Description



Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).


Input



Output



Sample Input



3 2 4 2 10 1000


Sample Output



1

24


高次幂取模，套公式搞定。

FZU 1759 Super A^B mod C_#include

(B>=phi(C))

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
LL n, m, x;
char s[N];

LL phi(LL x)
{
  LL res = 1;
  for (LL i = 2; i*i <= x; i++)
  {
    if (x%i) continue;
    res *= i - 1;
    for (x /= i; !(x%i); res *= i, x /= i);
  }
  return res * max(x - 1, 1LL);
}

void get()
{
  if (strlen(s) <= 10) { sscanf(s, "%lld", &x); return; }
  LL g = phi(m);
  x = 0;
  for (int i = 0; s[i]; i++) x = (x * 10 + s[i] - '0') % g;
  x = x + g;
}

int main()
{
  while (scanf("%lld%s%lld", &n, s, &m) != EOF)
  {
    LL ans = 1;   get();
    for (; x; x >>= 1)
    {
      if (x & 1) (ans *= n) %= m;
      (n *= n) %= m;
    }
    printf("%lld\n", ans % m);
  }
  return 0;
}


也可以直接用十进制快速幂，效率稍低

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define inone(x) scanf("%d", &x);
#define intwo(x,y) scanf("%d%d", &x, &y);
using namespace std;
typedef unsigned long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
int n, m;
char s[N];

int main()
{
  while (scanf("%d%s%d", &n, s, &m) != EOF)
  {
    int ans = 1;
    per(i, strlen(s) - 1, 0)
    {
      int res = n;
      rep(j, 1, 9)
      {
        if (s[i] - '0' == j) ans = 1LL * ans * res % m;
        res = 1LL * res * n % m;
      }
      n = res;
    }
    printf("%d\n", ans);
  }
  return 0;
}