题目大意：给定偶数张牌，问m次洗牌之后第l张牌是多少
x*2^m==l (mod n+1)
x=(n/2+1)^m*l mod n+1
快速幂+快速乘233
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MOD (n+1)
using namespace std;
typedef unsigned long long ll;
long long n,m,l;
ll Quick_Multiplication(ll x,ll y)
{
ll re=0;
while(y)
{
if(y&1) re+=x,re%=MOD;
x+=x,x%=MOD;y>>=1;
}
return re;
}
ll Quick_Power(ll x,ll y)
{
ll re=1;
while(y)
{
if(y&1) re=Quick_Multiplication(re,x);
x=Quick_Multiplication(x,x);y>>=1;
}
return re;
}
int main()
{
cin>>n>>m>>l;
cout<<Quick_Multiplication(Quick_Power(n/2+1,m),l)%MOD<<endl;
}
BZOJ 1965 AHOI2005 SHUFFLE 洗牌 数论
原文地址：http://blog.csdn.net/popoqqq/article/details/42261857
