Submission #1025757

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Source Code Expand

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>

#define FOR(i,k,n) for (int (i)=(k); (i)<(n); ++(i))
#define rep(i,n) FOR(i,0,n)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fst first
#define snd second
#define all(v) begin(v), end(v)
#define debug(x) cerr<< #x <<": "<<x<<endl
#define debug2(x,y) cerr<< #x <<": "<< x <<", "<< #y <<": "<< y <<endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<ll> vll;
typedef vector<vector<ll> > vvll;
template<class T> using vv=vector<vector< T > >;

int mod = 1e9 + 7;

int main() {
  ll l;
  scanf("%lld", &l);
  if (l <= 8) {
    printf("0\n");
    return 0;
  }
  ll ans = 0;
  ans += (l/3 - 2) % mod; // x, x+1, x+2, l >= 9
  ans %= mod;
  if (l <= 10) {
    printf("%lld\n", ans);
    return 0;
  }

  ll b3 = (l+1)/3;
  ll b3mod = b3 % mod;
  //debug(b3);
  //debug(((b3mod*(b3mod+1)/2) % mod + 3 + 3 * (mod - b3mod)) % mod);
  ans += ((b3mod*(b3mod+1)/2) % mod + 3 + 3 * (mod - b3mod)) % mod;
  ans %= mod;

  ll b4 = (l-3)/2;
  ll b4mod = b4 % mod;
  //debug(b4);
  //debug(((l-2)*(b4-b3) % mod + b3mod*(b3mod+1) % mod + (mod - (b4mod*(b4mod+1) % mod))) % mod);
  ans += (((l-2)%mod)*((b4-b3)%mod) % mod + b3mod*(b3mod+1) % mod + (mod - (b4mod*(b4mod+1) % mod))) % mod;
  ans %= mod;

  if (l <= 12) {
    printf("%lld\n", ans);
    return 0;
  }

  ll b1 = (l-1)/4;
  ll b1mod = b1 % mod;
  //debug(b1);
  //debug((b1mod*(b1mod+1)/2 + 1 + 2*(mod - b1mod)) % mod);
  ans += ((b1mod*(b1mod+1)/2) % mod + 1 + 2*(mod - b1mod)) % mod;
  ans %= mod;

  ll b2 = (l-4)/3;
  ll b2mod = b2 % mod;
  //debug(b2);
  //debug(((l-3)*(b2-b1) + b1mod*(b1mod+1)/2*3 + mod - (((b2mod*(b2mod+1))/2*3) % mod)) % mod);
  ans += (((l-3)%mod)*((b2-b1)%mod) % mod + (b1mod*(b1mod+1)/2*3) % mod + mod - (((b2mod*(b2mod+1))/2*3) % mod)) % mod;
  ans %= mod;

  printf("%lld\n", ans);
  return 0;
}

Submission Info

Judge Result