Apples and Ideas Time Limit: 2 Seconds      Memory Limit: 65536 KB

"If you have an apple and I have an apple and we exchange these apples then you and I will still each have one apple. But if you have an idea and I have an idea and we exchange these ideas, then each of us will have two ideas." - George Bernard Shaw

Now Alice has A apples and B ideas, while Bob has C apples and D ideas, what will they have if they exchange all things?

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The only line contains four integers A

, B, C, D (0 <= A, B, C, D <= 100) - as the problem described.

Output

For each test case, output two lines. First line contains two integers, indicating the number of Alice's apples and ideas; second line contains two integers, indicating the number of Bob's apples and ideas.

Sample Input

4
0 0 5 30
20 25 20 0
20 25 20 15
20 25 25 30

Sample Output

5 30
0 30
20 25
20 25
20 40
20 40
25 55
20 55

A. APPLES AND IDEAS

A,C=C,A; B=D=B+D;

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

int main() {
  int T; cin >> T;
  for (int _ = 0; _ < T; ++_ ) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    cout << c << " " << b + d << endl;
    cout << a << " " << b + d << endl;
  }
  return 0;
}

More Health Points Time Limit: 2 Seconds      Memory Limit: 65536 KB

Recently, Bright Luna is playing a massively multiplayer online role-playing game (MMORPG) called Legendary Age. Designers of this game prepared many mini-games during daily operations. Players can ignore these games without any loss. If you spend some time in them, you can always obtain some bonuses. This is one of the fascinating features ofLegendary Age.

One day, Bright Luna had a fight with other players. After that he came back to the city to recover his Health Points (HP). Surprisingly, the NPC gave him a special choice. The NPC can teleport Bright Luna to a colorful fantasy world with N dungeons. These dungeons are identified from 1 through N. Each dungeon has exactly a one-way exit leading to another dungeon which has a smaller identifier except for the dungeon 1, which has no exit.

There is a buff or debuff in each dungeon which can be described by an integer B_i. Bright Luna will get the i-th buff/debuff when he enters the i-th dungeon. When he leaves a dungeon or leaves the world, the buffs/debuffs he has obtained will change his HP by B_i. For example, If he already visited the dungeons 6 -> 4 -> 3, when he leaves the dungeon 3, his HP will be changed by (B₆ + B₄ + B₃). The buffs/debuffs will not expire until he completes the adventure.

The NPC had shown the world map to Bright Luna. Bright Luna can choose the dungeon he enters at the beginning and he can leave the world at any dungeon. Of course, he can refuse to take this adventure. Can you help Bright Luna to gain as many health points as possible?

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 ≤ N ≤ 10⁵).

The second line contains N integers B_i (|B_i| ≤ 10⁸).

The third line contains N - 1 integers F_i (2 ≤ i ≤ N, F_i < i) indicating the dungeons which the exit of dungeon i lead to.

Output

For each case, please output the maximum health points Bright Luna can gain.

Sample Input

4
4
1 4 3 2
1 1 1
4
1 2 -3 4
1 2 3
4
-4 2 -3 4
1 2 3
3
-1 -2 -3
1 1

Sample Output

9
12
8
0

Hint

In the first example, we can start from dungeon 2 and end at dungeon 1.

In the second example, we can start from dungeon 4 and end at dungeon 1.

In the third example, we can start from dungeon 4 and end at dungeon 2.

B. MORE HEALTH POINTS

考慮鏈上的版本(其實就是Bear and Bowling 4)，給出$n$個數a_1,a_2,...,a_na​1​​,a​2​​,...,a​n​​，求出\sum\limits_{i=l}^{r}{(i-l+1)a_i}​i=l​∑​r​​(i−l+1)a​i​​的最大值。令s_i=\sum\limits_{j=1}^{i}a_j, p_i=\sum\limits_{j=1}^{i}ja_js​i​​=​j=1​∑​i​​a​j​​,p​i​​=​j=1​∑​i​​ja​j​​，那麼就是求p_r-p_{l-1}-l(s_r-s_{l-1})p​r​​−p​l−1​​−l(s​r​​−s​l−1​​)的最大值。如果固定$r$，把式子變換一下就是找到一個$l$最大化p_r-ls_r-p_{l-1}+ls_{l-1}p​r​​−ls​r​​−p​l−1​​+ls​l−1​​，維護一個凸殼就可以搞了。推廣到樹上把鏈上用到的單調佇列可持久一下就好了，複雜度$O(n\log n)$。

