第31期：IT培训-领先公司的当前问题和挑战

问题

有1000个葡萄酒瓶。其中一瓶装有毒酒。老鼠喝了有毒的酒一小时后就死了。需要多少只最小的老鼠才能在一个小时内找出哪个瓶子含有毒药。

There are 5 pirates, they must decide how to distribute 100 gold coins among them. The pirates have seniority levels, the senior-most is A, then B, then C, then D, and finally the junior-most is E.

Rules of distribution are:

• The most senior pirate proposes a distribution of coins.
• All pirates vote on whether to accept the distribution.
• If the distribution is accepted, the coins are disbursed and the game ends.
• If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
• In case of a tie vote the proposer can has the casting vote

Rules every pirates follows:

• Every pirate wants to survive
• Given survival, each pirate wants to maximize the number of gold coins he receives.

What is the maximum number of coins that pirate A might get?

给定一个整数，任务是找到该数字的阶乘。

输入：输入
的第一行包含一个整数T，表示测试用例的数量。
每个测试用例的第一行是N，要找到其阶乘的数字

输出：
在单独的行中打印数字的阶乘。

约束：示例： 输入输出
1 ≤ T ≤ 100
1 ≤ N ≤ 1000



3
5
10
2


120
3628800
2

Given a Binary Tree, find diameter of it.
+The diameter of a tree is the number of nodes on the longest path between two leaves in the tree. The diagram below shows two trees each with diameter nine, the leaves that form the ends of a longest path are shaded (note that there is more than one path in each tree of length nine, but no path longer than nine nodes).



Input Format:
First line of input contains the number of test cases T. For each test case, there will be only a single line of input which is a string representing the tree as described below:

1. The values in the string are in the order of level order traversal of the tree where, numbers denotes node values, and a character “N” denotes NULL子级。

2.例如：

对于上面的树，字符串将为：1 2 3 NN 4 6 N 5 NN 7 N

输出格式：
对于每个测试用例，在新行中打印直径。

您的任务：
您需要完成将node作为参数并返回直径的函数直径（）。

约束：示例： 输入：输出：说明： Testcase1：树为 直径为3个长度。Testcase2：树是 直径为4个长度。
1 <= T <= 100
1 <= Number of nodes <= 100
1 <= Data of a node <= 100



2
1 2 3
10 20 30 40 60


3
4

传递

, .
+ — . , , , , ( , , ).

:
T. , , , :

1. , , “N” .

2. :

: 1 2 3 N N 4 6 N 5 N N 7 N

:
.

:
diameter(), node .

:
1 < = T <= 100
1 < = < = 100
1 < = <= 100

:
:
2
1 2 3
10 20 30 40 60

:
3
4

:
1: :

3.
2: :

4 .

How many ways are there to place a black and a white knight on an N * M chessboard such that they do not attack each other? The knights have to be placed on different squares. A knight can move two squares horizontally and one square vertically (L shaped), or two squares vertically and one square horizontally (L shaped). The knights attack each other if one can reach the other in one move.

Input:
The first line contains the number of test cases T. Each of the next T lines contains two integers N and M which is size of matrix.

Output:
For each testcase, print the required answer, i.e, number of possible ways to place knights.

Constraints:
1 <= T <= 100
1 <= N, M <= 105

Example:
Input:
3
2 2
2 3
4 5

Output:
12
26
312
Explanation:
Testcase 1: We can place a black and a white knight in 12 possible ways such that none of them attracts each other.

using namespace boost::multiprecision;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        cpp_int n, res = 1;
        cin >> n;
        while (n != 1)
        {
            res = res * n;
            n = n - 1;
        }
        cout << res << "\n";
    }
    return 0;
}

int find(Node* root, int &h)
{
    if (root == NULL) return 0;
    int lh = find(root->left, h);
    int rh = find(root->right, h);
    h = max(h, lh + rh + 1);
    return max(lh, rh) + 1;
}
int diameter(Node* root)
{
    int h = -1;
    find(root, h);
    return h;
}

#include <iostream>
#include <cstdint>

typedef unsigned __int128 uint128_t;

/**
 * O(1) solution to https://practice.geeksforgeeks.org/problems/black-and-white/0
 */
int main()
{
    unsigned t;
    uint64_t n, m, upper, lower;
    uint128_t res;
    
    std::cin >> t;
    while (t--)
    {
        std::cin >> n >> m;
        res = n*m; // total ways to choose first knight
        res *= (n*m-1); // now total ways to choose both knights
        if (n>1 && m>1)
            res -= 4*(2*n*m-3*n-3*m+4); // remove any collisions
        
        // the rest is just for printing a uint128_t
        upper = (uint64_t)(res/1000000000000000000);
        lower = (uint64_t)(res%1000000000000000000);
        if (upper)
        {
            std::cout << upper;
            uint64_t digitChecker = 100000000000000000;
            while (lower/digitChecker == 0 && digitChecker)
            {
                std::cout << 0;
                digitChecker /= 10;
            }
        }
        std::cout << lower << std::endl;
    }
    
	return 0;
}