时间:2015-01-30 来源:

CodeforcesRound#287(Div.2)题解 【编程语言】

A. Amr and Music time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.

/************************************************************************* > File Name: cf287d.cpp > Author: ALex > Mail: 405045132@qq.com > Created Time: 2015年01月24日 星期六 12时10分19秒 ************************************************************************/ #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; long long dp[1010][110][2]; long long f[1010]; int main() { int n, &n, 0, ans); } return 0; }

Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.

/************************************************************************* > File Name: cf287e.cpp > Author: ALex > Mail: 405045132@qq.com > Created Time: 2015年01月24日 星期六 12时51分10秒 ************************************************************************/ #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int N = 100010; const int inf = 0x3f3f3f3f; int dist[N]; int dp[N]; struct CODEFORCES { int E; int V; void clear() { V = -1; } }pre[N]; int head[N]; bool vis[N << 1]; bool vis_edge[N << 1]; int tot, m; struct node { int next; int to; int id; int sta; }edge[N << 1]; void addedge (int from, int sta, inf, 0, v, &n, -1, 0, 0, &u, &w); addedge (u, w, u, i); } spfa (); dfs (n); int ans = 0; for (int u = 1; u <= n; ++u) { for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].to; if (!vis_edge[edge[i].id] && edge[i].sta && !vis[edge[i].id]) { ++ans; vis[edge[i].id] = 1; } else if (vis_edge[edge[i].id] && !edge[i].sta && !vis[edge[i].id]) { ++ans; vis[edge[i].id] = 1; } } } memset (vis, sizeof(vis)); printf("%d\n", u, u, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.

水题,div切图排版沿着连线翻转网页切图制作,次数一定最少,网页切图制作所以答案就是d / (2 * r), gang has to blow up all other roads in country that don't lay on this path, they don't have to blow up it as it is already malfunctional.

A:水题psd切图html,排个序,psd切图html水题贪心吧算是

bin神教的web外包,设dp[i][j][0] 表示后i位,手机网页制作对k取模为j,且还没出现过整除的情况的数目,div页面dp[i][j][1]表示后i位div+css+js切图,且已经出现过整除的情况

Can you help Walter complete his task and gain the gang's trust?

In following m lines there are descriptions of roads. Each description consists of three integers x,?z (1?≤?x, ) meaning that there is a road connecting cities number x and y. If z?=?1, otherwise it is not.

First he gave Amr two positive integers n and k. Then he asked Amr, &h, ans); } return 0; }

if there are multiple optimal solutions output any. It is not necessary to use all days for studying.

Help Amr to achieve his goal in minimum number of steps.

Output In the first line output one integer m representing the maximum number of instruments Amr can learn.

Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1?≤?n?≤?2h, the player doesn't know where the exit is so he has to guess his way out!

最后枚举每一条边,承接网站前端判断即可

Output Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.

The gang is going to rob a bank! The bank is located in city 1. As usual, Walter is in charge of this operation, there should remain working only roads lying on some certain shortest past between city 1 and n.

Sample test(s) Input 4 10 4 3 1 2 Output 4 1 2 3 4 Input 5 6 4 3 1 1 2 Output 3 1 3 4 Input 1 3 4 Output 0 Note In the first test Amr can learn all 4 instruments.

In the third test Amr doesn't have enough time to learn the only presented instrument.

Character 'L' means "go to the left child of the current node"; Character 'R' means "go to the right child of the current node"; If the destination node is already visited, otherwise he moves to the destination node; If Amr skipped two consecutive commands, he returns to the parent of the current node; If he reaches an exit, if he follows this algorithm,?k, 1?≤?k?≤?100,?y,?y?≤?n, cities connected by a road and the new state of a road. z?=?1 indicates that the road between cities x and y should be repaired and z?=?0 means that road should be blown up.

Output Output a single integer — minimum number of steps required to move the center of the circle to the destination point.

Can you help Amr escape this embarrassing situation?

利用二进制思想psd切图html,表示第一层往左走web外包,第二层往右走,手机网页制作第三次往左.....

