C. Vanya and Label 题目链接:http://codeforces.com/contest/677/problem/C While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwise AND for these two words represented as a integers in base…

B. Vanya and Food Processor 题目链接:http://codeforces.com/contest/677/problem/B Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters…

A. Vanya and Fence 题目连接:http://codeforces.com/contest/677/problem/A Vanya and his friends are walking along the fence of height h and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exc…

题目链接: http://codeforces.com/contest/677/problem/D 题意: 让你求最短的从start->...->1->...->2->...->3->...->...->p的最短路径. 题解: 这题dp的阶段性还是很明显的,相同的值得方格为同一个阶段,然后求从阶段1->2->3...->p的阶段图最短路. 初始化所有a[x][y]==1的格子为起始点到(x,y)坐标的距离. 方程式为dp[x1][y1…

http://codeforces.com/contest/677/problem/E 题意:有n*n矩形,每个格子有一个值(0.1.2.3),你可以在矩形里画一个十字(‘+’形或‘x’形),十字的四条边需等长.问十字覆盖的格子的值累乘最大是多少? 思路: 1.防止溢出,在比较大小更新答案时用加法替换乘法:a*b==log(a)+log(b): 2.首先,遍历每个点,对于每个点,对8个方向dfs,直到越界或值为0:求出每个点各个方向的深度后,第二遍遍历时可以得到十字的长度,然后算出若以该点为中心…

http://codeforces.com/contest/677/problem/D 建颗新树,节点元素包含r.c.dis,第i层包含拥有编号为i的钥匙的所有节点.用i-1层更新i层,逐层更新到底层. 不使用就会超时的优化:用i-1层更新时不是所有节点都有必要用到,我们对i-1层排序,取前600节点更新下层. public class Main { private static final int c = 330,INF=Integer.MAX_VALUE/2,maxn=c*c*c+100,m…

D. Vanya and Treasure 题目连接: http://www.codeforces.com/contest/677/problem/D Description Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room located in the i-th row and j-th columns contains t…

C. Vanya and Label 题目连接: http://www.codeforces.com/contest/677/problem/C Description While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwise AND for these two words represented as a…

B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smash…

A. Vanya and Fence 题目连接: http://www.codeforces.com/contest/677/problem/A Description Vanya and his friends are walking along the fence of height h and they do not want the guard to notice them. In order to achieve this the height of each of the frien…

C. Vanya and Label time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwi…

A 弯腰 #include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<stack> #include<algorithm> using namespace std; #define clc(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f ; #define LL long long //…

题目大意: 给你一个n × m 的图,有p种宝箱, 每个点上有一个种类为a[ i ][ j ]的宝箱,a[ i ][ j ] 的宝箱里有 a[ i ][ j ] + 1的钥匙,第一种宝箱是没有锁的, 第p类宝箱只有一个且里面由宝藏,你现在在(1 ,1)问你最少需要多少步才能拿到宝藏. (n, m <= 300) 思路:这题真的好恶心啊...  我们考虑p类宝箱只能从p - 1类转移过来, 这样我们就能从第一类宝箱开始往后递推, 但是最坏的情况, p 类 和p - 1类,都有45000 个点, 那…

啊啊啊啊啊啊啊,真的是智障了... 这种题目,没有必要纠结来源.只要知道它的结果的导致直接原因?反正这句话就我听的懂吧... ">>"/"&" #include<cstdio> int main() { int ans=1; for(int i=0;i<=1;i++) printf("%d\n",ans&i); ans=0; for(int i=0;i<=1;i++) printf("…

菜菜菜!!!这么撒比的模拟题,听厂长在一边比比比了半天,自己想一想,然后纯模拟一下,中间过程检测一下,妥妥的就可以过. 题意:有N个东西要去搞碎,每个东西有一个高度,然后有一台机器支持里面可以达到的最大高度,东西可以连续放进去,只要不超过h就行了,每秒可以搞k高度,然后让你算时间 直接code: #include<cstdio> #include<vector> #include<string.h> #include<iostream> #include&l…

B. Vanya and Food Processor time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output ->  Link  <- cf上的题果然很靠思路,这让我想起了学长说的那句话:每A完CF上的一套题都有不同的收获. 这道题题意倒是不难理解:有一台机器,每次只能处理K cm的土豆,这台机器最多能放下hcm的土豆,如果剩下的土…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

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 =…

Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接快速幂 代码 #include<bits/stdc++.h> using namespace std; long long quickpow(long long m,long long n,long long k) { long long b = 1; while (n > 0) { if…