也可以樹分治。考慮每次分治中心$g$，一條合法的在原來有根樹上的路徑是這麼組成的：$g$往下標號一路遞減的路徑+$g$往下標號一路遞增的路徑。類似地，推一下合併需要的式子，會發現也是可以用凸殼維護的，複雜度O(n\log^2n)O(nlog​2​​n)。

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

class Solution {
  static const int MAXN = 100000 + 10;
  static const LL inf = 1ll << 60;
  vector<int> G[MAXN];
  int w[MAXN], n;

  struct Line {// m * x + b
    LL m, b;
    double inter(const Line &r) const {
      return (r.b - b) / (m - r.m);
    }
    inline LL eval(LL x) {return m * x + b;}
  } Q[MAXN];

  int rt, mins, total, top;
  int vs[MAXN], sz[MAXN], dep[MAXN];
  LL val[MAXN], ps[MAXN], ret;

  void getCenter(int u, int f = -1) {
    int mx = 0; sz[u] = 1;
    for (auto &v: G[u]) if (v != f && !vs[v]) {
      getCenter(v, u); sz[u] += sz[v];
      mx = max(mx, sz[v]);
    }
    mx = max(mx, total - sz[u]);
    if (mx < mins) mins = mx, rt = u;
  }
  int dfs1(int u, int d = 1) {
    ret = max(ret, val[d]);
    for (auto &v: G[u]) if (!vs[v] && u > v) {
      ps[v] = ps[u] + w[v];
      val[d + 1] = val[d] + ps[v];
      return dfs1(v, d + 1) + 1;
    }
    return 1;
  }
  void dfs2(int u, int d = 0, LL sum = 0) {
    sum += 1ll * d * w[u];
    // max(ps[u] * m + b + sum)
    int left = 0, right = top - 2;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (Q[mid].eval(ps[u]) >= Q[mid + 1].eval(ps[u])) right = mid;
      else left = mid + 1;
    }
    ret = max(ret, Q[left].eval(ps[u]) + sum);
    if (left + 1 < top) ret = max(ret, Q[left + 1].eval(ps[u]) + sum);
    for (auto &v: G[u]) if (!vs[v] && u < v) {
      ps[v] = ps[u] + w[v];
      dfs2(v, d + 1, sum);
    }
  }
  void solve(int u, int _tot) {
    total = _tot; mins = _tot * 2;
    getCenter(u); u = rt; vs[u] = 1; getCenter(u);
    val[1] = ps[u] = w[u];
    int md = dfs1(u);
    top = 0;
    for (int i = 1; i <= md; ++i) {
      Line now = (Line){i, val[i]};
      while (top >= 2 && Q[top - 2].inter(Q[top - 1]) >= Q[top - 1].inter(now)) --top;
      Q[top++] = now;
    }
    ps[u] = 0; dfs2(u);
    for (int i = 0; i <= md; ++i) val[i] = -inf;
    for (auto &v: G[u]) if (!vs[v]) {
      solve(v, sz[v]);
    }
  }

public:
  void run() {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", w + i); G[i].clear();
    }
    for (int i = 1; i < n; ++i) {
      int x; scanf("%d", &x); --x;
      G[x].push_back(i);
      G[i].push_back(x);
    }
    ret = 0;
    for (int i = 0; i < n; ++i) val[i] = -inf;
    memset(vs, 0, sizeof(vs[0]) * n);
    solve(0, n);
    printf("%lld\n", ret);
  }
} sol;

int main() {
  int T; scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) sol.run();
  return 0;
}

Defuse the Bomb Time Limit: 2 Seconds      Memory Limit: 65536 KB

The bomb is about to explode! Please defuse it as soon as possible!

There is a display showing a number from 1 to 4 on the bomb. Besides this, there are 4 buttons under the display. Each button is labeled by a number from 1 to 4. The numbers on the buttons are always distinct.

