【高次べき乗型取りの応用】HDU 3609 Up-up
KIDxの問題を解く報告のテーマはとても分かりやすいです:http://acm.hdu.edu.cn/showproblem.php?pid=3609はべき乗の公式を下げます:このスピードは前の10まで並ぶことができます:
#include <iostream>
using namespace std;
#define LL unsigned __int64
#define M 205
int phi[M];
int Euler (int n) // n
{
int i, res = n;
for (i = 2; i * i <= n; i++)
{
if (n % i == 0)
{
res = res - res/i;
while (n % i == 0)
n /= i;
}
}
if (n > 1)
res = res - res/n;
return res;
}
LL qmod (LL a, LL b, int c) //
{
LL res = 1;
for ( ; b; b >>= 1)
{
if (b & 1)
res = res * a % c;
a = a * a % c;
}
return res;
}
LL isok (LL a, LL b, int c) // a^b >=c
{
LL res = 1, i;
for (i = 0; i < b; i++)
{
res *= a;
if (res >= c)
return res;
}
return res;
}
LL upup (LL a, int k, int num) //
{
if (phi[num] == 1) return 1; // , 1,
if (k == 1) return a % phi[num];// a
LL b = upup (a, k-1, num+1); // a b
LL x = isok (a, b, phi[num]);
if (x >= phi[num]) // , , a
return qmod (a % phi[num], b, phi[num]) + phi[num];
else return x; // , a^b, a
}
int main()
{
LL a;
int k;
phi[0] = 100000000;
for (k = 1; k < M; k++)
phi[k] = Euler (phi[k-1]); //
while (~scanf ("%I64u%d", &a, &k))
{
printf ("%I64u
", upup (a, k, 0) % phi[0]);
}
return 0;
}