BZOJ1965洗牌（快速幂+快速乘+逆元）

思路：找规律题，可以发现，牌x移动后就会到2*x%(n+1)的位置，移动m次后就是，则

求2^m关于n+1的逆元就是2关于n+1的逆元。

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
typedef long long ll;
ll n, m, l, mod;
ll a, b, x, y;
ll exgcd(ll a, ll b) {
    if(!b) {
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b);
    ll t = x;
    x = y;
    y = t - (a / b) * y;
    return d;
}
ll mul(ll a, ll b, ll p) {
    ll ans = 0;
    for(ll i = b; i; i >>= 1, a = (a + a) % p)
        if(i & 1)
            ans = (ans + a) % p;
    return ans;
}
ll poww(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b & 1) {
            ans = mul(ans, a, mod);//快速乘防止爆ll
        }
        a = mul(a, a, mod);
        b >>= 1;
    }
    return ans;
}
int main() {
    cin >> n >> m >> l;
    mod = n + 1;
    exgcd(2, n + 1);//计算逆元
    while(x < 0)
        x += (n + 1);
    ll niyuan = x;
    niyuan = poww(niyuan, m);
    x = mul(niyuan, l, mod);
    cout << x;
    return 0;
}