SGU 117.Counting】的更多相关文章

时间限制: 0.25 sec. 空间限制: 4096 KB 题目大意: 给你n,m,k(都小于10001),和 n 个数,求这n个数中有多少个数的m次幂能够整除k.(即 n i^m % k==0). solution: 快速幂取余. 参考代码 #include <iostream> using namespace std; __int64 Quikpower(__int64 a,__int64 d,__int64 n){ __int64 k = ; ){ ) k = (k*a)%n; a=(a…
题目大意:求下面N个数里面有多少个数的M次方能整除K 代码如下: ======================================================== #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ; ; int QuickPow(int a, int b, int Mod) { ; while(b) { ) t = (t * a)…
SGU还是个不错的题库...但是貌似水题也挺多的..有些题想出解法但是不想写代码, 就写在这里吧...不排除是我想简单想错了, 假如哪位神犇哪天发现请告诉我.. 101.Domino(2015.12.16) 102.Coprimes 求φ(N). 1<=N<=10^4 按欧拉函数的公式直接算..O(N^0.5)(2015.12.16) #include<cstdio> #include<cstring> #include<algorithm> using n…
http://acm.sgu.ru/problemset.php?contest=0&volume=1 101 Domino 欧拉路 102 Coprime 枚举/数学方法 103 Traffic Lights 最短路 104 Little Shop of Flowers 动态规划 105 Div 3 找规律 106 The Equation 扩展欧几里德 107 987654321 Problem 找规律 108 Self-numbers II 枚举+筛法递推 109 Magic of Dav…
117. Counting time limit per test: 0.25 sec. memory limit per test: 4096 KB Find amount of numbers for given sequence of integer numbers such that after raising them to the M-th power they will be divided by K. Input Input consists of two lines. Ther…
SGU 解题报告(持续更新中...Ctrl+A可看题目类型): SGU101.Domino(多米诺骨牌)------------★★★type:图 SGU102.Coprimes(互质的数) SGU103.TrafficLights(交通灯)---------★★type:图 SGU104.LittleShopofFlowers(小花店)---★type:DP SGU105.Div3(整除3) SGU106.TheEquation(方程)-------------★type:数论 SGU107.…
1170 - Counting Perfect BST   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB BST is the acronym for Binary Search Tree. A BST is a tree data structure with the following properties. i)        Each BST contains a root node…
建议114:不要在构造函数中抛出异常 Java异常的机制有三种: Error类及其子类表示的是错误,它是不需要程序员处理也不能处理的异常,比如VirtualMachineError虚拟机错误,ThreadDeath线程僵死等. RunTimeException类及其子类表示的是非受检异常,是系统可能会抛出的异常,程序员可以去处理,也可以不处理,最经典的就是NullPointException空指针异常和IndexOutOfBoundsException越界异常. Exception类及其子类(不…
在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…
来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…
ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…
PHP使用openssl进行Rsa加密,如果要加密的明文太长则会出错,解决方法:加密的时候117个字符加密一次,然后把所有的密文拼接成一个密文:解密的时候需要128个字符解密一下,然后拼接成数据. 加密: /** * 加密 * @param $originalData * @return string|void */ /*function encrypt($originalData){ // if (openssl_private_encrypt($originalData, $encryptD…
#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…
1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…
Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…
Problem A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains. An example of a length 21 DNA string (whose alphabet contains the symbol…
// uva 11401 Triangle Counting // // 题目大意: // // 求n范围内,任意选三个不同的数,能组成三角形的个数 // // 解题方法: // // 我们设三角巷的最长的长度是c(x),另外两边为y,z // 则由z + y > x得, x - y < z < x 当y = 1时,无解 // 当y = 2时,一个解,这样到y = x - 1 时 有 x - 2个 // 解,所以一共是0,1,2,3....x - 2,一共(x - 2) * (x - 1…
当我们在使用JSONKit处理数据时,直接将文件拉进项目往往会报这两个错“JSONKit   does not support Objective-C Automatic Reference Counting(ARC)”,“ARC forbids Objective-C objects in struct”,这是由于JSONKit库未更新,不支持ARC机制.我们可以参照如下步骤解决:…
有使用JSonKit的朋友,如果遇到“JSonKit does not support Objective-C Automatic Reference Counting(ARC)”这种情况,可参照如下方法: 点击项目根目录->targets->Build Phases->JSONKit.m->添加“-fno-objc-arc”字段,在运行就OK了.…
Lake Counting Time Limit: 1000MS     Memory Limit: 65536K Total Submissions: 17917     Accepted: 9069 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <=…
水概率....SGU里难得的水题.... 495. Kids and Prizes Time limit per test: 0.5 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the…
Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each…
sgu 101 - Domino Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Description Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The bl…
http://acm.sgu.ru/problem.php?contest=0&problem=495 题意:N个箱子M个人,初始N个箱子都有一个礼物,M个人依次等概率取一个箱子,如果有礼物则拿出礼物放回盒子,如果没有礼物则不操作.问M个人拿出礼物个数的期望.(N,M<=100000) #include <cstdio> using namespace std; double mpow(double a, int n) { double r=1; while(n) { if(n&…
Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Total Submission(s): 2811    Accepted Submission(s): 827 Problem Description In this problem we consider a rooted tree with N vertices. The vertices a…
Counting Squares Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 946 Problem Description Your input is a series of rectangles, one per line. Each rectangle is sp…
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2610 Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith ele…
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=455 Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not gua…
转载地址:http://blog.csdn.net/xbl1986/article/details/7216668 Xcode是Version 4.2 Build 4D151a 根据Objective-c 2.0程序设计上的旧版本的代码会发生NSAutoreleasePool' is unavailable: not available in automatic reference counting mode的错误 需要手动关闭工程中ARC 工程中 Build Settings--->Apple…
Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Problem Description In this problem we consider a rooted tree with N vertices. The vertices are numbered from 1 to N, and vertex 1 represents the root…