Problem p02787 (C++)

Buggy submission (status: Time Limit Exceeded):
//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao Twitter/QQ:810396815
#pragma GCC optimize("O3")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>
#define IINF 10e8
#define INF 10e16
#define MOD 1000000007
#define mod 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x, y) (x > y ? x : y)
#define mmin(x, y) (x < y ? x : y)
#define chmax(x, y) x = mmax(x, y)
#define chmin(x, y) x = mmin(x, y)
#define all(x) (x).begin(), (x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
// using Int=int_fast64_t;
typedef long long int ll;
typedef long long int LL;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }
int dx[8] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dy[8] = {0, -1, 0, 1, 1, -1, 1, -1};
char dir[4] = {'u', 'l', 'd', 'r'};
ll cmp1(pair<ll, string> a, pair<ll, string> b) {
  if (a.fi != b.fi)
    return a.fi < b.fi;
  else
    return a.se < b.se;
}

//---------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main() { // 問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????统一类型
             // memset()!!!!!!!!!!!!       ペナを減らせ!!!!!!!!!!!!!
  cin.tie(0);
  ios::sync_with_stdio(false);
  //-------------------------------
  ll h, n;
  cin >> h >> n;
  vector<ll> a(n + 1), b(n + 1);
  ll amax = 0;
  for (ll i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    chmax(amax, a[i]);
  }
  ll dp1[100001];
  ll ans = INF;

  // 初期化 求最小值就这样 最大值反着来即可 不要多逼逼
  for (ll i = 0; i <= 10000 + amax; i++)
    dp1[i] = INF;
  dp1[0] = 0;

  // dp[i] 为了消灭i体力所需要的最小魔法值
  for (ll i = 1; i <= n; i++) {
    for (ll j = a[i]; j <= h + amax; j++) { // j可以直接从a[j]开始
      // chmin(dp1[j],dp1[j-a[i]]+b[i]);
      ll ncnt = j / a[i];
      for (int k = 0; k <= ncnt; k++)
        chmin(dp1[j], dp1[j - k * a[i]] + b[i] * k);
    }
  }

  // 打h到h+amax之间的最小ans 可以理解吧
  for (ll i = h; i <= h + amax; i++)
    chmin(ans, dp1[i]);
  cout << ans << endl;
  //-------------------------------
  return 0;
}
//---------------------------------------------------------------------------


Fixed submission:
//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao Twitter/QQ:810396815
#pragma GCC optimize("O3")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>
#define IINF 10e8
#define INF 10e16
#define MOD 1000000007
#define mod 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x, y) (x > y ? x : y)
#define mmin(x, y) (x < y ? x : y)
#define chmax(x, y) x = mmax(x, y)
#define chmin(x, y) x = mmin(x, y)
#define all(x) (x).begin(), (x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
// using Int=int_fast64_t;
typedef long long int ll;
typedef long long int LL;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }
int dx[8] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dy[8] = {0, -1, 0, 1, 1, -1, 1, -1};
char dir[4] = {'u', 'l', 'd', 'r'};
ll cmp1(pair<ll, string> a, pair<ll, string> b) {
  if (a.fi != b.fi)
    return a.fi < b.fi;
  else
    return a.se < b.se;
}

//---------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main() { // 問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????统一类型
             // memset()!!!!!!!!!!!!       ペナを減らせ!!!!!!!!!!!!!
  cin.tie(0);
  ios::sync_with_stdio(false);
  //-------------------------------
  ll h, n;
  cin >> h >> n;
  vector<ll> a(n + 1), b(n + 1);
  ll amax = 0;
  for (ll i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    chmax(amax, a[i]);
  }
  ll dp1[100001];
  ll ans = INF;

  // 初期化 求最小值就这样 最大值反着来即可 不要多逼逼
  for (ll i = 0; i <= 10000 + amax; i++)
    dp1[i] = INF;
  dp1[0] = 0;

  // dp[i] 为了消灭i体力所需要的最小魔法值
  for (ll i = 1; i <= n; i++) {
    for (ll j = a[i]; j <= h + amax; j++) { // j可以直接从a[j]开始
      chmin(dp1[j], dp1[j - a[i]] + b[i]);
      /*ll ncnt=j/a[i];
      for(int k=0;k<=ncnt;k++)
      chmin(dp1[j],dp1[j-k*a[i]]+b[i]*k);*///因为是从小到大遍历过来的所以不需要这个步骤
    }
  }

  // 打h到h+amax之间的最小ans 可以理解吧
  for (ll i = h; i <= h + amax; i++)
    chmin(ans, dp1[i]);
  cout << ans << endl;
  //-------------------------------
  return 0;
}
//---------------------------------------------------------------------------