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=10000010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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, -1, sizeof(flag));
    for (i = 2, plen = 0; i < nmax; i++) {
        if (flag[i]) {
            prime[plen++] = i;
        }
        for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
            if (i % prime[j] == 0) {
                break;
            }
        }
    }
}
int getPhi(int n) {
    int i, te, phi;
    te = (int) sqrt(n * 1.0);
    for (i = 0, phi = n; (i < plen) && (prime[i] <= te); i++) {
        if (n % prime[i] == 0) {
            phi = phi / prime[i] * (prime[i] - 1);
            while (n % prime[i] == 0) {
                n /= prime[i];
            }
        }
    }
    if (n > 1) {
        phi = phi / n * (n - 1);
    }
    return phi;
}
int cmpCphi(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0'));
        if (res > p) {
            return 1;
        }
    }
    return 0;
}
int getCP(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0')) % p;
    }
    return (int) res;
}
int modular_exp(int a, int b, int c) {
    LL res, temp;
    res = 1 % c, temp = a % c;
    while (b) {
        if (b & 1) {
            res = res * temp % c;
        }
        temp = temp * temp % c;
        b >>= 1;
    }
    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-1]--;
    if(ch[len-1]>='0')return;
    ch[len-1]='9';
    for(i=len-2;i>=0;i--){
        num=ch[i]-1;
        if(num>='0'){
            ch[i]=num;
            if(i==0 && ch[i]=='0')break;
            return;
        }
        ch[i]='9';
    }
    for(i=0;i<=len;i++){
        ch[i]=ch[i+1];
    }
}

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

        a=2,c=MOD;
        ans=solve(a % c, c, ch);
        printf("%d
",ans);
    }
    return 0;
}