Problem p02618 (C++) — buggy submission, verdict: Time Limit Exceeded

/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
                  　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

////多倍長整数, cpp_intで宣言
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;

#pragma gcc target("avx2")
#pragma gcc optimization("Ofast")
#pragma gcc optimization("unroll-loops")
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i <= (n); ++i)
#define rep2(i, n) for (int i = 2; i < (n); ++i)
#define repr(i, n) for (int i = n; i >= 0; --i)
#define reprm(i, n) for (int i = n - 1; i >= 0; --i)
#define printynl(a) printf(a ? "yes\\n" : "no\\n")
#define printyn(a) printf(a ? "Yes\\n" : "No\\n")
#define printYN(a) printf(a ? "YES\\n" : "NO\\n")
#define printim(a) printf(a ? "possible\\n" : "imposible\\n")
#define printdb(a) printf("%.50lf\\n", a)    // 少数出力
#define printLdb(a) printf("%.50Lf\\n", a)   // 少数出力
#define printdbd(a) printf("%.16lf\\n", a)   // 少数出力(桁少なめ)
#define prints(s) printf("%s\\n", s.c_str()) // string出力
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2LL) // 等差数列の和、初項,末項,項数
#define pb push_back
#define rpriq priq<int, vector<int>, greater<int>>
#define deg_to_rad(deg) (((deg) / 360.0L) * 2.0L * PI)
#define rad_to_deg(rad) (((rad) / 2.0L / PI) * 360.0L)
#define Please return
#define AC 0
#define manhattan_dist(a, b, c, d)                                             \\
  (abs(a - c) + abs(b - d)) /*(a, b) から (c, d) のマンハッタン距離 */
#define inf numeric_limits<double>::infinity();
#define linf numeric_limits<long double>::infinity()

using ll = long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1e9 + 7;
constexpr long double eps = 1e-6;
const long double PI = acosl(-1.0L);

using namespace std;

void scans(string &str) {
  char c;
  str = "";
  scanf("%c", &c);
  if (c == '\\n')
    scanf("%c", &c);
  while (c != '\\n' && c != -1 && c != ' ') {
    str += c;
    scanf("%c", &c);
  }
}

void scanc(char &str) {
  char c;
  scanf("%c", &c);
  if (c == -1)
    return;
  while (c == '\\n') {
    scanf("%c", &c);
  }
  str = c;
}

double acot(double x) { return PI / 2 - atan(x); }

ll LSB(ll n) { return (n & (-n)); }

/*-----------------------------------------ここからコード-----------------------------------------*/

int main() {

  int d;
  scanf("%d", &d);
  vector<ll> c(26);
  rep(i, 26) scanf("%lld", &c[i]);
  vector<vector<ll>> s(d, vector<ll>(26));
  rep(i, d) rep(j, 26) scanf("%lld", &s[i][j]);
  vector<ll> last(26);
  ll ans = 0, tmp;
  int idx1 = 1, idx2 = 1, idx3 = 1, idx4 = 1;
  vector<int> res(d);

  rep(i, d - 2) {
    ans = -INF;
    rep(j, 26) {
      rep(k, 26) {
        rep(l, 26) {
          rep(m, 26) {
            ll tmp = 0;
            tmp += s[i][j] + s[i + 1][k] + s[i + 2][l] + s[i + 3][m];
            int last1 = last[j], last2 = last[k], last3 = last[l],
                last4 = last[m];
            last[j] = i;
            rep(n, 26) tmp -= c[n] * (i - last[n]);
            last[k] = i + 1;
            rep(n, 26) tmp -= c[n] * (i + 1 - last[n]);
            last[l] = i + 2;
            rep(n, 26) tmp -= c[n] * (i + 2 - last[n]);
            last[m] = i + 3;
            rep(n, 26) tmp -= c[n] * (i + 3 - last[n]);
            if (tmp > ans) {
              ans = tmp;
              idx1 = j;
              idx2 = k;
              idx3 = l;
              idx4 = m;
            }
            last[j] = last1, last[k] = last2, last[l] = last3, last[m] = last4;
          }
        }
      }
    }
    last[idx1] = i, last[idx2] = i + 1, last[idx3] = i + 2, last[idx4] = i + 3;
    res[i] = idx1, res[i + 1] = idx2, res[i + 2] = idx3, res[i + 3] = idx4;
    i += 3;
  }
  ans = -INF;
  int idx = 1;
  rep(i, 26) {
    ll tmp = s[d - 1][i];
    last[i] = d - 1;
    rep(j, 26) tmp -= (c[j] * (d - 1 - last[j]));
    if (tmp > ans) {
      idx = i;
      ans = tmp;
    }
  }
  res[d - 1] = idx;
  for (const auto &aa : res)
    printf("%d\\n", aa + 1);

