ZhouRush

题意:将大数进行因式分解（分解质因数） 注意题目的数据范围 The numbers will fit in gcc'slong long int datatype. You may assume that there will be at most one factor more than 1000000. 参考网上的。。处理的确实妙极了。。http://blog.youkuaiyun.com/shankel

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=12&problem=966&mosmsg=Submission+received+with+ID+10537466 10025 - The ? 1 ? 2 ? ... ? n = k problem Tim

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=7&problem=514&mosmsg=Submission+received+with+ID+10536992 573 - The Snail Time limit: 3.000 seconds T

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=152 216 - Getting in Line Time limit: 3.000 seconds   Getting in Line Computer networking requires that t

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=82 146 - ID Codes Time limit: 3.000 seconds ID Codes It is 2084 and the year of Big Brother has fin

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=725 784 - Maze Exploration Time limit: 3.000 seconds Maze Exploration A maze of rectangular rooms i

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=532 591 - Box of Bricks Time limit: 3.000 seconds Box of Bricks Little Bob likes playing with his b

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=49 113 - Power of Cryptography Time limit: 3.000 seconds Power of Cryptography Background Current

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1361 10420 - List of Conquests Time limit: 3.000 seconds Problem B List of Conquests Input:standard input

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1047 10106 - Product Time limit: 3.000 seconds Product The Problem The problem is to multiply two i

函数名: atof 　　功 能: 把字符串转换成浮点数 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=406 465 - Overflow Time limit: 3.000 seconds Overflow Write a progra

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=365 424 - Integer Inquiry Time limit: 3.000 seconds Integer Inquiry One of the first users of BIT'

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=12&problem=955&mosmsg=Submission+received+with+ID+10538028 10014 - Simple calculations Time limit: 3.000

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=21&problem=1911&mosmsg=Submission+received+with+ID+10538116 10970 - Big Chocolate Time limit: 3.000 seco

就是按照给的公式做就行了。。  #include #include #include #include using namespace std; char s[1010]; int main() {  while(scanf("%s",s)!=EOF)  {   if(s[0]=='0')    break;   int len=strlen(s);   int i,j;   unsigned

题意：已知：方程 对于任意正整数k，则存在这个关系式： 其中，q(x)称为p(x)的商，r是余数。 给出p(x)的各项系数及k，求q(x)的各项系数，及r。 刚开始实在不会倒是头脑里面有个思路就是按照解方程的步骤求解。比如：题目中的 p(x) = x3 - 7x2+ 15x - 8 and k = 3 then q(x) = x2 - 4x + 3 and r = 1. 我们手动模拟可以发现要得

题意：就是上面给出N个点，下面给出M个点，上面的每一个点跟下面的给一个点都有连线，求连线的交点数的最大值。 公式不是自己想出来的。 摘自别人的bolg 因为一个交叉点的存在依赖于两条直线，四个顶点，因此用顶端中a个数字选出2个顶点共有a*(a-1)/2,同理于b。。 详细推理过程：http://hi.baidu.com/knowledgetime/blog/item/53fe431ddbe091f

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1108 10167 - Birthday Cake Time limit: 3.000 seconds Problem G. Birthday Cake Background Lucy and L

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=991 10050 - Hartals Time limit: 3.000 seconds Problem D: Hartals A social research organization has

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=92 156 - Ananagrams Time limit: 3.000 seconds Ananagrams Most crossword puzzle fans are used to anagrams-

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1415 10474 - Where is the Marble? Time limit: 3.000 seconds Where is the Marble? Raju and Meena lov

10098 - Generating Fast Time limit: 3.000 seconds Problem C Generating Fast, Sorted Permutation Input: Standard Input Output: Standard Output   Generating permutation has always been an important p

题意：在一个n*m的矩形中最少放置多少个声纳可以覆盖整个矩形，注意矩形的边界可以不要覆盖（因为水怪个头太大the border cells do not need to be controlled, because Nessy cannot hide there (she is too big).） 这题自己想的太高兴了1Y呵呵。。 就是自己手动模拟发现要使放置的声纳个数最小的话就是先考虑一行的结

题意：把一个球体切成n块后（按照表面积计算）可以额外获得的利益是百分之多少。。 球的表面积 ：4*π*r*r， 分别后多出的面积 ：n*π*r*r，利润 ：25*n ％ 。 按表面积分计算利润 初始球的表面积4*pi*r*r n=2；切一刀多了2个半圆面2*pi*r*r profit 50% n=3；多三个半圆面 profit 75% n=4；profit 100% 所以分成n份多n个半圆面。 #

题意： 给你一个个阶梯，要求起步和停止的步长为一，每次只能在上次的基础上增加一，减少一或者不变。求最少的步长。 直接模拟就好，分别从头尾开始增加步长。 #include #include using namespace std; int main() {  int t;  scanf("%d",&t);  while(t--)  {   int start,end;   scanf("%d%d

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1819 10878 - Decode the tape Time limit: 3.000 seconds Problem ADecode the tape Time Limit: 1 second  

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=951 10010 - Where's Waldorf? Time limit: 3.000 seconds Where's Waldorf? Given a m by n grid of letters, (

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=103 167 - The Sultan's Successors Time limit: 3.000 seconds The Sultan's Successors The Sultan of

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1039 10098 - Generating Fast Time limit: 3.000 seconds Problem C Generating Fast, Sorted Permutation Inp

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=945 10004 - Bicoloring Time limit: 3.000 seconds Bicoloring In 1976 the ``Four Color Map Theorem"

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=88 152 - Tree's a Crowd Time limit: 3.000 seconds Tree's a Crowd Dr William Larch, noted plant psyc

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=276 340 - Master-Mind Hints Time limit: 3.000 seconds Master-Mind Hints MasterMind is a game for tw

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1756   10815 - Andy's First Dictionary Time limit: 3.000 seconds Problem B: Andy's First Dictionary

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=15&problem=1302&mosmsg=Submission+received+with+ID+10379415 10361 - Automatic Poetry Time limit: 3.000 s

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=350 409 - Excuses, Excuses! Time limit: 3.000 seconds Excuses, Excuses! Judge Ito is having a probl

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=342 401 - Palindromes Time limit: 3.000 seconds   Palindromes A regular palindrome is a string of numbers

208 - Firetruck Time limit: 3.000 seconds Firetruck The Center City fire department collaborates with the transportation department to maintain maps of the city which reflects the current status of th

Input: standard input Output: standard output Time Limit: 1 second Memory Limit: 32 MB Your task is to write a program that can decide whether you can find an arithmetic expression consisting of five

1.Which is less then or equals 2^32-1。。。n不能定义为int。。。 2.in i'th walk he toggles only the switches whose serial is divisable by i. 第i次走的时候改变开关是那些只能被i整除的数。。。这样如果要最后一个开关是开的这能是完全平方数。因为一开始开关都是关的，要让他开只能改变奇数次

题意：求n！从有数起输出第一个不为0的数子。。 思路：就是求n!在处理一步就是结果了。。 n!求法白书p72... #include #include #include #include using namespace std; const int maxn=10010; int f[maxn]; int main() {  int i,j,n;  while(scanf("%d",&n)!=E