hdu 4704 Sum 费马小定理

题目链接

求2^n%mod的值， n<=10^100000。

费马小定理 如果a, p 互质， 那么a^(p-1) = 1(mod p)  然后可以推出来a^k % p = a^(k%(p-1))%p。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
ll pow(ll a, ll b) {
    ll ret = ;
    while(b) {
        if(b&)
            ret = ret*a%mod;
        a = a*a%mod;
        b>>=;
    }
    return ret;
}
int main()
{
    string s;
    while(cin>>s) {
        int len = s.size();
        ll num = ;
        for(int i = ; i<len; i++) {
            num = (num*+s[i]-'')%(mod-);
        }
        num--;
        num = (num+mod-)%(mod-);
        ll ans = pow(2LL, num)%mod;
        cout<<ans<<endl;
    }
    return ;
}