Problem p03388 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <stdio.h>
#include <vector>
using namespace std;
#define int long long
#define rep(s, i, n) for (int i = s; i < n; i++)
#define c(n) cout << n << endl;
#define ic(n)                                                                  \
  int n;                                                                       \
  cin >> n;
#define sc(s)                                                                  \
  string s;                                                                    \
  cin >> s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c, a, b) *min_element(c + a, c + b)
#define maxi(c, a, b) *max_element(c + a, c + b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int, int>
#define upp(a, n, x) upper_bound(a, a + n, x) - a;
#define low(a, n, x) lower_bound(a, a + n, x) - a;
#define UF UnionFind
#define pb push_back
// printf("%.12Lf\n",);
int keta(int x) {
  rep(0, i, 30) {
    if (x < 10) {
      return i + 1;
    }
    x = x / 10;
  }
}
int gcd(int x, int y) {
  if (x == 0 || y == 0)
    return x + y;
  int aa = x, bb = y;
  rep(0, i, 1000) {
    aa = aa % bb;
    if (aa == 0) {
      return bb;
    }
    bb = bb % aa;
    if (bb == 0) {
      return aa;
    }
  }
}
int lcm(int x, int y) {
  int aa = x, bb = y;
  rep(0, i, 1000) {
    aa = aa % bb;
    if (aa == 0) {
      return x / bb * y;
    }
    bb = bb % aa;
    if (bb == 0) {
      return x / aa * y;
    }
  }
}
bool prime(int x) {
  if (x == 1)
    return false;
  rep(2, i, sqrt(x) + 1) {
    if (x % i == 0 && x != i) {
      return false;
    }
  }
  return true;
}
int max(int a, int b) {
  if (a >= b)
    return a;
  else
    return b;
}
string maxst(string s, string t) {
  int n = s.size();
  int m = t.size();
  if (n > m)
    return s;
  else if (n < m)
    return t;
  else {
    rep(0, i, n) {
      if (s[i] > t[i])
        return s;
      if (s[i] < t[i])
        return t;
    }
    return s;
  }
}
string minst(string s, string t) {
  int n = s.size();
  int m = t.size();
  if (n < m)
    return s;
  else if (n > m)
    return t;
  else {
    rep(0, i, n) {
      if (s[i] < t[i])
        return s;
      if (s[i] > t[i])
        return t;
    }
    return s;
  }
}
string string_reverse(string s) {
  int n = s.size();
  string t;
  rep(0, i, n) t += s[n - i - 1];
  return t;
}
int min(int a, int b) {
  if (a >= b)
    return b;
  else
    return a;
}
int n2[41];
int nis[41];
int nia[41];
int mody[41];
int nn;
int com(int n, int y) {
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  int bunsi = 1, bunbo = 1;
  rep(0, i, y) bunsi = (bunsi * (n - i)) % mod;
  rep(0, i, y) bunbo = (bunbo * (i + 1)) % mod;
  mody[0] = bunbo;
  rep(1, i, 41) {
    bunbo = (bunbo * bunbo) % mod;
    mody[i] = bunbo;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = mod - 2;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn > n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  nis[0]++;
  rep(0, i, 41) {
    if (nis[i] == 1) {
      bunsi = (bunsi * mody[i]) % mod;
    }
  }
  return bunsi;
}
int gyakugen(int n, int y) {
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  mody[0] = y;
  rep(1, i, 41) {
    y = (y * y) % mod;
    mody[i] = y;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = mod - 2;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn > n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  nis[0]++;
  rep(0, i, 41) {
    if (nis[i] == 1) {
      n = (n * mody[i]) % mod;
    }
  }
  return n;
}
int yakuwa(int n) {
  int sum = 0;
  rep(1, i, sqrt(n + 1)) {
    if (n % i == 0)
      sum += i + n / i;
    if (i * i == n)
      sum -= i;
  }
  return sum;
}
int poow(int y, int n) {
  if (n == 0)
    return 1;
  n -= 1;
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  int yy = y;
  mody[0] = yy;
  rep(1, i, 41) {
    yy = (yy * yy) % mod;
    mody[i] = yy;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = n;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn >= n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  rep(0, i, 41) {
    if (nis[i] == 1) {
      y = (y * mody[i]) % mod;
    }
  }
  return y;
}
int minpow(int x, int y) {
  int sum = 1;
  rep(0, i, y) sum *= x;
  return sum;
}
int ketawa(int x, int sinsuu) {
  int sum = 0;
  rep(0, i, 80) {
    if (minpow(sinsuu, i) > x) {
      return sum;
    }
    sum += (x % minpow(sinsuu, i + 1)) / (minpow(sinsuu, i));
  }
  return sum;
}
double distance(double a, double b, double c, double d) {
  return sqrt((b - a) * (b - a) + (c - d) * (c - d));
}
int sankaku(int a) { return a * (a + 1) / 2; }
int sames(int a[1111111], int n) {
  int ans = 0;
  rep(0, i, n) {
    if (a[i] == a[i + 1]) {
      int j = i;
      while (a[j + 1] == a[i] && j <= n - 2)
        j++;
      ans += sankaku(j - i);
      i = j;
    }
  }
  return ans;
}
using Graph = vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
  depth[v] = d;
  oya[v] = p;
  for (auto nv : G[v]) {
    if (nv == p)
      continue;           // nv が親 p だったらダメ
    dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
  }
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
        memo[h][w] = 'x';

