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Rightmost DigitTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 33558 Accepted Submission (s): 12831Problem Descriptiongiven A positive integer N, you should output the most right digit of n^n.Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.Each test case

Title: http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblemproblemid=3138Algorithm idea: multiply 1-10 two number to produce 0, only 10x1=2x5,2x5. 200! =200x199x198......x2x 1=2x5x2x5x2x2x1, can be decomposed into prime number multiplication form 2 more than 5 (2*5=10) 200 the factorial can be decomposed out of the number of 200 the factorial end of the number of consecutive 0 That is : NUM=[200/5]+[200/5/5]+[200/5/5/5].Note : [x] indicates rounding to x . Code:#include #includestri

Transferred from: http://www.cnblogs.com/ranranblog/p/5845010.htmlThe pigs can fly under the Tuyere. The bull market in China today is a "missed seven years". Give you an opportunity to review the history of a stock, known for a continuous n-day price movement, in the length of an integer array of n, the array of elements I (prices[i]) represents the stock price of the first day. Suppose you don't have a st

The title means: X is double,n int, and x's n-th square is obtained.Thinking Analysis: Direct seeking, attention to critical conditions1 classSolution {2 Public:3 DoubleMypow (DoubleXintN) {4 if(x==1.0)returnx;5 Else if(x==-1.0){6 if(n%2==0)return 1.0;7 Else return-1.0;8 }9 Doubleans=1.0;Ten int

First, we all know that we can create n elements as a maximum (large | small) Heap and O (n ). The following common practice is to remove an element from the top of the heap and take it K times. Then, the first k elements are taken out. But the problem is that after each step removes an element, it takes the log (n) Time to maximize (large | small) the heap. So

# include # include Knuth (intint m) { int) time (0)); for (int0; i ) { if (rand ()% (N-i) m) { Endl; M--; }} }To facilitate interpretation, assume that n equals 10,m equals 5:The remainder range of the first rand ()% (n-0) is 0~9, which is likely to be less than M (=5) and can be output i=0; then i++,m--The r

Reprinted please indicate the source, thank youHttp://blog.csdn.net/acm_cxlove/article/details/7854526By --- cxlove Question: There is a point set that asks how many points a unit circle can cover. Http://poj.org/problem? Id = 1981 N ^ 3. A circle with the most vertices must have at least two points on the circle. Of course n> = 2 and the result is greater than 2 In this case, enumerate the two poin

LeetCode 135. Candy (O (n) time O (n) and O (1) space), leetcodecandy Greedy algorithms promote local optimization to global optimization. Here we will introduce two implementation methods of O (n) time O (n) and O (1) space. O (n) time O (

Longest Palindromic Substring [leetcode] O (n ^ 2) DP and O (n) algorithms, longestsubstring Two methods are implemented. One is DP, which is made of loops. recursion is easier. string longestPalindrome(string s) { int n = s.size(); bool dp[1001][1001]; int maxl = 1; int maxs = 0; for (int i =

Today I encountered a pen question: there are two arrays A and B, with the length of M and N respectively. Find the same element in the two arrays with the number of comparisons not greater than m + n. I didn't do it at the time. Now I am giving C #, which makes up a little regret. Using system; using system. collections. generic; using system. LINQ; using system. text; namespace sortab {class program {stat

Code A: Copy Code code as follows: Echo ' helloEcho ' world! '; ?> Output Helllo world! Code B: Copy Code code as follows: Echo ' hello\n ';//unix system using \n;windows system \ r \ n Echo ' world! '; ?> Output helloworld! Analysis : browsers do not recognize \ n or \ r \