POJ 1091 跳蚤 扩展欧拉函数 or 容斥原理
POJ 1091 好题！
题意：要使得gcd(a1,a2,a3...,an,m) = 1 (1<=a[x] <= m)
容斥原理：首先预处理出m所有的素因子，组合总数就是m^n，P(x) 表示由x个素因子乘积得到的数为gcd的
总的组合数，所以由容斥原理可知ans = m^n - P(1) + P(2) - P(3) + ...
扩展欧拉函数 ： 参考自http://scturtle.is-programmer.com/posts/19402.html
欧拉函数
的概率解释:
1/p看做从1~n中某数有n的因子p的概率,则1-1/p是没有因子p的概率,最前面的n是指1~n的数的个数,由乘法公式可见F的意义
此题设函数F(n,m)为公因子为1的排列的个数,易见当m为质数时 
普遍看,总排列个数为
,取p|m,则1/p为1~m的某数有因子p的概率,
为前n个数都有p因子的概率,即这n+1个数有公因子p的概率
所以由乘法公式见:
这位大牛的解释真的很犀利啊，原本不大清楚欧拉函数具体的意思，这下子大牛把欧拉函数拓展了~~~对欧拉函数的理解深刻了好多，膜拜！
这题我是用高精度写的，好像用long long也能过，是因为数据太弱了，其实必须要用高精度了。。。我还是不改成long long了
容斥原理
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std ;
const int maxn = 200 ;
char s1[maxn] , s2[maxn] , s3[maxn] ;
int max ( int a , int b ) { return a > b ? a : b ; }
struct bign
{
short s[maxn] , len ;
bign () { memset ( s ,0 , sizeof ( s ) ) ; len = 1 ; }
bign operator = ( const char *num)
{
len = strlen ( num ) ;
for ( int i = 0 ; i < len ; i ++ ) s[i] = num[len-i-1] - '0' ;
return *this ;
}
bool operator < ( const bign &b ) const
{
if ( len != b.len ) return len < b.len ;
for ( int i = len - 1 ; i >= 0 ; i -- )
if ( s[i] != b.s[i] ) return s[i] < b.s[i] ;
return false ;
}
bool operator > ( const bign &b ) const { return b < *this ; }
bool operator <= ( const bign &b ) const { return !( b < *this ) ; }
bool operator >= ( const bign &b ) const { return !( *this < b ) ; }
bool operator != ( const bign &b ) const { return ( *this < b ) || ( b < *this ) ; }
bool operator == ( const bign &b ) const { return !( *this < b ) && !( b < *this ) ; }
bign operator = ( int num )
{
char s[maxn] ;
sprintf ( s , "%d" , num ) ;
*this = s ;
return *this ;
}
bign ( const char *num ) { *this = num ; }
bign ( int num ) { *this = num ; }
string str () const
{
string res ;
res = "" ;
for ( int i = 0 ; i < len ; i ++ ) res = ( char ) ( s[i] + '0' ) + res ;
if ( res == "" ) res = '0' ;
return res ;
}
bign operator + ( const bign& b ) const
{
bign c ;
c.len = 0 ;
for ( int i = 0 , g = 0 ; g || i < max ( len , b.len ) ; i ++ )
{
int x = g ;
if ( i < len ) x += s[i] ;
if ( i < b.len ) x += b.s[i] ;
c.s[c.len++] = x % 10 ;
g = x / 10 ;
}
return c ;
}
bign operator += ( const bign& b )
{
*this = *this + b ;
return *this ;
}
bign operator - ( const bign& b ) const
{
bign t1 , t2 ;
bign t3 ;
t3.len = 0 ;
if ( *this < b ) t1 = b , t2 = *this ;
else t1 = *this , t2 = b ;
for ( int i = 0 ; i < t2.len ; i ++ )
{
int g = t1.s[i] - t2.s[i] ;
if ( g < 0 )
{
g += 10 ;
int j = i ;
while ( t1.s[j+1] == 0 ) t1.s[j+1] = 9 , j ++ ;
t1.s[j+1] -- ;
}
t3.s[t3.len++] = g ;
}
for ( int i = t2.len ; i < t1.len ; i ++ ) t3.s[t3.len++] = t1.s[i] ;
for ( int i = t3.len - 1 ; i >= 0 ; i -- )
if ( t3.s[i] == 0 ) t3.len -- ;
else break ;
return t3 ;
}
bign operator -= ( const bign& b )
{
*this = *this - b ;
return *this ;
}
bign operator * ( const bign& b ) const
{
bign d ;
d.len = 0 ;
for ( int i = 0 ; i < len ; i ++ )
{
bign c ;
c.len = 0 ;
for ( int k = 0 ; k < i ; k ++ )
c.s[c.len++] = 0 ;
for ( int j = 0 , g = 0 ; g || j < b.len ; j ++ )
{
