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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=5005;
//Code begin...

LL mult_mod(LL a, LL b, LL c){
    a%=c; b%=c;
    LL ret=0, tmp=a;
    while (b){
        if (b&1) {
            ret+=tmp;
            if (ret>c) ret-=c;
        }
        tmp<<=1;
        if (tmp>c) tmp-=c;
        b>>=1;
    }
    return ret;
}
LL pow_mod(LL a, LL n, LL mod){
    LL ret=1, temp=a%mod;
    while (n) {
        if (n&1) ret=mult_mod(ret,temp,mod);
        temp=mult_mod(temp,temp,mod);
        n>>=1;
    }
    return ret;
}
LL extend_gcd(LL a, LL b, LL &x, LL &y){
    if (a==0&&b==0) return -1;
    if (b==0) {x=1; y=0; 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(2,m,n+1); b=n+1;
    d=extend_gcd(a,b,x,y); x=x*l/d; mod=b/d;
    x=(x%mod+mod)%mod;
    printf("%lld\n",x);
    return 0;
}