Bestcode #6-1003 HDU 4983 goffi and GCD [Euler's function]

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4983

Give N and K to find the conditions that meet

Gcd (n? A, n) × gcd (n? B, n) = NK .

A and B.

First, we can obtain the gcd (n, x) <n, so when K is greater than 2, there is no solution. When k is equal to 2, A = n, B = n to get n ^ 2, only one group.

N = 1.

This was the first time I thought about this question. Divide a number into a prime factor.

/* ===================================================== === The prime factor is decomposed into a positive integer, input: N output: Tot number of different prime factors a [] represents the complexity of the number of I prime factors: O (SQRT (n )) ========================================================== = */# define maxn 1111int A [maxn], B [maxn], TOT = 0; void factor (int n, int A [], int B [], Int & ToT) {int TMP, now; TMP = (INT) (SQRT (n) + 1); Tot = 0; now = N; For (INT I = 2; I <= TMP; I ++) if (now % I = 0) {A [tot] = I, B [tot ] = 0; while (now % I = 0) {B [tot] ++; now/= I;} tot ++;} If (now! = 1) A [tot] = now, B [tot ++] = 1 ;}

For example, 800 can be divided into 2 ^ 5*5 ^ 2, and the approximate number can be obtained based on the obtained quality factor:

1	800
2	400
4	200
8	100
16	50
32	25
5	160
10	80
20	40

40	20
80	10
160	5
25	32
50	16
100	8
200	4
400	2
800	1

For each appointment: for example ~ In 800, the number of dikes with the 800 public divisor 2 is: Euler (800/2 = 400,

For example, in this group: 5*160 = 800 in 1 ~ In 800, the number of public approx. 5 with 800 is Euler (800/5 = 160), ranging from 1 ~ In 800, the number of public appointments with 800 is 160 Euler (800/160 = 5.

Therefore, the total number in the 5*160 group is Erler (5) * Euler (160.

Then I encountered a problem. After finding out its prime factor, I didn't know how to find all the dikes, and then I didn't know how to do it. Then I found that it could be directly violent and SQRT (n) complexity. Time is acceptable.

Note that whether N is equal to 1. If n is equal to 1, 1 is output directly.

#include<stdio.h>#include<algorithm>#include<iostream>#include<vector>#include<string.h>#define LL long longusing namespace std;LL n,k;LL euler(LL x) {    LL res = x;    for (LL i = 2; i <= x / i; i++) if (x % i == 0) {        res = res / i * (i - 1);        while(x % i == 0) x /= i;    }    if (x > 1) res = res / x * (x - 1);    return res;}const LL mod=1000000007;int main(){    while(scanf("%I64d%I64d",&n,&k)!=EOF)    {       if(n==1) {printf("1\n");  continue;}       if(k>2)  {printf("0\n");  continue;}       if(k==2) {printf("1\n");  continue;}       else       {           LL ans=0;           for(LL i=1;i*i<=n;++i)           {               if(n%i==0)               {                   LL x=n/i;                   if(i*i==n) ans+=euler(x)*euler(i);                   else ans+=euler(x)*euler(i)*2;                   ans%=mod;               }           }           printf("%I64d\n",ans);       }    }    return 0;}

I wanted to directly create a 1000000000/2 table and found that the table was too big to be typed ......

Later, let's look at the problem solution and find a map table used by a person. It's awesome to think about it. The time complexity is halved.

map<int,int> mp;int phi(int n){  int i,mul,nn;  if (n==1) return 1;  if (mp.find(n)!=mp.end()) return mp[n];  for (i=2;i*i<=n;i++)    if (n%i==0)    {      nn=n;      nn/=i;      mul=(i-1);      while (nn%i==0)      {        nn/=i;        mul*=i;      }      mp[n]=phi(nn)*mul;      return phi(nn)*mul;    }  mp[n]=n-1;  return n-1;}

Note that there must be an operation in the main function:

mp.clear();

Gcd (n? A, n) × gcd (n? B, n) = NK .

Bestcode #6-1003 HDU 4983 goffi and GCD [Euler's function]