if ( j < b.len ) g += s[i]* b.s[j];
c.s[c.len++] = g % 10 ;
g = g/10 ;
}
d += c ;
}
for ( int i = d.len - 1 ; i >= 0 ; i -- )
if ( d.s[i] == 0 ) d.len -- ;
else break ;
if ( d.len == 0 ) d = 0 ;
return d ;
}
bign operator *= ( const bign &b )
{
*this = *this * b ;
return *this ;
}
bign operator / ( const bign &b ) const
{
if ( *this < b ) return 0 ;
bign t1 = *this ;
bign t2 = b ;
bign c = 10 ;
bign ans ;
ans.len = 0 ;
int j ;
while ( t1 > t2 )
{
bign t3 ; t3 = 0 ;
for ( int i = t1.len - 1 ; i >= 0 ; i -- )
{
t3 *= c ; t3.s[0] = t1.s[i] ;
if ( t3 >= t2 ) { j = i ; break ; }
}
bign x , y , z ;
for ( int i = 1 ; i <= 9 ; i ++ )
{
x = i ;
y = i + 1 ;
if ( t2 * i <= t3 && t2 * y > t3 )
{
ans.s[ans.len++] = i ;
break ;
}
}
z = t2 * x ;
z = t3 - z ;
while ( z < t2 && j > 0 )
{
z *= c ;
z.s[0] = t1.s[--j] ;
if ( z < t2 )
ans.s[ans.len++] = 0 ;
}
if ( j >= 0 )
{
y.len = 0 ;
for ( int i = 0 ; i < j ; i ++ )
{
z *= c ;
y.s[y.len++] = t1.s[i] ;
}
z += y ;
}
t1 = z ;
}
reverse ( ans.s , ans.s + ans.len ) ;
return ans ;
}
};
istream& operator >> ( istream &in , bign& x )
{
string s ;
in >> s ;
x = s.c_str() ;
return in ;
}
ostream& operator << ( ostream &out , const bign &x )
{
out << x.str() ;
return out ;
}
int f[111],d[111], m ,n, top; // top 表示总的素因子数
bign temp , change;
void rongchi(int now,int num,int tot) // now 当前处理到的素因子 num 已经有的素因子数 tot表示总共需要的素因子数
{
int i;
if(num == tot)
{
int nn = m;
for(i = 0;i < tot; i++)
nn = nn/d[i];
bign mm = nn;
for(i = 0;i < n-1 ;i++)
mm = mm*nn;
change = change+mm;
}
else
for(i = now;i < top ;i++)
{
d[num] = f[i];
rongchi(i+1,num+1,tot);
}
}
int main()
{
int i;
int tot = 0;
scanf("%d%d",&n,&m);
int cur = m;
for(i = 2;i*i <= m ;i ++)
if(m%i == 0)
{
f[tot++] = i; // 处理出素因子
while(m%i==0)
m /= i;
}
if(m != 1)
f[tot++] = m;
top = tot;
m = cur;
bign ans = m;
for(i = 0;i < n-1; i++)
ans = ans*m;
temp = ans;
for(i = 0;i < top; i++) // 就是筛选素因子 其实就可以直接二进制压缩状态来做，以前做的就不改了
{
change = 0;
rongchi(0,0,i+1);
if(i&1)
ans = ans+change;
else
ans = ans-change;
}
cout<<ans<<endl;
}
扩展欧拉函数
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std ;
const int maxn = 200 ;
char s1[maxn] , s2[maxn] , s3[maxn] ;
int max ( int a , int b ) { return a > b ? a : b ; }
struct bign
{
short s[maxn] , len ;
bign () { memset ( s ,0 , sizeof ( s ) ) ; len = 1 ; }
bign operator = ( const char *num)
{
len = strlen ( num ) ;
for ( int i = 0 ; i < len ; i ++ ) s[i] = num[len-i-1] - '0' ;
return *this ;
}
bool operator < ( const bign &b ) const
{
if ( len != b.len ) return len < b.len ;
for ( int i = len - 1 ; i >= 0 ; i -- )
if ( s[i] != b.s[i] ) return s[i] < b.s[i] ;
return false ;
}
bool operator > ( const bign &b ) const { return b < *this ; }
bool operator <= ( const bign &b ) const { return !( b < *this ) ; }
bool operator >= ( const bign &b ) const { return !( *this < b ) ; }
bool operator != ( const bign &b ) const { return ( *this < b ) || ( b < *this ) ; }
bool operator == ( const bign &b ) const { return !( *this < b ) && !( b < *this ) ; }
bign operator = ( int num )
{
char s[maxn] ;
sprintf ( s , "%d" , num ) ;
