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Problem Description There must be many A + B problems in our HDOJ , now a new one is coming. Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too. Easy ? AC it ! Input The input contains several test cases, please process to the end of the file. Each case consists of two hexadecimal integers A and B in a line seperated by a blank. The length of A and B is less than 15. Output For each test case,print the sum of A and B in hexadecimal in one line. Sample Input +A -A +1A 12 1A -9 -1A -12 1A -AA Sample Output 0 2C 11 -2C -90 解

题目 An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree. Input Specification: Each input file contains one test case. For each case, the first line contains a positive integer

After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it N rows with M trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the j -th tree in the i -th row would have the coordinates of ( i ,  j ). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation. The burning began in K points simultaneously, which means that initially K trees started to burn. Every minute the fire gets from the burning trees to

设dp_i为所求答案，每次选择因数的概率相同，设i有x个因数，dp_i=sum(1/x*x_j)+1,(x_j表示第j个因数)，那我们就预处理每个数的因数即可，T=10000，需要预处理出答案 #include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; typedef pair<int,int> pii; const int maxn = 1e5+5; vector<int> factor[maxn]; double dp[maxn]; void get_factor() { for(int i = 1; i < maxn; ++i) { for(int j = i; j < maxn; j += i) factor[j].push_back(i); } } void prework() { for(int i = 2; i < maxn; ++i) { int siz = factor[i].size(); for(int j = 0; j < siz-1; ++j) { dp[i] += 1.0/siz * dp[factor[i][j]]; } dp[i]++; dp[i] *= (double)siz/(siz-1); } } void run

题意：题目要求对n个数中的某些进行任意多次的+1操作，使得对于每个元素(除开第一个)有a[i-1]!=a[i],1<i<=n，由于每个a[i]对应一个b[i],没增加1，那么费用就相应增加1*b[i]，问使得在修改为满足条件a[i-1]!=a[i]的情况下最小的费用； 这个题我也是看了题解才明白的，由于每次增加1，所以最多一个元素可以增加2次1就能保证左相邻是不想等的了； 比如一个序列:1,1,1；那么我增加第二个1一次即可，如果是2,1,1,3,3；那么增加第一个1两次才行，所以这点就很重要了，一个元素只能增加0/1/2次；所以我需要把所有的情况全部枚举出来，这我感觉就是dp的思路了，就像01背包一样的感觉；01背包是对于这个物品我想么选，要么不选，所以需要枚举所有情况；而这道题是当前元素增加0次，1次或者2次，都是多种情况；注意题目中1e18，所以我需要设置无穷大为1e18多一点；对于dp那么我就需要枚举三种情况；具体一点就是：对于第一个篱笆，他有0,1,2三种情况，那么对应的第二个篱笆也是0,1,2三种情况，所以我需要两个for来枚举，由于一共有n的篱笆，所以我外面还需要一个for，并且在a[i-1]+j!=a[i]+k的情况下才能计算； 所以dp定义为dp[i][k]当修到第i个篱笆的时候增加k次的最小费用； # include <bits/stdc++.h> using

题意 题意翻译 对于一个字符串 \(s\) ，我们定义其 美丽值 \(f(s)\) 为满足下列两个条件的正整数 \(i\) 的个数： \(1\leq i<|s|\) \(s\) 长度为 \(i\) 的前缀与后缀相等，即 \(\forall j\in N,1\leq j\leq i\) ，均有 \(s_j=s_{|s|-i+j}\) 给定正整数 \(n(1\leq n\leq 10^5),k(1\leq k\leq 10^9)\) 。设字符集大小为 \(k\) ，请求出所有长度为 \(n\) 的字符串 \(s\) 的 \(f(s)^2\) 的期望值。 做法 长度为 \(x\) 的 \(border\) 等价于 \(n-x\) 非严格周期 另 \(g_i(s)\) 为字符串 \(s\) 有周期 \(i\) 的概率， \(f(s)^2=\sum\limits_{i=1}^{n-1}E(g_{i,j}(s))\) ， \(g_{i,j}(s)\) 表示字符串 \(s\) 同时具有 \(i\) 与 \(j\) 周期的概率 结论1 ：有计算概率公式： \[g_{i,j}(s)=\dfrac{k^{max(i+j-n,(i,j))}}{k^n}\] Periodicity Lemma 然后开始推式子，下面关于 即 ，省略了一些没必要的东西，就是那种不会影响计算的东西 即计算 \(\sum

题目链接： Codeforces - Team Rocket Rises Again 很明显如果我们先跑一个最短路出来，我们把最短路上面的边单独拿出来建图。 然后就可以得到边全是最短路上面的边的新图。 根据题目定义我们可以发现，这个是支配树。 跑一次支配树取max即可。 AC代码： # pragma GCC optimize("-Ofast","-funroll-all-loops") # include <bits/stdc++.h> //#define int long long using namespace std ; typedef long long LL ; const LL inf = 0x3f3f3f3f3f3f3f3f ; const int N = 8e5 + 10 , M = N << 1 ; int n , m , s , vis [ N ] , mx ; LL d [ N ] ; int head [ N ] , nex [ M ] , w [ M ] , to [ M ] , tot ; inline void ade ( int a , int b , int c ) { to [ ++ tot ] = b ; nex [ tot ] = head [ a ] ; w [ tot ] = c ; head [ a ] = tot ; } inline

描述 There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). 输入 Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). 输出 Print the word “yes” if 3 divide evenly into F(n). Print the word “no” if not. 样例输入 0 1 2 3 4 5 样例输出 no no yes no no no 题目来源 HDOJ 分析：斐波那契递推。 代码： #include<bits/stdc++.h> using namespace std; int f[1000001]; void Fibo() { f[0]=7;f[1]=11; for (int i=2;i<1000001;i++) f[i]=(f[i-1]+f[i-2])%3; } int main() { Fibo(); int n; while(cin>>n) { if (f[n]%3==0) cout<<“yes”<<endl; else cout<<“no”<<endl; } return 0; } 来源：

一、jquery操作符号：jquery()，简写为$(); 二、基本查找： 1、根据给定的ID匹配一个元素。如果选择器中包含特殊字符，可以用两个斜杠转义 例如：html码：<div id="myDiv">hello world</div> jquery码：$("#myDiv"); 结果：[ <div id="myDiv">id="myDiv"</div> ] 注意： 查找含有特殊字符的元素 html码：<span id="foo:bar"></span> <span id="foo[bar]"></span> <span id="foo.bar"></span> jquery码：#foo\\:bar #foo\\[bar\\] #foo\\.bar 2、根据给定的 class 类匹配元素。html码：<div class="myClass">div class="myClass"</div> jquery码： $(".myClass"); 结果： [ <div class="myClass">div class="myClass"</div>] 3、根据给定的元素名匹配所有元素：html码：<div>DIV1</div><div>DIV2</div> jquery码：$("div"); 结 果： [ <div>DIV1</div>, <div>DIV2</div> ] 4

题目描述 There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). Input Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). Output Print the word “yes” if 3 divide evenly into F(n). Print the word “no” if not. Sample Input 0 1 2 3 4 5 Sample Output no no yes no no no 解题思路 一开始做这道题时，直接用的笨办法，没有去找规律，后来才知道这道题是有规律的，其实只要满足n除以4余2这个条件，就是yes，否则，就是no。 AC代码 # include <bits/stdc++.h> using namespace std ; int main ( ) { int n ; while ( cin >> n ) { if ( n % 4 == 2 ) { cout << "yes" << endl ; } else { cout << "no" << endl ; } } return 0 ;