ACM_快速幂

11的二进制为1011，即11== 2º×1+2¹×1 +2²×0 +2³×1
快速幂：a¹¹= a^1*a^2*a^8

//快速幂 
#include<iostream>
using namespace std;
typedef long long ll;

ll pow_quick (ll a,ll b){
    ll ans=1,base=a;
    while (b!=0)
    {
        if (b&1==1)//末位为1（否则为0） 
            ans*=base;       
        base*=base;
        b>>=1;
    }
    return ans;
}

int main()//计算2^11
{
    ll a=2,b=11,ans;
    ans=pow_quick(a,b);
    cout<<ans<<endl;
} 

关于base*=base这一步：因为 base*base==base2，下一步再乘，就是base2*base2==base4，然后同理 base4*base4=base8，由此可以做到base–>base2–>base4–>base8–>base16–>base32…….指数正是 2^i ，再看上面的例子，a¹¹= a1*a2*a8，这三项就可以完美解决。

//快速幂取模
#include<iostream>
using namespace std;

typedef long long ll;
#define mod 1000000007
//对底数和每个乘法取模
ll pow_mod (ll a,ll b){
    ll ans=1,base=a;
    a%=mod;
    while (b!=0)
    {
        if (b&1==1)
            ans=(ans*base)%mod;       
        base=(base*base)%mod;
        b>>=1;
    }
    return ans%=mod;
}

int main()
{
    ll n,m,ans; 
    cin>>n>>m;
    //对底数和每个乘法取模
    m%=mod;
    //计算m^n−m∗(m−1)^(n−1)
    ans=pow_mod(m,n)-(m*pow_mod(m-1,n-1))%mod;
    //（结果+mod）然后再取模，否则可能有负值
    cout<<(ans+mod)%mod<<endl;
}

附：超大数快速幂。
http://acm.fzu.edu.cn/problem.php?pid=1759
Problem 1759 Super A^B mod C

Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4 2 10 1000

Sample Output

1 24

Source

FZU 2009 Summer Training IV–Number Theory

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include<string>
#include<cmath>
using namespace std;
typedef __int64 ll;
int phi(int x)
{
  int i,j;
  int num = x;
  for(i = 2; i*i <= x; i++)
  {
    if(x % i == 0)
    {
      num = (num/i)*(i-1);
      while(x % i == 0)
      {
        x = x / i;
      }
    }
  }
  if(x != 1) num = (num/x)*(x-1);
  return num;
}
ll quickpow(ll m,ll n,ll k)
{
  ll ans=1;
  while(n)
  {
    if(n&1) ans=(ans*m)%k;
    n=(n>>1);
    m=(m*m)%k;
  }
  return ans;
}
char tb[1000015];
int main()
{
  ll a,nb;
  int c;
  while(scanf("%I64d%s%d",&a,tb,&c)!=EOF)
  {
    int PHI=phi(c);
    ll res=0;
    for(int i=0;tb[i];i++)
    {
      res=(res*10+tb[i]-'0');
      if(res>c)break;
    }
    if(res<=PHI)
    {
      printf("%I64d\n",quickpow(a,res,c));
    }
    else
    {
      res=0;
      for(int i=0;tb[i];i++)
      {
        res=(res*10+tb[i]-'0')%PHI;
      }
      printf("%I64d\n",quickpow(a,res+PHI,c));
    }
  }
  return 0;
}

别问我，大神的代码，我看不懂。