*this = s ;
return *this ;
}
bign ( const char *num ) { *this = num ; }
bign ( int num ) { *this = num ; }
string str () const
{
string res ;
res = "" ;
for ( int i = 0 ; i < len ; i ++ ) res = ( char ) ( s[i] + '0' ) + res ;
if ( res == "" ) res = '0' ;
return res ;
}
bign operator + ( const bign& b ) const
{
bign c ;
c.len = 0 ;
for ( int i = 0 , g = 0 ; g || i < max ( len , b.len ) ; i ++ )
{
int x = g ;
if ( i < len ) x += s[i] ;
if ( i < b.len ) x += b.s[i] ;
c.s[c.len++] = x % 10 ;
g = x / 10 ;
}
return c ;
}
bign operator += ( const bign& b )
{
*this = *this + b ;
return *this ;
}
bign operator - ( const bign& b ) const
{
bign t1 , t2 ;
bign t3 ;
t3.len = 0 ;
if ( *this < b ) t1 = b , t2 = *this ;
else t1 = *this , t2 = b ;
for ( int i = 0 ; i < t2.len ; i ++ )
{
int g = t1.s[i] - t2.s[i] ;
if ( g < 0 )
{
g += 10 ;
int j = i ;
while ( t1.s[j+1] == 0 ) t1.s[j+1] = 9 , j ++ ;
t1.s[j+1] -- ;
}
t3.s[t3.len++] = g ;
}
for ( int i = t2.len ; i < t1.len ; i ++ ) t3.s[t3.len++] = t1.s[i] ;
for ( int i = t3.len - 1 ; i >= 0 ; i -- )
if ( t3.s[i] == 0 ) t3.len -- ;
else break ;
return t3 ;
}
bign operator -= ( const bign& b )
{
*this = *this - b ;
return *this ;
}
bign operator * ( const bign& b ) const
{
bign d ;
d.len = 0 ;
for ( int i = 0 ; i < len ; i ++ )
{
bign c ;
c.len = 0 ;
for ( int k = 0 ; k < i ; k ++ )
c.s[c.len++] = 0 ;
for ( int j = 0 , g = 0 ; g || j < b.len ; j ++ )
{
if ( j < b.len ) g += s[i]* b.s[j];
c.s[c.len++] = g % 10 ;
g = g/10 ;
}
d += c ;
}
for ( int i = d.len - 1 ; i >= 0 ; i -- )
if ( d.s[i] == 0 ) d.len -- ;
else break ;
if ( d.len == 0 ) d = 0 ;
return d ;
}
bign operator *= ( const bign &b )
{
*this = *this * b ;
return *this ;
}
bign operator / ( const bign &b ) const
{
if ( *this < b ) return 0 ;
bign t1 = *this ;
bign t2 = b ;
bign c = 10 ;
bign ans ;
ans.len = 0 ;
int j ;
while ( t1 > t2 )
{
bign t3 ; t3 = 0 ;
for ( int i = t1.len - 1 ; i >= 0 ; i -- )
{
t3 *= c ; t3.s[0] = t1.s[i] ;
if ( t3 >= t2 ) { j = i ; break ; }
}
bign x , y , z ;
for ( int i = 1 ; i <= 9 ; i ++ )
{
x = i ;
y = i + 1 ;
if ( t2 * i <= t3 && t2 * y > t3 )
{
ans.s[ans.len++] = i ;
break ;
}
}
z = t2 * x ;
z = t3 - z ;
while ( z < t2 && j > 0 )
{
z *= c ;
z.s[0] = t1.s[--j] ;
if ( z < t2 )
ans.s[ans.len++] = 0 ;
}
if ( j >= 0 )
{
y.len = 0 ;
for ( int i = 0 ; i < j ; i ++ )
{
z *= c ;
y.s[y.len++] = t1.s[i] ;
}
z += y ;
}
t1 = z ;
}
reverse ( ans.s , ans.s + ans.len ) ;
return ans ;
}
};
istream& operator >> ( istream &in , bign& x )
{
string s ;
in >> s ;
x = s.c_str() ;
return in ;
}
ostream& operator << ( ostream &out , const bign &x )
{
out << x.str() ;
return out ;
}
int f[11111];
int main()
{
int n,m,i;
int tot = 0;
scanf("%d%d",&n,&m);
int cur = m;
for(i = 2;i*i <= m ;i ++)
if(m%i == 0)
{
f[tot++] = i;
while(m%i==0)
m /= i;
}
if(m != 1)
f[tot++] = m;
m = cur;
bign ans = m;
for(i = 0;i < n-1; i++)
ans = ans*m;
//cout<<ans<<endl;
for(i = 0;i < tot; i++)
{
bign d = f[i];
for(int j = 0;j < n-1; j++)
d = d*f[i];
ans = ans*(d-1);
ans = ans/d;
}
cout<<ans<<endl;
}