POJ2720 Last Digits

嘟嘟嘟

一道题又写了近两个点……

这道题直接暴力快速幂肯定会爆（别想高精），所以还是要用一点数学知识的～

有一个东西叫欧拉降幂公式，就是：

\(x ^ y \equiv x ^ {y \ \ mod \ \ \varphi(p) + \varphi(p)} (mod \ \ p)\)

然后对于那些爆了的答案，就可以用这个公式递归求解。

然而这样还是会\(TLE\)，因此采用打表法：对于已经算过的答案\(f_{b, i}\)，将这个数记录下来，从而减少运算复杂度。

#include<cstdio>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<vector>

#include<stack>

#include<queue>

using namespace std;

#define enter puts("")

#define space putchar(' ')

#define Mem(a, x) memset(a, x, sizeof(a))

#define rg register

typedef long long ll;

typedef double db;

const int INF = 0x3f3f3f3f;

const db eps = 1e-8;

const int maxn = 105;

inline ll read()

{

  ll ans = 0;

  char ch = getchar(), last = ' ';

  while(!isdigit(ch)) last = ch, ch = getchar();

  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();

  if(last == '-') ans = -ans;

  return ans;

}

inline void write(ll x)

{

  if(x < 0) x = -x, putchar('-');

  if(x >= 10) write(x / 10);

  putchar(x % 10 + '0');

}

int b, n, m;  //f[m] = b ^ f[m - 1]

ll f[maxn][maxn], a[10];

ll check(ll b, ll y)

{

  if(y == -1) return -1;

  ll ret = 1;

  for(int i = 1; i <= y; ++i)

    {

      ret *= b;

      if(ret > (ll)1e14) return -1;

    }

  return ret;

}

void init()

{

  a[0] = 1;

  for(int i = 1; i <= 7; ++i) a[i] = a[i - 1] * 10;

  Mem(f, -1);

  for(int i = 0; i < maxn; ++i) f[0][i] = 0, f[1][i] = 1;

  for(int i = 2; i < maxn; ++i)

    {

      f[i][0] = 1;

      for(int j = 1; j < maxn; ++j)

	{

	  f[i][j] = check(i, f[i][j - 1]);

	  if(f[i][j] == -1) break;

	}

    }

}

ll quickpow(ll a, ll b, ll mod)

{

  a %= mod;

  ll ret = 1;

  for(; b; b >>= 1, a = a * a % mod)

    if(b & 1) ret = ret * a % mod;

  return ret;

}

ll phi(int n)

{

  ll ret = n;

  for(int i = 2; i * i <= n; ++i)

    {

      if(n % i == 0)

	{

	  ret = ret / i * (i - 1);

	  while(n % i == 0) n /= i;

	}

    }

  if(n > 1) ret = ret / n * (n - 1);

  return ret;

}

ll calc(int b, int m, int mod)

{

  if(mod == 1) return 0;

  if(!m) return 1;

  if(f[b][m] != -1) return f[b][m] % mod;

  int g = phi(mod);

  return quickpow(b, calc(b, m - 1, g) + g, mod);

}

void print(ll x, int n)

{

  if(!n) return;

  print(x / 10, n - 1);

  putchar(x % 10 + '0');

}

int main()

{

  init();

  while(scanf("%d", &b) && b)

    {

      m = read(); n = read();

      if(f[b][m] == -1) f[b][m] = calc(b, m, a[7]);

      print(f[b][m], n), enter;

    }

  return 0;

}

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