BZOJ 1965 洗牌

答案((n/2+1)^m*l)%(n+1)。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
long long n,m,l;
long long mul(long long a,long long b)
{
    long long d=(long long)floor(a*(long double)b/(n+1)+0.5);
    long long ret=a*b-d*(n+1);
    if (ret<0) ret+=n+1;
    return ret;
}
long long f_pow(int a,int b)
{
    int ans=1,base=a;
    while (b)
    {
        if (b&1) ans=mul(ans,base)%(n+1);
        base=mul(base,base)%(n+1);
        b>>=1;
    }
    return ans%(n+1);
}
int main()
{
    scanf("%lld%lld%lld",&n,&m,&l);
    printf("%lld\n",mul(l,f_pow(n/2+1,m))%(n+1));
    return 0;
}