hdu 4549 欧拉定理+各种快速幂

来源：互联网发布：win7 nls数据怎么修复编辑：程序博客网时间：2024/05/22 03:21
与hdu 3221类似......不想多说
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#define MOD 1000000007using namespace std;typedef long long LL;struct mat {    LL m[3][3];    mat ( )    {        memset ( m , 0 , sizeof ( m ) );    }};mat multi ( mat a , mat b , LL mod ){    mat c;    for ( LL i = 1 ; i < 3 ; i++ )        for ( LL j = 1 ; j < 3 ; j++ )            if ( a.m[i][j] )                for ( LL k = 1 ; k < 3 ; k++ )                    c.m[i][k] = (c.m[i][k] + a.m[i][j]*b.m[j][k]%mod )%mod;    return c;    }mat quickMulti ( mat a , LL n , LL mod ){    mat ans;    for ( int i = 1 ; i < 3 ; i++ ) ans.m[i][i] = 1;    while ( n )    {        if ( n&1 ) ans = multi ( ans , a , mod  );        a = multi ( a , a , mod );        n >>= 1;    }    return ans;}LL pow ( LL num , LL n ){    LL ans = 1;    while ( n )    {        if ( n&1 ) ans = ans * num % MOD;        num = num*num%MOD;        n >>= 1;    }    return ans;}LL euler ( LL x ){    LL res = x;    for ( int i = 2 ; i*i<=x ; i++ )    {        if ( x%i ) continue;        res -= res/i;        while ( x%i==0 ) x /= 1;    }     if ( x > 1 ) res -= res/x;    return res;}LL index ( LL n  , LL mod ){    LL  a = 1 , b = 0 ,temp , i ;    for ( i = 1 ; i <= n ; i++ )    {        temp = a + b;        b = a;        a = temp;        if ( a >= mod ) break;    }    if ( i > n ) return a;    mat ma ,mb;    ma.m[1][1] = mb.m[1][1] = mb.m[2][1] = mb.m[1][2]=1;    mb = quickMulti ( mb , n , mod );    ma = multi ( ma , mb , mod );    return ma.m[1][1]%mod + mod;}int main ( ){    LL phi = euler ( MOD );    LL a,b,n;    while ( ~scanf ( "%lld%lld%lld" , &a , &b , &n ) )    {        if ( n == 0 )        {            printf ( "%lld\n" , a%MOD );            continue;        }        if ( n == 1 )        {            printf ( "%lld\n" , b%MOD );            continue;        }        if ( a == 0 || b == 0 )        {            printf ( "0\n" );            continue;        }        LL ans = pow ( a , index(n-2 , phi ))*pow( b , index ( n-1 , phi ))%MOD;         printf ( "%lld\n" , ans );    }}
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