n=read (); theLL ans=0; -          for(intI=1, j;i<=n;i=j+1){ thej=n/(n/i); theLL sd=2*sphi[n/i]-1; theans= (ans + sd* (pow_m (x,i) *sn (x,j-i+1)%mod-(j-i+1))%mod)%MOD;94         } theprintf"%i64d\n", ans); the     } the     return 0;98}

HDU5780 gcd Euler function