castlevania puzzle

http://poj.org/problem?id=3678Test instructions: Very naïve topic to see the english problem directly =#include 　　Just screw it up. = ="POJ" 3678 katu Puzzle

, dynamic planning many topics to do problems in the way there is common: first we have to modify the operation to design the state of the problem, the state must be no Overlap exists individually (that is, for each element in the dynamic planning array corresponds to only one state), and we are able to get the answer through a linear shift. Some states are obvious, easy to design, such as "chorus Formation", "Turtle Chess", "energy necklace" feeling to be able to see the approximate model, but

Topic linksGive a string number, take out a number i the price is a[i-1]*a[i]*a[i+1], the first number and the last number can not be taken, to take away the remaining number of the minimum value of the price.#include #include#include#include#include#include#include#includeSet>#includestring>#includeusing namespacestd;#definePB (x) push_back (x)#definell Long Long#defineMK (x, y) make_pair (x, y)#defineLson L, M, rt#defineMem (a) memset (a, 0, sizeof (a))#defineRson m+1, R, rt#defineMem1 (a) mem

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Chars[ -];5 intN;6 intFlag;7 intans;8 9 voidDFS (intLevel//possibility of Dfs search +-= appearingTen { One A if(level==n-1) - { - intI=0, l=a[0],r=0; the - while(s[i]!='='i1) { - - if(s[i]=='+') l+=a[i+1]; + Else if(s[i]=='-') l-=a[i+1]; - +i++; A at } - - while(i1) { - - if(s[i]=='+') r+=a[i+1]; - Else if(s[i]=='-') r-=a[i+1]; in Elser=a[i+1]; - toi++; + - } the *

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=4708Test data:51 9 9) 1 11 9 9) 9 99 9 9) 9 99 9 9) 9 11 1 9) 9 181 11#include 2#include 3#include 4 using namespacestd;5 Long Longa[ the][ the];6 Long LongStep,ans;7 voidSolve1 (intTintX1,intY1,intX2,intY2,intX3,intY3,intX4,inty4)8 {9 inti;Ten Long LongW; Onet = t1;//The range is wrong, WA to the end of the game is not right, alas A for(i=0; i) - { -W = a[x1][y1] + a[x2][y2]+a[x3][y3]+A[x4][y4]; they1++; -x2++; -y3--; -x4--;

Topic Links:http://hihocoder.com/problemset/problem/1165Test instructionsSelect two numbers from N to make them the number of factors that multiply.Analysis:For a number, we decompose it into a factor of element.x= (P1^A1) * (P2^A2) *....* (Pn^an);Then the number of factors for x is (a1+1) * (a2+1) *...* (an+1);Then we can preprocess the number of factors from 1 to 100000, and thenThe input data is ordered by the number of factors, and then the first 200 is enumerated.The number of combinations,

This problem, simple math problem.For the function, f (x) =5*x^13+13*x^5+k*a*x, enter K, for any x, there is a number a, so that f (x) is divisible by 65.for any x. So when x = 1 o'clock, f (x) = + A * k, the satisfaction is divisible by 65.i.e. (+ A * k)% 65 = 0.So, everything is very simple.The following is the code for the AC:#include Hangzhou Electric Acm1098--ignatius ' s puzzle

Problem DescriptionTorry Love maths from childhood. One day, the teacher told him, like 2, 3, 5, 7 ... Such numbers are called prime numbers. Torry suddenly thought of a problem, the first 10, 100, 1000, 10000 ... What is the product of a prime number? He told the teacher the question. The teacher was stunned and couldn't answer it. So Torry turned to the programmer, and asked you to calculate the product of the first n prime numbers. However, considering that you are only in touch with programm

minimum number of steps required to convert from "01234567" to another sequence A voidBFs () { at //Initialize -Queueque; -Que.push (P ("01234567",0)); -dp["01234567"] =0; - //Width First search - while(!Que.empty ()) { inP p =Que.front (); - Que.pop (); to strings =P.first; + intCur =P.second; - for(inti =0; I 4; i + +){ the //constructs the next Exchange * intNext = cur +Op[i]; $ stringstr =s;Panax Notoginseng swap (Str

1. Title Description: Click to open the link2. Problem-Solving ideas: The topic is the maze, the typical BFS solution. But first you should determine what the state is. According to test instructions description: The coordinates of the Sieve + the front and top surface of the sieve constitute a state. In this case, the subject is somewhat similar to the implicit graph search. So how does the state gradually expand? Since the sieve can be scrolled in four directions at a time, it is possible to e