        // 八方向を探索
        for (int dh = -1; dh <= 1; ++dh) {
                for (int dw = -1; dw <= 1; ++dw) {
                        if(abs(0-dh)+abs(0-dw)==2)continue;
                        int nh = h + dh, nw = w + dw;

                        // 場外アウトしたり、0 だったりはスルー
                        if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
                        if (memo[nh][nw] == 'x') continue;

                        // 再帰的に探索
                        dfs(nh, nw);
                }
        }
}*/
int XOR(int a, int b) {
  if (a == 0 || b == 0) {
    return a + b;
  }
  int ni = 1;
  rep(0, i, 41) {
    n2[i] = ni;
    ni *= 2;
  }
  rep(0, i, 41) nis[i] = 0;
  for (int i = 40; i >= 0; i -= 1) {
    if (a >= n2[i]) {
      nis[i]++;
      a -= n2[i];
    }
    if (b >= n2[i]) {
      nis[i]++;
      b -= n2[i];
    }
  }
  int sum = 0;
  rep(0, i, 41) sum += (nis[i] % 2 * n2[i]);
  return sum;
}
// int ma[1024577][21];
// for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

  UnionFind(int N) : par(N) { // 最初は全てが根であるとして初期化
    for (int i = 0; i < N; i++)
      par[i] = i;
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x)
      return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x);        // xの根をrx
    int ry = root(y);        // yの根をry
    if (rx == ry)
      return; // xとyの根が同じ(=同じ木にある)時はそのまま
    par[rx] =
        ry; // xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
};
signed main() {
  ic(n) rep(0, i, n) {
    ic(a) ic(b) int t = a * b - 1;
    int l = 0, r = mod;
    int p;
    while (r - l >= 1) {
      int i = (l + r) / 2;
      if (i * i <= t && (i + 1) * (i + 1) > t) {
        p = i;
        break;
      } else if (i * i <= t)
        l = i;
      else
        r = i + 1;
    }
    if (t >= p * p + p)
      p *= 2;
    else
      p += p - 1;
    if (a != b)
      p -= 1;
    c(p)
  }
}


