What is N?
Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=4335
Mean:
给你三个数b、P、M,让你求有多少个n满足下式。
analyse:
看到数据被吓到了,没半点思路,后来看了解题报告,方法竟然是暴力!
当然暴力是有条件的。
有这样一个公式:
A^x = A^(x % Phi(C) + Phi(C)) (mod C) (x>=Phi(C))
这个公式的具体证明原来在aekdycoin的百度空间有,但是随着百度空间被转移(百度作死,流失了好多优质的文章==),这篇文章的完整版也流失了。
我们就当这个公式是定理吧!
当n!<Phi(C)时,此时我们暴力解决就可。
当n!大于phi(P)的时候,就需要用上面的降幂公式了。
方法还是暴力,n!%phi(p)会出现0,这是必然的,至少n>=phi(p)为0,
那么(n+1)!%phi(p)也为0,这便出现了重复,转变为n^(phi(p))%p==b的问题了。
固定了指数,根据鸽巢原理,余数是循环的,那么只要找出p个的结果,之后通过循环节求解便可以了。
Trick:当P为1的时候,b为0,这时候答案是m+1,不过m可能为2^64-1,如果加1的话就会溢出,巨坑。
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-25-23.41
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef __int64( LL);
typedef unsigned __int64( ULL);
const double eps( 1e-8);
LL get_eular( LL m)
{
LL ret = 1;
for( LL i = 2; i * i <= m; i ++)
if( m % i == 0)
{
ret *= i - 1;
m /= i;
while( m % i == 0)
{
m /= i;
ret *= i;
}
}
if( m > 1) ret *= m - 1;
return ret;
}
long long Quickpow( long long a , long long b , long long m)
{
long long ans = 1;
while(b)
{
if(b & 1) { ans =( ans * a) % m ,b --; }
b /= 2 , a = a * a % m;
}
return ans;
}
LL b ,p , m , ring [ 100010 ];
int main()
{
int t , Cas = 0;
scanf( "%d" , & t);
while( t --)
{
scanf( "%I64u %I64u %I64u" , &b , &p , & m);
if(p == 1)
{
if( m == 18446744073709551615ULL)
printf( "18446744073709551616 \n ");
else
printf( "%I64u \n " , m + 1);
continue;
}
LL i = 0 , phi = get_eular(p ), fac = 1 , ans = 0;
for( i = 0; i <= m && fac <= phi; i ++)
{
if( Quickpow( i , fac ,p) ==b)
ans ++;
fac *= i + 1;
}
fac = fac % phi;
for(; i <= m && fac; i ++)
{
if( Quickpow( i , fac + phi ,p) ==b)
ans ++;
fac =( fac *( i + 1)) % phi;
}
if( i <= m)
{
LL cnt = 0;
for( int j = 0; j <p; j ++)
{
ring [ j ] = Quickpow( i + j , phi ,p);
if( ring [ j ] ==b)
cnt ++;
}
LL idx =( m - i + 1) /p;
ans += cnt * idx;
LL remain =( m - i + 1) %p;
for( int j = 0; j < remain; j ++)
if( ring [ j ] ==b)
ans ++;
}
printf( "Case #%d: %I64u \n " , ++ Cas , ans);
}
return 0;
} /
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-25-23.41
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef __int64( LL);
typedef unsigned __int64( ULL);
const double eps( 1e-8);
LL get_eular( LL m)
{
LL ret = 1;
for( LL i = 2; i * i <= m; i ++)
if( m % i == 0)
{
ret *= i - 1;
m /= i;
while( m % i == 0)
{
m /= i;
ret *= i;
}
}
if( m > 1) ret *= m - 1;
return ret;
}
long long Quickpow( long long a , long long b , long long m)
{
long long ans = 1;
while(b)
{
if(b & 1) { ans =( ans * a) % m ,b --; }
b /= 2 , a = a * a % m;
}
return ans;
}
LL b ,p , m , ring [ 100010 ];
int main()
{
int t , Cas = 0;
scanf( "%d" , & t);
while( t --)
{
scanf( "%I64u %I64u %I64u" , &b , &p , & m);
if(p == 1)
{
if( m == 18446744073709551615ULL)
printf( "18446744073709551616 \n ");
else
printf( "%I64u \n " , m + 1);
continue;
}
LL i = 0 , phi = get_eular(p ), fac = 1 , ans = 0;
for( i = 0; i <= m && fac <= phi; i ++)
{
if( Quickpow( i , fac ,p) ==b)
ans ++;
fac *= i + 1;
}
fac = fac % phi;
for(; i <= m && fac; i ++)
{
if( Quickpow( i , fac + phi ,p) ==b)
ans ++;
fac =( fac *( i + 1)) % phi;
}
if( i <= m)
{
LL cnt = 0;
for( int j = 0; j <p; j ++)
{
ring [ j ] = Quickpow( i + j , phi ,p);
if( ring [ j ] ==b)
cnt ++;
}
LL idx =( m - i + 1) /p;
ans += cnt * idx;
LL remain =( m - i + 1) %p;
for( int j = 0; j < remain; j ++)
if( ring [ j ] ==b)
ans ++;
}
printf( "Case #%d: %I64u \n " , ++ Cas , ans);
}
return 0;
} /