quickly calc pow(i, n) since i in [1~n]

#include <bits/stdc++.h>
using namespace std;

#define inf (0x3f3f3f3f)
typedef long long int LL;
const int mod = 1e9 + ;
const int maxn =  + ;
LL f[maxn];
int prime[maxn];
bool check[maxn];
int n;

int qp(int a, int b) {
  LL result = ;
  LL base = a;
  while (b) {
    if (b & ) {
      result *= base;
      result %= mod;
    }
    b >>= ;
    base *= base;
    base %= mod;
  }
  return result;
}

void init_prime() {
  f[] = ;
  int total = ;
  for (int i = ; i <= n; ++i) {
    if (!check[i]) {
      prime[++total] = i;
      f[i] = qp(i, n);
    }
    for (int j = ; j <= total; ++j) {
      if (i * prime[j] > n) break;
      check[i * prime[j]] = true;
      f[i * prime[j]] = f[prime[j]] * f[i] % mod;
      if (i % prime[j] == ) break;
    }
  }
}

void work() {
  scanf("%d\n", &n);
  init_prime();
  LL result = ;
  for (int i = ; i <= n; ++i) {
    result ^= f[i];
  }
  printf("%lld\n", result);
}

int main(int argc, char *argv[]) {
  // freopen("data.txt", "r", stdin);
  work();
  return ;
}

https://ac.nowcoder.com/acm/contest/392/C