BZOJ 1965 洗牌(扩展欧几里得)

时间：2023-03-10 07:15:53
BZOJ 1965 洗牌(扩展欧几里得)

容易发现，对于牌堆里第x张牌，在一次洗牌后会变成2*x%(n+1)的位置。

于是问题就变成了求x*2^m%(n+1)=L，x在[1,n]范围内的解。

显然可以用扩展欧几里得求出。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

const int N=;

//Code begin...

LL mult_mod(LL a, LL b, LL c){

    a%=c; b%=c;

    LL ret=, tmp=a;

    while (b){

        if (b&) {

            ret+=tmp;

            if (ret>c) ret-=c;

        }

        tmp<<=;

        if (tmp>c) tmp-=c;

        b>>=;

    }

    return ret;

}

LL pow_mod(LL a, LL n, LL mod){

    LL ret=, temp=a%mod;

    while (n) {

        if (n&) ret=mult_mod(ret,temp,mod);

        temp=mult_mod(temp,temp,mod);

        n>>=;

    }

    return ret;

}

LL extend_gcd(LL a, LL b, LL &x, LL &y){

    if (a==&&b==) return -;

    if (b==) {x=; y=; return a;}

    LL d=extend_gcd(b,a%b,y,x);

    y-=a/b*x;

    return d;

}

int main ()

{

    LL n, m, l, a, b, d, x, y, mod;

    scanf("%lld%lld%lld",&n,&m,&l);

    a=pow_mod(,m,n+); b=n+;

    d=extend_gcd(a,b,x,y); x=x*l/d; mod=b/d;

    x=(x%mod+mod)%mod;

    printf("%lld\n",x);

    return ;

}

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