There are 5 defusing stages in total. Pressing the correct button can progress the bomb to the next defusing stage. The number on the display and the number on each button may be different in different stages. The bomb will be defused only when all 5 defusing stages get passed. Pressing the incorrect button will cause the bomb to explode immediately. Be careful!

Here is the detailed bomb defusing manual. Button positions are ordered from left to right.

Stage 1:

• If the display is 1, press the button in the second position.
• If the display is 2, press the button in the second position.
• If the display is 3, press the button in the third position.
• If the display is 4, press the button in the fourth position.

Stage 2:

• If the display is 1, press the button labeled "4".
• If the display is 2, press the button in the same position as you pressed in stage 1.
• If the display is 3, press the button in the first position.
• If the display is 4, press the button in the same position as you pressed in stage 1.

Stage 3:

• If the display is 1, press the button with the same label you pressed in stage 2.
• If the display is 2, press the button with the same label you pressed in stage 1.
• If the display is 3, press the button in the third position.
• If the display is 4, press the button labeled "4".

Stage 4:

• If the display is 1, press the button in the same position as you pressed in stage 1.
• If the display is 2, press the button in the first position.
• If the display is 3, press the button in the same position as you pressed in stage 2.
• If the display is 4, press the button in the same position as you pressed in stage 2.

Stage 5:

• If the display is 1, press the button with the same label you pressed in stage 1.
• If the display is 2, press the button with the same label you pressed in stage 2.
• If the display is 3, press the button with the same label you pressed in stage 4.
• If the display is 4, press the button with the same label you pressed in stage 3.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. For each test case:

There are 5 lines. Each line contains 5 integers D, B₁, B₂, B₃, B₄ indicating the number on the display and the numbers on the buttons respectively. The i-th line correspond to the i-th stage.

Output

For each test case, output 5 lines. The i-th line contains two integers indicating the position and the label of the correct button for the i-th stage.

Sample Input

1
4 2 1 3 4
2 2 4 3 1
4 3 1 4 2
4 3 4 2 1
2 3 1 2 4

Sample Output

4 4
4 1
3 4
4 1
2 1

Hint

Keep talking with your teammates and nobody explodes!

C. DEFUSE THE BOMB

按照題目意思模擬。

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;

vector<PII> res;
int b[5][5], a[5];

int main() {
  int T; scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) {
    int r[5];
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        cin >> b[i][j];
      }
    }
    // stage 1
    int *x = b[0], d = x[0];
    if (d <= 2) r[0] = 2;
    else r[0] = d;
    // stage 2
    x = b[1], d = x[0];
    for (int i = 1; i < 5; ++i) a[x[i]] = i;
    if (d == 1) r[1] = a[4];
    else if (d == 3) r[1] = 1;
    else r[1] = r[0];
    // stage 3
    x = b[2], d = x[0];
    for (int i = 1; i < 5; ++i) a[x[i]] = i;
    if (d == 1) r[2] = a[b[1][r[1]]];
    else if (d == 2) r[2] = a[b[0][r[0]]];
    else if (d == 4) r[2] = a[4];
    else r[2] = d;
    // stage 4
    x = b[3], d = x[0];
    if (d == 1) r[3] = r[0];
    else if (d == 2) r[3] = 1;
    else r[3] = r[1];
    // stage 5
    x = b[4], d = x[0];
    for (int i = 1; i < 5; ++i) a[x[i]] = i;
    if (d <= 2) r[4] = a[b[d - 1][r[d - 1]]];
    else r[4] = a[b[6 - d][r[6 - d]]];
    for (int i = 0; i < 5; ++i) {
      cout << r[i] << " " << b[i][r[i]] << endl;
    }
  }
  return 0;
}

The Lucky Week Time Limit: 2 Seconds      Memory Limit: 65536 KB

Edward, the headmaster of the Marjar University, is very busy every day and always forgets the date.

There was one day Edward suddenly found that if Monday was the 1st, 11th or 21st day of that month, he could remember the date clearly in that week. Therefore, he called such week "The Lucky Week".

But now Edward only remembers the date of his first Lucky Week because of the age-related memory loss, and he wants to know the date of the N-th Lucky Week. Can you help him?

