8 puzzle problem game

Surface:You are now asked to write a program to simulate the roll of the dice. The dice do not slide or beat, only on the table in the four-direction scrolling, that is, east. At the beginning of each game, the player makes the dice with a nominal value of one, north, west. The relative sides of the dice are all 7. Your program can accept a series of input commands, the command content is north,south,east,west. As North, the dice roll northward, the n

Title Link:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1511The main topic: In a 8*8 board, give you a starting position and an end location, but also give you a trap location, ask you from the beginning to bypass the trap to the end of the shortest distance. (You can walk up and down or you can walk diagonally)Problem-solving ideas: can be directly with the

The game suit began to serve, the result reported disorderly seven or eight strokes of error, first CCS that IP has a problem, I gave directly commented out, and then reported Keyvaluedictcache in the IPs set problems, are reported format error, the results of my breakpoint is clearly digital results in judging the number that is still an error, the result I want to estimate Meter is a tm coding

Problem Description8 Ball is a game of billiards rules. The table has 7 red balls, 7 yellow ball and a black ball, and of course a white ball. For the subject, we use the following simplification rules: Red, yellow two players take turns with the white ball to hit their own color of the ball, if the color of the 7 balls in all, then the player can play black ball, if the hit in the count he wins. If you put

= 2.The previous point, because of the distance limit, may only jump to index = 1, not to index = 2, 3. So the second-to-last step is set at index = 1 to get the most solution.1 PackageAlgorithms.greedy;2 3 Public classJump {4 Public Static voidMain (string[] strs) {5 int[] A = {2, 3, 1, 1, 4};6 System.out.println (Jump (A));7 }8 9 Public Static intJumpint[] A) {Ten if(A = =NULL|| A.length = = 0) { One

Description Small h is a person who likes to think, so he is also very happy to challenge various thinking topics. Today, his friend small L came to him to play, while small l also brought a simple game, small l gave a piece of paper divided into WxH cell, and then told Small H, now requires small h with wxh of paper (not allowed to rotate) on the sheet, ask the small H up to put how many pieces of paper, So that no two pieces of paper overlap.Of cour

output in the simplest fractional form, the first is the numerator and the second is the denominator."Input Sample"32 3 5"Output Example"22 3the sample Tips "There are 6 possible travel examples:[2, 3, 5]: The distance to travel: |2–0| + |3–2| + |5–3| = 5;[2, 5, 3]: |2–0| + |5–2| + |3–5| = 7;[3, 2, 5]: |3–0| + |2–3| + |5–2| = 7;[3, 5, 2]: |3–0| + |5–3| + |2–5| = 8;[5, 2, 3]: |5–0| + |2–5| + |3–2| = 9;[5, 3, 2]: |5–0| + |3–5| + |2–3| =

0X3FFFFFFusing namespacestd;intN;intC[n];//the travel expenses of door i-1 to gate I are CIintM[n];//minimum tolls for road before gate IintT[n];//every time the door opensintMain () {scanf ("%d",N); for(intI=1; i1; i++) {scanf ("%d",C[i]); if(i==1) m[i]=C[i]; ElseM[i]=min (m[i-1],c[i]); } for(intI=0; i) {scanf ("%d",T[i]); } inttt=0;//Current Moment intI=0; Long Longret=0; while(iN) {i++; TT++; RET+=(Long Long) (C[i]); intTmp=t[i]-tt;//how long till the door is open?

the swollen level AI and water level bi for the flood of the first I, and count how many bridges have been flooded at least k times. The initial water level is 1, and the swollen water level of each flood must be greater than the water level of the last flood.InputThe input file contains up to 25 sets of test data. The first behavior of each group of data is three integers n, m, K (1OutputFor each set of data, output the number of bridges that have been flooded at least k times.Sample Input2 2

=L8Z022v3MVV7Ba0N13ghO9geTjmCsueAHgLFqG0Igw5TH3SraQG7smVuAkuC0fqWj1_dlBC_0OSRlnbJG6qq0KProve:1) When the n2) assuming that n3) then n=p (k+1), because P (k+1) is the Fibonacci number, can be expressed as the first two Fibonacci number P (k-1), p (k) and the sum of the Yi Certificate p (k-1) >p (k)/3, that a 1th times the number is necessarily less thanP (k), from the assumption that P (k-1) is a must lose state, this must lose the state actually has a meaning, that is, B must be able to get P (k

" Packagechapter1youxizhilechinesechess;/*** Chinese Chess Generals problem * "solution two" *@authorDELL **/ Public classChineseChess2 { Public Static voidMain (string[] args) {//byte i = Bayi;//While (i!=0) {//if (i/9%3!=i%9%3)//System.out.printf ("a=%d, b=%d\n", i/9+1,i%9+1);//i--;// } bytei = 1; while(i!=80){ if(i/9%3!=i%9%3) System.out.printf ("A=%d, b=%d\n", i/9+1,i%9+1); I++; } }}"Solution three"Some say it is the m

The problem defines a new sort algorithm, which is to put a number in a sequence of sequences, if its right number is smaller than itYou can move to the right of the past until the number on the right is larger than it.Easy to get, if the simulation is O (n^2) efficiency, certainly notThinking about it, this problem can be translated intoFor each element in this sequence, the right side of the string is les

Divides a string into n strings each with a string length of MSample Input12 5//N mKlmbbileaySample OutputKlmbbIleay1# include 2# include 3# include 4# include 5# include string>6# include 7# include 8# include 9# define LLLong LongTen using namespacestd; One A Chars[ -] ; - - intMain () the { - //freopen ("In.txt", "R", stdin); - intT; -scanf"%d", T); + while(t--) - { + intN, M; Ascanf"%d%d", n,m); at intI, J; -

Title Link: http://acm.zzuli.edu.cn/zzuliacm/problem.php?cid=1158pid=0Sample Input 2 6 4 2 1 Sample Outputnoyes************************************************1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9 Ten using namespacestd; One A #defineN 5500 - #defineINF 0x3f3f3f3f - the intMain () - { - intT,n,k,ans; - +scanf"%d", T); - + while(t--) A { atscanf"%d%d", n,k); - - if(k==1) - {

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5122Problem-Solving report: Define a sorting algorithm, each round can be randomly find a number, the number and the subsequent smaller than this number of exchanges, has been judged, until there is no smaller than this number, so called a round, now given a sequence of length n, you ask, at least through how many rounds, you can make this sequence into order.Because we can only compare with the number of the following, so I just need to coun