1090 sdr

http://www.lydsy.com/JudgeOnline/problem.php?id=1090Casually yy.Set F[I,J] Indicates the minimum length of the I~jF[i, J]=min{j-i+1, F[i,k]+f[k+1, J], Count[x]+2+f[i, I+x-1]}, where count[x] represents the number of bits of x, and the last side of the transition that condition is that i~j are concatenated together in length X.Then the last transfer of violence 233 can be water over ...#include 　　 DescriptionThe definition of folding is as follows: 1. A string can be seen as its own folding

-root[l,r]:=i; -F[l,r]:=dfs (l,i-1) *dfs (i+1, R) +F[i,i]; the End; - End; - exit (F[l,r]); - End; + - procedureprint (l,r:longint); + begin A ifL>r Thenexit; atWrite (Root[l,r],' ');//output pre-order traversal -Print (l,root[l,r]-1); -Print (root[l,r]+1, R); - End; - - begin in read (n); -Fillchar (F,sizeof (f), Byte (-1)); to fori:=1 toN Doread (f[i,i]); + fori:=1 toN Do - begin thef[i,i-1]:=1;//If the tree does not exist fill 1 *root[i,i]:=i;//the root of a

A tree DP, the equation is in the program, pay attention to the sequence and initialization of the loop.#include #include #include using namespace Std;Long Long dp[35][35],n,w[35],c[35][35];void find (Long long left,long long right){if (left>right) return;printf ("%d", c[left][right]);if (left!=right){Find (LEFT,C[LEFT][RIGHT]-1);Find (C[left][right]+1,right);}Return}int main (){Memset (Dp,0,sizeof (DP));Memset (C,0,sizeof (c));scanf ("%lld", n);For (Long long i=1;iscanf ("%lld", w[i]);For (Long

#include the "iostream" #include "memory.h" #include "Cstdio" using namespace Std;int grap[500][500];int ans;int dis[500]; BOOL Visited[500];void Prime (int n) {int source = 1;for (int i = 1; I 1090. Highways

,sizeof(been)); for(LL i=pos+1; i) { if(been[i]==true)Continue; Vector:: Iterator it; if(Binary_search (Vv.begin () +pos+1, Vv.end (), K-Vv[i])) It= Lower_bound (Vv.begin () +pos+1, Vv.end (), K-Vv[i]); Else Continue; if(It!=vv.end () *it!=Vv[i]) {All_sign=true; if(*it>Vv[i]) printf ("%lld%lld%lld\n",-k,vv[i],*it); Elseprintf ("%lld%lld%lld\n",-k,*It,vv[i]); Been[it-vv.begin ()] =true; } }}intMain () {LL temp,i; scanf ("%lld",N); for(i=0; i) {scanf ("%lld",temp); Vv

#include#includeUsingnamespace Std;Constint max=100010;int depest=0,times=0;vectorInt> Child[max];child[father [Child] two-dimensional array representation tree structurevoid DFS (int root,int depth) {if (child[root].size () = =0)The leaf node begins to judge the depth {if (depth>depest) {depest=depth; times=1; }Elseif (depth==depest)The same as the deepest depth, the number of times plus 1 ++times;Return }Forint i=0; I1);}int main () {int root=-1,n;Double p,r; scanf"%d%lf%lf", n,p,r); R/=100;in

/*based on the nature of the middle sequence traversal and the quasi-recursive processing of the subtree well, it's a stone merge. However, the boundary is not always written and the cycle sequence is written in the memory .... The optimal value of each interval is separated by that point, that is, who is the root of the subtree and then recursively outputs to note that the left subtree is handed to the right subtree .*/#include#include#include#defineMAXN 35using namespacestd;intN,f[maxn][maxn],

A+b for Input-output Practice (II)Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 73325 Accepted Submission (s): 47653Problem Descriptionyour task is to Calculate A + B.Inputinput contains an integer n on the first line, and then n lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line. Outputfor each pair of input integers A and B are should output the sum of a and B in one line, a

The main topic: give a string, without changing the contents of the string can be folded, specifically see in the title. Ask how long the shortest string can be folded into.Idea: Data range is only 100, how violent how to engage. The first is an interval DP, set F[I][J] for the string from I start to j the shortest can be folded into how short. To use the fold in the body of the method, in fact, only need to force enumeration of the paragraph folded into a few paragraphs, and then a hash to dete

//////////////////////////////////////// //////////// 1090 mathematical questions// Obtain the circumference of the circle determined by the three points.// First use the cosine theorem cos α = (B ^ 2 + C ^ 2-A ^ 2)/(2bc) and then calculate sin α, then the diameter d = A/sin α, perimeter L = D π# Include # Include # Include Using namespace STD;# Define PI 3.141592653589793 Int main (){Double X1, Y1, X2, Y2, X3, Y3;Double wer1, wer2, wer3;Double A,

51nod 1090 3 count and 0 amp; 51nod 1267 4 count and 0 (marked as binary) Question meaning: The sum of 3 is 0: An unordered array with the length of N is given. The elements in the array are integers, positive and negative values include 0, and they are not equal to each other. Find the combination of all three numbers with and = 0. If there is No such combination, No Solution is output. If there are multiple, the minimum number of three numbers is

-1. All the numbers in a line is separated by a space.Output Specification:for all test case, print on one line the highest price we can expect from some retailers, accurate up to 2 decimal pl Aces, and the number of retailers that sell at the highest. There must be one space between the numbers. It is guaranteed, the price would not be exceed ten.Sample Input:9 1.80 1.001 5 4 4-1 4 5 3 6Sample Output:1.85 2AC Code:In fact, the height of the tree and the height of the number of leaf nodes, direc

1090.Highest Price in Supply Chain (25) pat-al-1090 2017-03-03 DFS and BFS should all be implemented, this time the DFS /** * pat-al-1090 * 2017-03-03 * Cpp version * author:fenglian_s */ #include -fin-

1090. Highest price in Supply Chain (a) time limit MSMemory Limit 65536 KBCode length limit 16000 BStandard author CHEN, Yue A supply chain is a network of retailers (retailer), distributors (reseller), and suppliers (suppliers)--everyone involved in moving a produ CT from supplier to customer. Starting from one root supplier, everyone on the chain buys products from one ' s supplier in a price P and sell or Distribu Te them in a and is r% higher than

Description: Give you a string of paths, such:A \ B \ cA \ D \ eB \ CSTD \You can draw out the directory structure contained in these paths. The subdirectories are directly listed under the parent directory, which is one cell to the right

// The radius can be easily obtained using the cosine and sine formula! # Include "iostream" # include "cmath" # include "cstdio" using namespace STD; const double Pi = 3.141592653589793; int main () {double X1, Y1, X2, Y2, x3, Y3; double A2, B2, C2,

Use merge sort to find the reverse order logarithm, only need to add cnt = n1-i in the proper place of the native merge sort OK. Well understood. # Include # Include Using namespace std;Const int MAXM = 10003;Const int INF = 0x7fffffff-1;Int cnt;//

The question is to find the longest side. When the Minimum Spanning Tree is generated, use a variable to save the maximum value.   // Prim Algorithm # Include # include # define INF 1000000 # define max 200 using namespace STD; int main () {int

A 0 1 string requires you to change all to 0. The change rule is that when the nth string is changed, the n-1 string must be 1, and all the values before n-1 must be 0, n> = 3; 1st can be changed at will, and 2nd must be changed when 1st is set to 1.

Problem DescriptionYour task is to Calculate A + B. InputInput contains an integer n in the The "the", and then n lines follow. Each line consists of the a pair of integers a and b, separated by a space, one pair of integers per line. OutputFor