Fixed submission:
#include <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <stdio.h>
#include <vector>
using namespace std;
#define int long long
#define rep(s, i, n) for (int i = s; i < n; i++)
#define c(n) cout << n << endl;
#define ic(n)                                                                  \
  int n;                                                                       \
  cin >> n;
#define sc(s)                                                                  \
  string s;                                                                    \
  cin >> s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c, a, b) *min_element(c + a, c + b)
#define maxi(c, a, b) *max_element(c + a, c + b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int, int>
#define upp(a, n, x) upper_bound(a, a + n, x) - a;
#define low(a, n, x) lower_bound(a, a + n, x) - a;
#define UF UnionFind
#define pb push_back
// printf("%.12Lf\n",);
int keta(int x) {
  rep(0, i, 30) {
    if (x < 10) {
      return i + 1;
    }
    x = x / 10;
  }
}
int gcd(int x, int y) {
  if (x == 0 || y == 0)
    return x + y;
  int aa = x, bb = y;
  rep(0, i, 1000) {
    aa = aa % bb;
    if (aa == 0) {
      return bb;
    }
    bb = bb % aa;
    if (bb == 0) {
      return aa;
    }
  }
}
int lcm(int x, int y) {
  int aa = x, bb = y;
  rep(0, i, 1000) {
    aa = aa % bb;
    if (aa == 0) {
      return x / bb * y;
    }
    bb = bb % aa;
    if (bb == 0) {
      return x / aa * y;
    }
  }
}
bool prime(int x) {
  if (x == 1)
    return false;
  rep(2, i, sqrt(x) + 1) {
    if (x % i == 0 && x != i) {
      return false;
    }
  }
  return true;
}
int max(int a, int b) {
  if (a >= b)
    return a;
  else
    return b;
}
string maxst(string s, string t) {
  int n = s.size();
  int m = t.size();
  if (n > m)
    return s;
  else if (n < m)
    return t;
  else {
    rep(0, i, n) {
      if (s[i] > t[i])
        return s;
      if (s[i] < t[i])
        return t;
    }
    return s;
  }
}
string minst(string s, string t) {
  int n = s.size();
  int m = t.size();
  if (n < m)
    return s;
  else if (n > m)
    return t;
  else {
    rep(0, i, n) {
      if (s[i] < t[i])
        return s;
      if (s[i] > t[i])
        return t;
    }
    return s;
  }
}
string string_reverse(string s) {
  int n = s.size();
  string t;
  rep(0, i, n) t += s[n - i - 1];
  return t;
}
int min(int a, int b) {
  if (a >= b)
    return b;
  else
    return a;
}
int n2[41];
int nis[41];
int nia[41];
int mody[41];
int nn;
int com(int n, int y) {
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  int bunsi = 1, bunbo = 1;
  rep(0, i, y) bunsi = (bunsi * (n - i)) % mod;
  rep(0, i, y) bunbo = (bunbo * (i + 1)) % mod;
  mody[0] = bunbo;
  rep(1, i, 41) {
    bunbo = (bunbo * bunbo) % mod;
    mody[i] = bunbo;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = mod - 2;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn > n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  nis[0]++;
  rep(0, i, 41) {
    if (nis[i] == 1) {
      bunsi = (bunsi * mody[i]) % mod;
    }
  }
  return bunsi;
}
int gyakugen(int n, int y) {
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  mody[0] = y;
  rep(1, i, 41) {
    y = (y * y) % mod;
    mody[i] = y;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = mod - 2;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn > n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  nis[0]++;
  rep(0, i, 41) {
    if (nis[i] == 1) {
      n = (n * mody[i]) % mod;
    }
  }
  return n;
}
int yakuwa(int n) {
  int sum = 0;
  rep(1, i, sqrt(n + 1)) {
    if (n % i == 0)
      sum += i + n / i;
    if (i * i == n)
      sum -= i;
  }
  return sum;
}
int poow(int y, int n) {
  if (n == 0)
    return 1;
  n -= 1;
  int ni = 1;
  for (int i = 0; i < 41; i++) {
    n2[i] = ni;
    ni *= 2;
  }
  int yy = y;
  mody[0] = yy;
  rep(1, i, 41) {
    yy = (yy * yy) % mod;
    mody[i] = yy;
  }
  rep(0, i, 41) nis[i] = 0;
  nn = n;
  for (int i = 40; i >= 0; i -= 1) {
    if (nn >= n2[i]) {
      nis[i]++;
      nn -= n2[i];
    }
  }
  rep(0, i, 41) {
    if (nis[i] == 1) {
      y = (y * mody[i]) % mod;
    }
  }
  return y;
}
int minpow(int x, int y) {
  int sum = 1;
  rep(0, i, y) sum *= x;
  return sum;
}
int ketawa(int x, int sinsuu) {
  int sum = 0;
  rep(0, i, 80) {
    if (minpow(sinsuu, i) > x) {
      return sum;
    }
    sum += (x % minpow(sinsuu, i + 1)) / (minpow(sinsuu, i));
  }
  return sum;
}
double distance(double a, double b, double c, double d) {
  return sqrt((b - a) * (b - a) + (c - d) * (c - d));
}
int sankaku(int a) { return a * (a + 1) / 2; }
int sames(int a[1111111], int n) {
  int ans = 0;
  rep(0, i, n) {
    if (a[i] == a[i + 1]) {
      int j = i;
      while (a[j + 1] == a[i] && j <= n - 2)
        j++;
      ans += sankaku(j - i);
      i = j;
    }
  }
  return ans;
}
using Graph = vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
  depth[v] = d;
  oya[v] = p;
  for (auto nv : G[v]) {
    if (nv == p)
      continue;           // nv が親 p だったらダメ
    dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
  }
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
        memo[h][w] = 'x';

