Hdoj 2582 F (n) (yy+ search pattern)

Test instructions

GCD (n) =gcd (c[n][1],c[n][2],......, c[n][n-1])
F (n) = GCD (3) +GCD (4) +...+GCD (i) +...+GCD (n).
F (N).

Ideas

At the beginning of the GCD () to eliminate the formula to see, found that gcd (n) seems to be equal to N, and then Python forget the second example found the wrong, and then consider the possible quality factors will make the GCD smaller, hit a table, and then found a total of three cases:
1. If n is a prime number, GCD (n) = n
2. If n is a unique mass factor, GCD (n) = P
3. If n has multiple mass factors, GCD (n) = 1
Then just enumerate the number of the 1e6 number of qualitative factors to be good.

Code

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace STD;#define LL Long Long#define LOWBIT (x) ((x) & ( -X))#define Lson L, Mid, RT << 1#define Rson Mid + 1, R, RT << 1|1#define MP (A, B) Make_pair (A, b)Const intINF =0x3f3f3f3f;Const intMAXN =1e6+7;Const DoubleEPS =1e-8;Const DoublePI =ACOs(-1.0); LL SUM[MAXN];BOOLVIS[MAXN]; vector<int>PrimevoidInit () { for(inti =2; I * I <= MAXN; i++)if(!vis[i]) {Prime.push_back (i); for(intj = i *2; J <= Maxn; J + = i) vis[j] =1; }}intSolveintN) {intResintCNT =0; for(inti =2; I * I <= N; i++)if(n% i = =0) {cnt++, res = i; while(n% i = =0) n/= i;if(CNT >1)return 1; }if(N >1) cnt++;if(CNT >1)return 1;Else returnRes;}intMain () {//freopen ("In.txt", "R", stdin);    //freopen ("OUT.txt", "w", stdout);    intN    Init (); sum[2] =0; for(inti =3; I <= MAXN; i++)//Prime is N, a mass factor is p, and several qualitative factors are 1        if(!vis[i]) sum[i] = sum[i-1] + i;ElseSum[i] = sum[i-1] + solve (i); while(Cin>> N) {cout<< Sum[n] << Endl; }return 0;}

Hdoj 2582 F (n) (yy+ search pattern)