pythonCalculate A * B. 输入 Each line will contain two integers A and B. Process to end of file. Note: the length of each integer will not exceed 50000.

时间: 2023-02-12 19:12:47 浏览: 105
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Automatic_Fingerprint_Authentication_System.zip_Cropping_64.m_ST

可以使用 Python 的内置函数 int() 将输入的字符串转换为整数，然后使用乘法运算符 * 计算 A * B。 示例代码如下： ``` while True: try: a, b = map(int, input().split()) print(a * b) except: break ``` 这段代码会不断读取输入的两个整数并相乘，直到输入结束。
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sort(solution.begin(), solution.end(), greater()); for (int i = 0; i < solution.size(); i++) { cout [i] ; } cout ; } // main.cpp #include #include #include "lib.h" using namespace std; int ...

Copil Copac is given a list of n−1 edges describing a tree of n vertices. He decides to draw it using the following algorithm: Step 0 : Draws the first vertex (vertex 1 ). Go to step 1 . Step 1 : For every edge in the input, in order: if the edge connects an already drawn vertex u to an undrawn vertex v , he will draw the undrawn vertex v and the edge. After checking every edge, go to step 2 . Step 2 : If all the vertices are drawn, terminate the algorithm. Else, go to step 1 . A reading is defined as the number of times Copil Copac performs step 1 . Find the number of readings needed by Copil Copac to draw the tree. Input Each test contains multiple test cases. The first line of input contains a single integer t (1≤t≤104 ) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (2≤n≤2⋅105 ) — the number of vertices of the tree. The following n−1 lines of each test case contain two integers ui and vi (1≤ui,vi≤n , ui≠vi ) — indicating that (ui,vi) is the i -th edge in the list. It is guaranteed that the given edges form a tree. It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output the number of readings Copil Copac needs to draw the tree.

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Let 𝑃 be a set of 𝑛 points on the 𝑥-axis and each of the points is colored with one of the colors 1,2, . . . , 𝑘. For each color 𝑖 of the 𝑘 colors, there is at least one point in 𝑃 which is colored with 𝑖. For a set 𝑃 ′ of consecutive points from 𝑃, if both 𝑃 ′ and 𝑃 ∖ 𝑃 ′ contain at least one point of each color, then we say that 𝑃 ′ makes a double rainbow. See the below figure as an example. The set 𝑃 consists of ten points and each of the points is colored by one of the colors 1, 2, 3, and 4. The set 𝑃 ′ of the five consecutive points contained in the rectangle makes a double rainbow. Given a set 𝑃 of points and the number 𝑘 of colors as input, write a program that computes and prints out the minimum size of 𝑃 ′ that makes a double rainbow. Input Your program is to read from standard input. The input starts with a line containing two integers 𝑛 and 𝑘 (1 ≤ 𝑘 ≤ 𝑛 ≤ 10,000), where 𝑛 is the number of the points in 𝑃 and 𝑘 is the number of the colors. Each of the following 𝑛 lines consists of an integer from 1 to 𝑘, inclusively, and the 𝑖-th line corresponds to the color of the 𝑖-th point of 𝑃 from the left. Output Your program is to write to standard output. Print exactly one line. The line should contain the minimum size of 𝑃 ′ that makes a double rainbow. If there is no such 𝑃 ′ , print 0. The following shows sample input and output for two test cases.

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用c++解决下述问题：描述 Akko is learning magic in Luna Nova Magical Academy. It is said that there are 10^5 types of different curses in the magic world. It is too hard for Akko to memorize all of them. Fortunately, Proffesor Meridies says that she can help Akko make a Curse Dictionary. When Akko hears a curse, the Dictionary will tell her the effect of the curse. Also, when Akko wants to do something, the Dictionary will find the curse for her. If there is no such curse, the Dictionary will be silence. Can you write a program to reproduce the Curse Dictionary? 输入 Each test case will contain no more than 10^5 curses. Each curse will be like "[name] effect". The name and effect of the curse will only contain lowercase letters. There will be only one space between ']' and the string after it. The length of the name will not exceed 20 and the length of the effect will not exceed 80. The end of the dictionary is a line of "END". After all curses there will be an integer N (1<=N<1000), followed by N queries. Every query will be a "[curse name]" or a "curse effect". There will be only one test case in each input file. 输出 For each test case, you must output the effect of the curse by its name or the name of the curse by its effect. If there is no curse satisfying the condition, print "silence".

= "END") { string name = curse.substr(1, curse.find(']') - 1); string effect = curse.substr(curse.find(']') + 2); curses[name] = effect; } // 处理查询请求 int n; cin >> n; getchar(); // 读入...

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