BZOJ1965: [Ahoi2005]SHUFFLE 洗牌

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1965

找规律可知，设答案为x，有x*2^m%(n+1)=L

然后快速幂+逆元就可以了。

#include<cstring>
#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define low(x) (x&(-x)) 
#define maxn 505
#define inf int(1e9)
#define mm 1000000007
#define ll long long
using namespace std;
ll n,m,p,x,y,l,a;
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
ll Pow(ll x,ll y){
    ll ans=1;
    while (y){
        if (y&1) ans=ans*x%p;
        x=x*x%p;
        y/=2;
    }
    return ans;
}
ll exgcd(ll a,ll b,ll &x,ll &y){
    ll d;
    if (b==0) {
        x=1; y=0;
        return a;
    }
    d=exgcd(b,a%b,y,x);
    y-=a/b*x;
}
int main(){
    n=read(); m=read(); l=read(); p=n+1;
    a=Pow(2,m); 
    exgcd(a,p,x,y);
    x=(x%p+p)%p;
    printf("%lld\n",x*l%p);
    return 0;
}