BZOJ1965 [Ahoi2005]SHUFFLE 洗牌

x * 2^m = l (mod n + 1)，故

x = l * (2^m)^-1 (mod n + 1)

只需要求一下逆元什么的就做完了，注意乘法要用"快速加"的方法。。。否则会爆long long

/**************************************************************
    Problem: 1965
    User: rausen
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:804 kb
****************************************************************/
 
#include <cstdio>
 
using namespace std;
typedef long long ll;
 
ll n, m, l, p;
 
ll mul(ll x, ll y, ll mod) {
  ll res = 0;
  while (y) {
    if (y & 1) (res += x) %= mod;
    (x <<= 1) %= mod;
    y >>= 1;
  }
  return res;
}
 
ll pow(ll x, ll y, ll mod) {
  ll res = 1;
  while (y) {
    if (y & 1) res = mul(res, x, mod);
    x = mul(x, x, mod);
    y >>= 1;
  }
  return res;
}
 
int main() {
  scanf("%lld%lld%lld", &n, &m, &l);
  p = n / 2 + 1;
  p = pow(p, m, n + 1);
  printf("%lld
", mul(p, l, n + 1));
  return 0;
}