time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Vanya got an important task - he should enumerate books in the library and label each book with its number. Each of the n books should be assigne…

题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制数字,求出其长len,当len为奇数时,第一位为1,后面的位数如果都为0,则输出len,如果有一个不为0,则输出len+1: 当len为偶数时,则输出len.(之所以这样输出是因为题目给定4的次幂是从0开始的) #include<iostream> #include<string> #…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of w milliliters and 2n tea cups,…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n…

time limit per test4 seconds memory limit per test512 megabytes inputstandard input outputstandard output Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many tri…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The mobile application store has a new game called "Subway Roller". The protagonist of the game Philip is located in one end of the tunnel a…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output There are n beacons located at distinct positions on a number line. The i-th beacon has position ai and power level bi. When the i-th beacon is…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The process of mammoth's genome decoding in Berland comes to its end! One of the few remaining tasks is to restore unrecognized nucleotides in a…

[UOJ#33][UR #2]树上GCD(长链剖分,分块) 题面 UOJ 题解 首先不求恰好,改为求\(i\)的倍数的个数,最后容斥一下就可以解决了. 那么我们考虑枚举一个\(LCA\)位置,在其两棵不同的子树中选择两个点,那么贡献就是这两段的\(gcd\). 那么发现要统计的东西类似于\(u\)的子树中,深度为\(d\)的点的个数,这个可以很容易的用长链剖分来维护,那么维护出这个数组之后就可以\(O(\log {dep})\)的对于贡献进行计算.然而这个复杂度是假的,因为你每次都需要一次\(O…

33 [程序 33 杨辉三角] 题目:打印出杨辉三角形(要求打印出 10 行如下图) 程序分析: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 package cskaoyan; public class cskaoyan33 { @org.junit.Test public void pascalsTriangle() { int[][] arr = new int[6][6]; for (int i = 0; i < 6; i++) { arr[i]…