Output In the first line output one integer k, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.

Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.

Decimal representation of x (without leading zeroes) consists of exactly n digits; There exists some integer y?>?0 such that: ; decimal representation of y is a suffix of decimal representation of x. As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.

Input Input consists of two integers h, 1?≤?n?≤?2h).

Amr has a circle of radius r and center in point (x,?y').

Walter discovered that there was a lot of paths that satisfied the condition of being shortest possible so he decided to choose among them a path that minimizes the total number of affected roads (both roads that have to be blown up and roads to be repaired).

/************************************************************************* > File Name: cf287b.cpp > Author: ALex > Mail: 405045132@qq.com > Created Time: 2015年01月24日 星期六 00时05分53秒 ************************************************************************/ #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; int main () { double r, y1, y2; while (~scanf("%lf%lf%lf%lf%lf", &x1, &x2, (int)ceil(d / (2 * r))); } return 0; }

C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.

The second line contains n integers ai (1?≤?ai?≤?100), x, x' y' (1?≤?r?≤?105,?y,?y'?≤?105), coordinates of original center of the circle and coordinates of destination center of the circle respectively.

E. Breaking Good time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is really difficult even for experienced gamers.

Walter William, wants to join a gang called Los Hermanos (The Brothers). The gang controls the whole country which consists of n cities with m bidirectional roads connecting them. There is no road is connecting a city to itself and for any two cities there is at most one road between them. The country is connected, it is possible to reach any city from any other city using the given roads.

The roads aren't all working. There are some roads which need some more work to be performed to be completely functioning.

In the second test other possible solutions are: {2,?5} or {3,?5}.

First of all the path which they are going to use on their way back from city 1 to their headquarters n must be as short as possible, since it is important to finish operation as fast as possible.

Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:

B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Amr loves Geometry. One day he came up with a very interesting problem.

If the chosen path has some roads that doesn't work they'll have to repair those roads before the operation.

可以发现,psd切图html如果相邻2个二进制位相异web外包,那么多走了一个节点,div切图排版如果相同网页切图制作,就多走了一颗子树,网页切图制作子树上节点数目也很好求(二叉树),于是我们只要对目标叶子节点的每一位进行访问,div+css+js切图与上次走的方向比较承接网站前端,?m (2?≤?n?≤?105, the number of cities and number of roads respectively.

/************************************************************************* > File Name: cf287a.cpp > Author: ALex > Mail: 405045132@qq.com > Created Time: 2015年01月23日 星期五 23时49分38秒 ************************************************************************/ #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; struct node { int day; int d; }aaa[110]; int cmp (node a, k; while (~scanf("%d%d", &k)) { for (int i = 1; i <= n; ++i) { scanf("%d", aaa + n + 1, cnt); if (cnt == 0) { continue; } printf("%d", aaa[i].d); } printf("\n"); } return 0; }

Sample test(s) Input 2 0 0 0 4 Output 1 Input 1 1 1 4 4 Output 3 Input 4 5 6 5 6 Output 0 Note In the first sample test the optimal way is to put a pin at point (0, no matter).

You may output roads in any order. Each affected road should appear exactly once. You may output cities connected by a single road in any order. If you output a road, k (1?≤?n?≤?100, the number of instruments and number of days respectively.

In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.

比D题简单,承接网站前端维护最短路的同时维护num[u]表示1->u的最短路上psd切图html,同时记录上一步的节点

Sample test(s) Input 2 1 1 2 0 Output 1 1 2 1 Input 4 4 1 2 1 1 3 0 2 3 1 3 4 1 Output 3 1 2 0 1 3 1 2 3 0 Input 8 9 1 2 0 8 3 0 2 3 1 1 4 1 8 7 0 1 5 1 4 6 1 5 7 0 6 8 0 Output 3 2 3 0 1 5 0 6 8 1 Note In the first test the only path is 1?-?2

最后统计即可

Sample test(s) Input 1 2 1000 Output 4 Input 2 2 1000 Output 45 Input 5 3 1103 Output 590 Note A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.

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