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. For each test case:

The only line contains four integers Y, M, D and N (1 ≤ N ≤ 10⁹) indicating the date (Y: year, M: month, D: day) of the Monday of the first Lucky Week and the Edward's query N.

The Monday of the first Lucky Week is between 1st Jan, 1753 and 31st Dec, 9999 (inclusive).

Output

For each case, print the date of the Monday of the N-th Lucky Week.

Sample Input

2
2016 4 11 2
2016 1 11 10

Sample Output

2016 7 11
2017 9 11

D. THE LUCKY WEEK

可以觀察到1753年1月1日是週一，然後我們知道大概再過2800年就會重複。於是預處理出1753.01.01後面2800年的所有lucky week。之後根據迴圈節算一下就好了。

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

vector<tuple<int, int, int>> p;

int day(int y, int m, int d) {
  int tm = m >= 3 ? (m - 2) : (m + 10);
  int ty = m >= 3 ? y : (y - 1);
  return (ty + ty / 4 - ty / 100 + ty / 400 + (int)(2.6 * tm - 0.2) + d) % 7;
}

void run() {
  int y, m, d, n; cin >> y >> m >> d >> n;
  y -= 1753; n += y / 2800 * p.size();
  y %= 2800;
  n += lower_bound(p.begin(), p.end(), make_tuple(y, m, d)) - p.begin() - 1;
  auto res = p[n % p.size()];
  y = get<0>(res);
  m = get<1>(res);
  d = get<2>(res);
  y += n / p.size() * 2800 + 1753;
  cout << y << " " << m << " " << d << endl;
}

int main() {
  cerr << day(1753, 1, 1) << endl;
  for (int y = 0; y < 2800; ++y) {
    for (int m = 1; m <= 12; ++m) {
      if (day(y + 1753, m,  1) == 1) p.push_back(make_tuple(y, m,  1));
      if (day(y + 1753, m, 11) == 1) p.push_back(make_tuple(y, m, 11));
      if (day(y + 1753, m, 21) == 1) p.push_back(make_tuple(y, m, 21));
    }
  }
  int T; scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) run();
  return 0;
}

Modulo Query Time Limit: 2 Seconds      Memory Limit: 65536 KB

One day, Peter came across a function which looks like:

• F(1, X) = X mod A₁.
• F(i, X) = F(i - 1, X) mod A_i, 2 ≤ i ≤ N.

Where A is an integer array of length N, X is a non-negative integer no greater than M.

Peter wants to know the number of solutions for equation F(N, X) = Y, where Y is a given number.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers N and M (2 ≤ N ≤ 10⁵, 0 ≤ M ≤ 10⁹).

The second line contains N integers: A₁, A₂, ..., A_N (1 ≤ A_i ≤ 10⁹).

The third line contains an integer Q (1 ≤ Q ≤ 10⁵) - the number of queries. Each of the following Q lines contains an integer Y_i (0 ≤ Y_i ≤ 10⁹), which means Peter wants to know the number of solutions for equation F(N, X) = Y_i.

Output

For each test cases, output an integer S = (1 ⋅ Z₁ + 2 ⋅ Z₂ + ... + Q ⋅ Z_Q) mod (10⁹ + 7), where Z_i is the answer for the i-th query.

Sample Input

1
3 5
3 2 4
5
0
1
2
3
4

Sample Output

8

Hint

The answer for each query is: 4, 2, 0, 0, 0.

E. MODULO QUERY

可以觀察到一個區間對一個數取模後的結果也可以用區間來表示，並且這些區間的左端點都是0。於是直接用一個map存每次取模之後的區間和這個區間出現次數，要取模的時候，找出所有右端點大於等於當前模數的所有區間，暴力算一下結果即可。對於查詢，二分下位置，求個字尾和就好了。下面分析下複雜度為什麼是對的。

眾所周知：一個數對一堆數取模，最多會有$\log$次值的改變。對於這題每對一個新數取模，最多隻會增加一個區間。考慮當前區間是[0,v_i][0,v​i​​]，要對$x$取模，那麼對於v_i < xv​i​​<x的區間沒有變化，對於v_i \ge xv​

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