期望计算的套路: 1.定义:算出所有测试值的和,除以测试次数. 2.定义:算出所有值出现的概率与其乘积之和. 3.用前一步的期望,加上两者的期望距离,递推出来. 题意: 一个树,dfs遍历子树的顺序是随机的.所对应的子树的dfs序也会不同.输出每个节点的dfs序的期望   思路: 分析一颗子树: 当前已知节点1的期望为1.0 ->anw[1]=1.0 需要通过节点1递推出节点2.4.5的期望值 1的儿子分别是2.4.5,那么dfs序所有可能的排列是6种: 1:1-2-4-5  (2.4.5节点的…

D. Puzzles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. C…

D. Puzzles Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. Cities and roads of USC form a rooted tree (Barney's not sure why it is rooted). Root of the tree is the city…

题目链接:http://codeforces.com/problemset/problem/697/D 给你一个有规则的二叉树,大概有1e18个点. 有两种操作:1操作是将u到v上的路径加上w,2操作是求u到v上的路径和. 我们可以看出任意一个点到1节点的边个数不会超过64(差不多就是log2(1e18)),所以可以找最近相同祖节点的方式写. 用一条边的一个唯一的端点作为边的编号(比如1到2,那2就为这条边的编号),由于数很大,所以用map来存. 进行1操作的时候就是暴力加w至u到LCA(u,v…

2018-03-16 http://codeforces.com/problemset/problem/697/C C. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersect…

A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time.…

B. Barnicle time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the di…

A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time.…

B. Barnicle time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the di…

A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time.…

闲来无事一套CF啊,我觉得这几个题还是有套路的,但是很明显,这个题并不难 A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in…

A. Puzzles Time Limit: 2 Sec  Memory Limit: 60 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373 Description The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. S…

起初误以为到每个叶子的概率一样于是.... /* CodeForces 839C - Journey [ DFS,期望 ] | Codeforces Round #428 (Div. 2) */ #include <bits/stdc++.h> using namespace std; const int N = 100005; int n; vector<int> G[N]; double dp[N], val[N]; bool vis[N]; void dfs(int u, i…

题目传送门 /* 题意:这题就是求b+1到a的因子个数和. 数学+DP:a[i]保存i的最小因子,dp[i] = dp[i/a[i]] +1;再来一个前缀和 */ /************************************************ Author :Running_Time Created Time :2015-8-1 14:08:34 File Name :B.cpp ************************************************…

题目传送门 /* 题意:求一个点为根节点,使得到其他所有点的距离最短,是有向边,反向的距离+1 树形DP:首先假设1为根节点,自下而上计算dp[1](根节点到其他点的距离),然后再从1开始,自上而下计算dp[v], 此时可以从上个节点的信息递推出来 */ #include <cstdio> #include <cstring> #include <cmath> #include <vector> using namespace std; ; const in…

Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solving different tasks. Today he found one he wasn't able to solve himself, so he asks you to help. Vasiliy is given n strings consisting of lowercase Engl…

http://codeforces.com/contest/353/problem/C Codeforces Round #205 (Div. 2)C #include<stdio.h> ]; ]; ]; int Max(int a,int b){ if(a<b)return b; else return a; } int main(){ int n; while(scanf("%d",&n)!=EOF){ int i; dp[]=; ;i<=n;i++…

题目传送门 /* 题意:b+1,b+2,...,a 所有数的素数个数和 DP+埃氏筛法:dp[i] 记录i的素数个数和,若i是素数,则为1:否则它可以从一个数乘以素数递推过来 最后改为i之前所有素数个数和,那么ans = dp[a] - dp[b]: 详细解释:http://blog.csdn.net/catglory/article/details/45932593 */ #include <cstdio> #include <algorithm> #include <cs…

题目传送门 /* 题意:每棵树给出坐标和高度,可以往左右倒,也可以不倒 问最多能砍到多少棵树 DP:dp[i][0/1/2] 表示到了第i棵树时,它倒左或右或不动能倒多少棵树 分情况讨论,若符合就取最大值更新,线性dp,自己做出来了:) */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <iostream> using na…

题目传送门 /* 贪心/数学:还以为是BFS,其实x1 + 4 * k = x2, y1 + 4 * l = y2 */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN][MAXN]; int main(void) //Codeforces Round #212 (Div. 2)…

题目传送门 /* 数学/暴力:只要一个数的最后三位能被8整除,那么它就是答案:用到sprintf把数字转移成字符读入 */ #include <cstdio> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <vector> using namespace std; ; const int INF = 0x3f3…

题目传送门 /* 题意:选择a[k]然后a[k]-1和a[k]+1的全部删除,得到点数a[k],问最大点数 DP:状态转移方程:dp[i] = max (dp[i-1], dp[i-2] + (ll) i * cnt[i]); 只和x-1,x-2有关,和顺序无关,x-1不取,x-2取那么累加相同的值,ans = dp[mx] */ #include <cstdio> #include <algorithm> #include <cstring> #include <…

Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~ The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the follow…

题目传送门 /* 数学:这题一直WA在13组上,看了数据才知道是计算cost时超long long了 另外不足一个区间的直接计算个数就可以了 */ #include <cstdio> #include <cmath> #include <iostream> #include <algorithm> #include <cstring> using namespace std; typedef unsigned long long ull; int…

题目传送门 /* DP:先用l,r数组记录前缀后缀上升长度,最大值会在三种情况中产生: 1. a[i-1] + 1 < a[i+1],可以改a[i],那么值为l[i-1] + r[i+1] + 1 2. l[i-1] + 1 3. r[i+1] + 1 //修改a[i] */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF…

题目传送门 /* DP:从1到最大值,dp[i][1/0] 选或不选,递推更新最大值 */ #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll dp[MAXN][]; ll cnt[MAXN]; ll work(…

Pythagorean Triples 题目链接: http://codeforces.com/contest/707/problem/C Description Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such th…

题目传送门 /* 水题:求总数字个数,开long long竟然莫名其妙WA了几次,也没改啥又对了:) */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <string> #include <queue> #incl…

Codeforces Round #622 (Div. 2) B. Different Rules 题意: 你在参加一个比赛,最终按两场分赛的排名之和排名,每场分赛中不存在名次并列,给出参赛人数 n 和你两场分赛的排名 x, y,问你最终名次最小和最大可能是多少. 思路: 以8人为例: x + y = 2,最小第一名,最大第一名:               1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1               x + y = 3,最小第一名,最大第二名.…

Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…