xinxiaxindong的博客

两只青蛙在网上相识了，它们聊得很开心，于是觉得很有必要见一面。它们很高兴地发现它们住在同一条纬度线上，于是它们约定各自朝西跳，直到碰面为止。可是它们出发之前忘记了一件很重要的事情，既没有问清楚对方的特征，也没有约定见面的具体位置。不过青蛙们都是很乐观的，它们觉得只要一直朝着某个方向跳下去，总能碰到对方的。但是除非这两只青蛙在同一时间跳到同一点上，不然是永远都不可能碰面的。为了帮助这两只乐观的青蛙，你

DescriptionGiven a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N. Two integers are said to be co-prime or relatively prime if they

DescriptionGiven 5 integers: a, b, c, d, k, you’re to find x in a…b, y in c…d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large,

Description给出一个数N，求1至N中，有多少个数不是2 3 5 7的倍数。 例如N = 10，只有1不是2 3 5 7的倍数Input输入1个数N(1 <= N <= 10^18)。Output输出不是2 3 5 7的倍数的数共有多少。Sample Input10Sample Output1Hint题意题解：容斥原理 奇加偶减 求1~n中一些数的倍数的个数 如果是求m与1~n中互质的数的个

DescriptionWhile skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed that the telephone number of his brother-in-law H. Smith had the following peculiar

DescriptionThere are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see. If two trees and Sherlock are in one

DescriptionHOHO，终于从Speakless手上赢走了所有的糖果，是Gardon吃糖果时有个特殊的癖好，就是不喜欢将一样的糖果放在一起吃，喜欢先吃一种，下一次吃另一种，这样；可是Gardon不知道是否存在一种吃糖果的顺序使得他能把所有糖果都吃完？请你写个程序帮忙计算一下。 Input第一行有一个整数T，接下来T组数据，每组数据占2行，第一行是一个整数N（0Output对于每组数据，输出一

杭电6053TrickGCDTime Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2187 Accepted Submission(s): 846Problem DescriptionYou are given an array A , a

Description有一个序列是这样定义的：f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. 给出A，B和N，求f(n)的值。Input输入3个数：A,B,N。数字之间用空格分割。(-10000 <= A, B <= 10000, 1 <= N <= 10^9)Output输出f(n)的值。Sample Input3

Descriptionqwb又遇到了一道题目：有一个序列，初始时只有两个数x和y，之后每次操作时，在原序列的任意两个相邻数之间插入这两个数的和，得到新序列。举例说明：初始：1 2 操作1次：1 3 2 操作2次：1 4 3 5 2 …… 请问在操作n次之后，得到的序列的所有数之和是多少？Input多组测试数据，处理到文件结束（测试例数量<=50000）。输入为一行三个整数x，y，n，相邻两个

Description小Kevin从没有听说过单词“无限”。于是他请导师给他解释这个词。他的导师知道“无限”是一个非常大的数。为了展示无限可能有多大, 他的导师给了他一个挑战: 求从1到 N的数的和。和数开始变得非常大，并且已经超出了“long long int”的表示范围。于是，这一课非常清晰。现在，他的导师引入了取余的概念并让他只保留余数而非巨大的数。 然后, 导师给了他一个计算的公式:Inpu

Descriptionqwb同时也是是之江学院的志愿者，暑期要前往周边地区支教，为了提高小学生的数学水平。她把小学生排成一排，从左至右从1开始依次往上报数。玩完一轮后，他发现这个游戏太简单了。于是他选了3个不同的数x,y,z；从1依次往上开始报数，遇到x的倍数、y的倍数或z的倍数就跳过。如果x=2,y=3,z=5；第一名小学生报1，第2名得跳过2、3、4、5、6，报7；第3名得跳过8、9、10，报1

DescriptionIt’s said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first mystery.Aladdin was about to enter t

Joseph likes taking part in programming contests. His favorite problem is, of course, Joseph’s problem. It is stated as follows. There are n persons numbered from 0 to n - 1 standing in a circle. The

Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students, so he asked his assistant Bi-Shoe to go

Find the result of the following code:long long pairsFormLCM( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) for( int j = i; j <= n; j++ ) if( lcm(i, j) == n

题目大意.n个人报数从左报到右再从右报到左 知道所有人都报到k的倍数思路找规律发现 k*a-（2*n-2）*b = c 要求c取到1到n所有数 这就要求gcd（k,2*n-2）==1 否则取不到#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<stack>#include<queue>