HDU-4704 Sum 大数幂取模

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4704

　　题意：求a^n%m的结果，其中n为大数。

　　S(1)+S(2)+...+S(N)等于2^(n-1)，第一次多校都出过吧。然后就是一个裸的大数幂了。。

　　关于大数的A^B mod C推荐看AC神的两篇文章<如何计算A^B mod C>,<计算a^(n!) mod c>...

　　当然，这个还以一个更简单的方法，由费马小定理：a^(p-1)=1(mod p)，那么a^n=1(mod p)可以转化为：2^(n%(1e9+7-1)) % 1e9+7...

 //STATUS:C++_AC_15MS_1360KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 //#include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 #define nnum 1000005
 #define nmax 31625
 int flag[nmax], prime[nmax];
 int plen;
 void mkprime() {
     int i, j;
     memset(flag, -, sizeof(flag));
     for (i = , plen = ; i < nmax; i++) {
         if (flag[i]) {
             prime[plen++] = i;
         }
         for (j = ; (j < plen) && (i * prime[j] < nmax); j++) {
             flag[i * prime[j]] = ;
             if (i % prime[j] == ) {
                 break;
             }
         }
     }
 }
 int getPhi(int n) {
     int i, te, phi;
     te = (int) sqrt(n * 1.0);
     for (i = , phi = n; (i < plen) && (prime[i] <= te); i++) {
         if (n % prime[i] == ) {
             phi = phi / prime[i] * (prime[i] - );
             while (n % prime[i] == ) {
                 n /= prime[i];
             }
         }
     }
     if (n > ) {
         phi = phi / n * (n - );
     }
     return phi;
 }
 int cmpCphi(int p, char *ch) {
     int i, len;
     LL res;
     len = strlen(ch);
     for (i = , res = ; i < len; i++) {
         res = (res *  + (ch[i] - ''));
         if (res > p) {
             return ;
         }
     }
     return ;
 }
 int getCP(int p, char *ch) {
     int i, len;
     LL res;
     len = strlen(ch);
     for (i = , res = ; i < len; i++) {
         res = (res *  + (ch[i] - '')) % p;
     }
     return (int) res;
 }
 int modular_exp(int a, int b, int c) {
     LL res, temp;
     res =  % c, temp = a % c;
     while (b) {
         if (b & ) {
             res = res * temp % c;
         }
         temp = temp * temp % c;
         b >>= ;
     }
     return (int) res;
 }

 int solve(int a, int c, char *ch) {
     int phi, res, b;
     phi = getPhi(c);
     if (cmpCphi(phi, ch)) {
         b = getCP(phi, ch) + phi;
     } else {
         b = atoi(ch);
     }
     res = modular_exp(a, b, c);
     return res;
 }

 void getch(char ch[])
 {
     int i,j,num,len=strlen(ch);
     ch[len-]--;
     if(ch[len-]>='')return;
     ch[len-]='';
     for(i=len-;i>=;i--){
         num=ch[i]-;
         if(num>=''){
             ch[i]=num;
             if(i== && ch[i]=='')break;
             return;
         }
         ch[i]='';
     }
     for(i=;i<=len;i++){
         ch[i]=ch[i+];
     }
 }

 int main() {
  //   freopen("in.txt", "r", stdin);
     int a, c;
     int ans;
     char ch[nnum];
     mkprime();
     while (~scanf("%s",ch)) {
         getch(ch);

         a=,c=MOD;
         ans=solve(a % c, c, ch);
         printf("%d\n",ans);
     }
     return ;
 }