        // 八方向を探索
        for (int dh = -1; dh <= 1; ++dh) {
                for (int dw = -1; dw <= 1; ++dw) {
                        if(abs(0-dh)+abs(0-dw)==2)continue;
                        int nh = h + dh, nw = w + dw;

                        // 場外アウトしたり、0 だったりはスルー
                        if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
                        if (memo[nh][nw] == 'x') continue;

                        // 再帰的に探索
                        dfs(nh, nw);
                }
        }
}*/
int XOR(int a, int b) {
  if (a == 0 || b == 0) {
    return a + b;
  }
  int ni = 1;
  rep(0, i, 41) {
    n2[i] = ni;
    ni *= 2;
  }
  rep(0, i, 41) nis[i] = 0;
  for (int i = 40; i >= 0; i -= 1) {
    if (a >= n2[i]) {
      nis[i]++;
      a -= n2[i];
    }
    if (b >= n2[i]) {
      nis[i]++;
      b -= n2[i];
    }
  }
  int sum = 0;
  rep(0, i, 41) sum += (nis[i] % 2 * n2[i]);
  return sum;
}
// int ma[1024577][21];
// for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

  UnionFind(int N) : par(N) { // 最初は全てが根であるとして初期化
    for (int i = 0; i < N; i++)
      par[i] = i;
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x)
      return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x);        // xの根をrx
    int ry = root(y);        // yの根をry
    if (rx == ry)
      return; // xとyの根が同じ(=同じ木にある)時はそのまま
    par[rx] =
        ry; // xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
};
signed main() {
  ic(n) rep(0, i, n) {
    ic(a) ic(b) if (a == 1 && b == 1) { c(0) continue; }
    int t = a * b - 1;
    int l = 0, r = mod;
    int p;
    while (r - l >= 1) {
      int i = (l + r) / 2;
      if (i * i <= t && (i + 1) * (i + 1) > t) {
        p = i;
        break;
      } else if (i * i <= t)
        l = i;
      else
        r = i + 1;
    }
    if (t >= p * p + p)
      p *= 2;
    else
      p += p - 1;
    if (a != b)
      p -= 1;
    c(p)
  }
}