(inti =0; I ) to { + intL =i; - while(L Tmp[i]) the++l; *Z *= l-i +1; $i = L-1;Panax Notoginseng } - returnZ; the } + A intFactors (intN) the { + if(1= = N)return 1; - intCNT =2; $ for(intj =2; J * J ) $ { - if(N% J = =0) - { thecnt++; - if(J * J! =N)Wuyicnt++; the } - } Wu returnCNT; - } About $ intMain () { - //freopen ("Input.txt", "R", stdin); - intT; - while(cin>>t) A { +

:32768 KB problem Descriptionto a "01" string of length n (n InputThe first line is an integer t (1 Outputminimum cost. Sample Input2121100120011Sample Output03Title Analysis: Two strategies, cross-change, adjacent, first the cost of the adjacent exchange is fixed, so a 1 and a 0, to exchange the cost of min (y, x * len)First processing all 1 positions and the number of CNT, we want to let the former CNT is 1, then from 0 to Cnt-1 enumeration, encounter a 0 then take the last one 1 and it exchan

Title Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1836Given a list of integers (A1, A2, ..., an), and a positive integer M, please find the number of positive integers that is Not greater than M and dividable by any integer from the given list.InputThe input contains several test cases.For the test case, there is and lines. The first line contains n (1 OutputThe For each test is in the input, and the result in a is output.Sample Input3 22 3 73 62 3 7Sample Output14 Aut

It is thought that the violence of the solution time complexity should be very high, maybe the cycle will be relatively small barThe better harvest is this Floyd.　　#include Set>#include#include#includeusing namespaceStd;typedefLong LongLL;Const intMAXN =100005;Const intINF =0x3f3f3f3f; LL POW10 (intN) {LL ans=1; for(inti =1; iTen; returnans;} LL m; ll next (ll K) {k*=K; while(k>=m) k/=Ten; returnK;}intMain () {//freopen ("In.txt", "R", stdin); intT;SCANF ("%d",T); while(t--) {LL n,k; CIN>>

PHP Pixel Puzzle, Niu Ge is a programmer who claims to be a designer! When you get an image, you just want to unload it in eight parts, one pixel and one pixel to split it and see what color values the image contains. This php class floated away from Niu Ge! SetImage ($ img); // You Need to initialize the pixel size to be displayed $ pro = $ px-> getProperty (); if (! $ Px-> error () {$ palete = $ px-> image2palete (); echo"Attributes Image:"; Ech

........nine in the string array A, we have to find out who the numbers in the numbers are in the input string (num). Very simple, the a array to go through the search for the corresponding string of the subscript and then return.int tra (char num[]) { int i; for (i=0;iWe're almost done with the procedure here.Put in a complete code:#include Next, the second method:We can use the map in the STL to map a string string and an int number.Map of the use of people can surf the internet Baidu, I

].x=sx,q[head].y=sy,map[sx][sy]=1;While (Head!=tail){int nx=q[head].x,ny=q[head].y;for (int i=1;iif (map[nx+vx[i]][ny+vy[i]]==0){map[nx+vx[i]][ny+vy[i]]=map[nx][ny]+1;Q[tail].x=nx+vx[i],q[tail].y=ny+vy[i];check (tail), tail++;}head++;}}int main (){freopen ("maze.in", "R", stdin);freopen ("Maze.out", "w", stdout);scanf ("%d%d", n,m);for (int i=1;ifor (int j=1;j{scanf ("%d", map[i][j]);map[i][j]= (!map[i][j])? 0:-1;}for (int i=1;iscanf ("%d%d%d%d", sx,sy,tx,ty);BFS ();printf ("No answer!");return

thing, mark it as 1. If we use a COR array tag, then the cor[i][j]=1 indicates that the block I is rotated clockwise after J times and the Block I pattern of the target State. After this preprocessing, the judgment function is not difficult to write. Just look at all I, whether there is a cor[i][st[i]]=1 (St[i] Indicates the first state value component).At this point, we successfully solved the 4 large implicit graph search problem, the remaining find the shortest is the ordinary BFS can be ach

d): M_pos (P), m_direction (d) {} void Move (int x) {if (m_direction > 0) M_pos + = x; else M_pos-= x; } void Turnback () {m_direction *=-1; } int m_pos; int m_direction;}; Ant Firstmeet (const qlist After spending a few minutes writing here, stop to think about giving the ant ID, doing all the things in all directions ... Consider the situation where not only a couple of ants meet at the same time, etc. ... I suddenly wake up .... This is not a program topic, if you have a l