  Please AC;
}

Error: TLE

Fixed submission:

/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
                  　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

////多倍長整数, cpp_intで宣言
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;

#pragma gcc target("avx2")
#pragma gcc optimization("Ofast")
#pragma gcc optimization("unroll-loops")
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i <= (n); ++i)
#define rep2(i, n) for (int i = 2; i < (n); ++i)
#define repr(i, n) for (int i = n; i >= 0; --i)
#define reprm(i, n) for (int i = n - 1; i >= 0; --i)
#define printynl(a) printf(a ? "yes\\n" : "no\\n")
#define printyn(a) printf(a ? "Yes\\n" : "No\\n")
#define printYN(a) printf(a ? "YES\\n" : "NO\\n")
#define printim(a) printf(a ? "possible\\n" : "imposible\\n")
#define printdb(a) printf("%.50lf\\n", a)    // 少数出力
#define printLdb(a) printf("%.50Lf\\n", a)   // 少数出力
#define printdbd(a) printf("%.16lf\\n", a)   // 少数出力(桁少なめ)
#define prints(s) printf("%s\\n", s.c_str()) // string出力
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2LL) // 等差数列の和、初項,末項,項数
#define pb push_back
#define rpriq priq<int, vector<int>, greater<int>>
#define deg_to_rad(deg) (((deg) / 360.0L) * 2.0L * PI)
#define rad_to_deg(rad) (((rad) / 2.0L / PI) * 360.0L)
#define Please return
#define AC 0
#define manhattan_dist(a, b, c, d)                                             \\
  (abs(a - c) + abs(b - d)) /*(a, b) から (c, d) のマンハッタン距離 */
#define inf numeric_limits<double>::infinity();
#define linf numeric_limits<long double>::infinity()

using ll = long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1e9 + 7;
constexpr long double eps = 1e-6;
const long double PI = acosl(-1.0L);

using namespace std;

void scans(string &str) {
  char c;
  str = "";
  scanf("%c", &c);
  if (c == '\\n')
    scanf("%c", &c);
  while (c != '\\n' && c != -1 && c != ' ') {
    str += c;
    scanf("%c", &c);
  }
}

void scanc(char &str) {
  char c;
  scanf("%c", &c);
  if (c == -1)
    return;
  while (c == '\\n') {
    scanf("%c", &c);
  }
  str = c;
}

double acot(double x) { return PI / 2 - atan(x); }

ll LSB(ll n) { return (n & (-n)); }

/*-----------------------------------------ここからコード-----------------------------------------*/

int main() {

  int d;
  scanf("%d", &d);
  vector<ll> c(26);
  rep(i, 26) scanf("%lld", &c[i]);
  vector<vector<ll>> s(d, vector<ll>(26));
  rep(i, d) rep(j, 26) scanf("%lld", &s[i][j]);
  vector<ll> last(26);
  ll ans = 0, tmp;
  int idx1 = 1, idx2 = 1, idx3 = 1, idx4 = 1;
  vector<int> res(d);

  rep(i, d - 2) {
    ans = -INF;
    rep(j, 26) {
      rep(k, 26) {
        rep(l, 26) {
          rep(m, 26) {
            ll tmp = 0;
            tmp += s[i][j] + s[i + 1][k] + s[i + 2][l] + s[i + 3][m];
            int last1 = last[j], last2 = last[k], last3 = last[l],
                last4 = last[m];
            last[j] = i;
            rep(n, 26) tmp -= c[n] * (i - last[n]);
            last[k] = i + 1;
            rep(n, 26) tmp -= c[n] * (i + 1 - last[n]);
            last[l] = i + 2;
            rep(n, 26) tmp -= c[n] * (i + 2 - last[n]);
            last[m] = i + 3;
            rep(n, 26) tmp -= c[n] * (i + 3 - last[n]);
            if (tmp >= ans) {
              ans = tmp;
              idx1 = j;
              idx2 = k;
              idx3 = l;
              idx4 = m;
            }
            last[j] = last1, last[k] = last2, last[l] = last3, last[m] = last4;
          }
        }
      }
    }
    last[idx1] = i, last[idx2] = i + 1, last[idx3] = i + 2, last[idx4] = i + 3;
    res[i] = idx1, res[i + 1] = idx2, res[i + 2] = idx3, res[i + 3] = idx4;
    i += 3;
  }
  ans = -INF;
  int idx = 1;
  rep(i, 26) {
    ll tmp = s[d - 1][i];
    last[i] = d - 1;
    rep(j, 26) tmp -= (c[j] * (d - 1 - last[j]));
    if (tmp > ans) {
      idx = i;
      ans = tmp;
    }
  }
  res[d - 1] = idx;
  for (const auto &aa : res)
    printf("%d\\n", aa + 1);

  Please AC;
}