movable ruler

After opening the IE8/IE9, we press the F12 key of the keyboard, open Developer tools, click Tools on the menu bar, select Show Color Picker, or press Ctrl+k to open the show Color picker when you click the Developer Tools area. This will absorb the color of any place in the page. After you draw the color, you can get the hexadecimal encoding of the color by clicking Copy and close on the color picker. Select Show ruler under Tools opt

The rulers in Word are by default in characters, and in practical applications it is not good for us to calculate the size of the paper. In many cases, such as printing documents, papers, pictures and so on need to specify the size, to be in Word in advance to set a long width, then we need to change the ruler unit to centimeters, to facilitate our confirmation size. 1. The unit of measure in the current document is a character. Click the Office Butt

Boring countTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 932 Accepted Submission (s): 382Problem descriptionyou is given a string S consisting of lowercase letters, and your task is counting the number of subst The ring, the number of each lowercase, the substring is no more than K.Inputin The first line there was an integer T, indicates the number of test cases.For each case, the first line contains a string which only consist of lowercase

24-segment magic ruler, which of the following exquisite patterns can be folded What are the exquisite patterns that can be stacked with the 24-segment magic ruler? The ball, the turtle, the cross, the three-leaf flower, the triangle, the ox lord, and the notes are quite familiar to everyone. I am more interested in this issue. First, we will show the 24 magic tricks on the Internet with exquisite patter

Topic Links:http://poj.org/problem?id=3320Test instructionsA book has P page, each page has a knowledge point AI, there are different two pages on the same knowledge points,Ask for a minimum number of consecutive pages of the book, can be all the knowledge points covered.Ruler extraction:The method of finding the minimum interval for satisfying a condition is called a ruler by repeatedly pushing the beginning and ending of the interval.The degree of c

O (n) to enumerate all segments with a length of K, each violent transfer. During the transfer, only one number is inserted from the end and one number is deleted from the front. Calculate the current Max and Min values. These operations are O (logn) with the Multiset (distinct element is rewritable. 1 #include [Ruler acquisition method] [Multiset] bzoj1342 [baltic2007] sound mute Problem

Test instructionsIn a sequence of length n, find the shortest length sequence so that it is greater than or equal to s;Ideas:pointer, water problem;AC Code:#include /*#include */using namespacestd;#definefor (i,j,n) for (int i=j;i#defineRiep (n) for (int i=1;i#defineRIOP (n) for (int i=0;i#defineRJEP (n) for (int j=1;j#defineRJOP (n) for (int j=0;j#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong Longll;templateclassT>voidRead (tnum) { CharCH;BOOLf=false; for(Ch=getchar (); ch'0'|| Ch>

A Bit FunTime limit:5000/2500 MS (java/others) Memory limit:32768/32768 K (java/others)Problem descriptionthere is n numbers in a array, as A0, A1 ..., an-1, and another number M. We define a function f (i, j) = ai|ai+1|ai+2| ... | Aj. Where "|" is the bit-or operation. (I The problem is really simple:please count the number of the different pairs of (I, J) where F (i, J) Inputthe first line has a number T (T For each test case, first line contains, numbers n and M. (1 Outputfor every case, you

Title Link: Http://codeforces.com/problemset/problem/660/CRuler, each time you encounter 0 of the time to fill a 1, until the completion or out of bounds. Each time a 0 point is recycled from left to right. The recording path is jammed with two pointers, each updated.1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten#include One#include A#include -#include -#include the#include -#include Set> -#include -#include + - using namespacestd; + A Con

POJ3320 Jessica ' s Reading problemSet is used to count the number of all non-repeating knowledge points, and map is used to maintain the number of occurrences of each knowledge point on the interval [s,t], which is a good representation of the flexible application of map　　#include #include#include#include#include#include#include#includeSet>#includestring>#includeusing namespacestd;Const intINF =0x3f3f3f3f; typedefLong Longll;Const intMax_p =1000010;intP;inta[max_p];intMain () {scanf ("%d", q);

A-Fried chicken want to lose weight of stringTime limit:MS Memory Limit:65535KB 64bit IO Format:SubmitStatusPracticenbut 1576DescriptionOnce upon a time, there was a string of long, long strings consisting of n lowercase letters.One day when it was looking in the mirror, he felt too fat, so he wanted to lose weight. When you lose weight, you can constantly remove the first or last character.After it wants to lose weight, for a slim factor m, the slim factor must contain ' a ', ' B ', ' C ' ...

) (-X) the #defineRead () freopen ("A.txt", "R", stdin) * #defineWrite () freopen ("B.txt", "w", stdout); $ #defineMAXN 1000000000Panax Notoginseng #defineN 510 - #defineMoD 1000000000 the using namespacestd; + Charstr[100010]; A intMain () the { + //Read (); -scanf"%s", str); $ intL=strlen (str); $mapChar,int>m; - ints=0, t=0, num=0, res=inf; - while(1) the { - while(t -)Wuyi { the if(m[str[t++]]++==0) num++; - //printf ("%d\n", num

Title: http://poj.org/problem?id=3320Test instructions: Given an array of n elements, find the shortest interval so that the element type within the interval equals the element type of the entire array.Analysis: The beginning of the violent enumeration interval x, and then find the smallest y, so that the interval [x, y] satisfies the condition, X shifted to the X ', now Y ' certainly not on the left of Y. Both map and hash can be used.Map Code:#include Hash code:#include POJ 3320 Jessica ' s Re

, t=1; thell sum=0; - for(;;) { - - while(sumN) { +sum=sum+t*T; -t++; + } A if((t1) * (t1) >N) at Break; - if(sum==N) { -Ans.push_back (Make_pair (s,t-1)); - } -sum-=s*s; -s++; in - } toll m=ans.size (); +printf"%i64d\n", m); - for(intI=0; i){ thell l=Ans[i].first; *ll r=Ans[i].second; $printf"%i64d", r-l+1);Panax Notoginseng for(LL j=l;j){ -printf"%i64d", j); the }

Poj3320 (ruler acquisition) and poj3320 N number. The minimum interval overwrites all the different numbers in n number. Solution: AC code: Import java. util. hashMap; import java. util. hashSet; import java. util. map; import java. util. imports; import java. util. set; public class Main {/*** @ param args */static int n; static Set 　　

Jessica‘s text-book. The second line contains P non-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type. Output Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book. Sample Input 51 8 8 8 1 Sample output 2 AC cod

Subsequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9050 Accepted: 3604 DescriptionA sequence of n positive integers (InputThe first line is the number of test cases. The program have to read the numbers N and S, separated by a interval, from the first line. The numbers of the sequence is given in the second line of the "test case", separated by intervals. The input would finish with the end of file.OutputThe the

A series integer a0,a1......an-1 and integer s with a given length of N. The minimum value of the length of a continuous sub-sequence with a sum not less than S is calculated. If the solution does not exist, the output is 0.Input n=10S=15a=[5,1,3,5,10,7,4,9,2,8]Output2 [5,10]functionsolve () {varRes=n+1; varS=0,t=0,sum=0; for(;;) { while(tS) {Sum+=a[t++]; } if(Sum Break} res=math.min (res,t-s); Sum-=a[s++]; } if(res>N) { //solution does not existRes=0; } returnRes;}Do the f