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;
}