Codeforces Round #198 (Div. 2) 昨天看到奋斗群的群赛,好奇的去做了一下, 大概花了3个小时Ak,我大概可以退役了吧 那下面来稍微总结一下 A. The Wall Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being mad…

http://codeforces.com/contest/341 赛后做的虚拟比赛,40分钟出了3题,RP爆发. A计数问题 我们可以对每对分析,分别对每对<a, b>(a走到b)进行统计,那么这对<a, b>产生的期望为distance(a, b)/n (把这一对选出来以后相当于一个点,那么分子distance(a, b)*(n-1)!,分母n!,     (n-1)被约掉了.) 这样的算法是O(n^2)的,问题转化为统计所有对<a, b> 的距离.我们可以对输入的…

题目链接:http://codeforces.com/contest/340/problem/E E. Iahub and Permutations time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is so happy about inventing bubble sort graphs that he's sta…

D. Iahub and Xors   Iahub does not like background stories, so he'll tell you exactly what this problem asks you for. You are given a matrix a with n rows and n columns. Initially, all values of the matrix are zeros. Both rows and columns are 1-based…

A.The Wall 题意:两个人粉刷墙壁,甲从粉刷标号为x,2x,3x...的小块乙粉刷标号为y,2y,3y...的小块问在某个区间内被重复粉刷的小块的个数. 分析:求出x和y的最小公倍数,然后做一个一维的区间减法就可以了. #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long L…

比赛时,开了大号去做,算了半天发现不会做A,囧.于是跑去看B,发现很水?于是很快敲完了,但是A不会,没敢交.于是去看C,一直找规律啊,后来总算调了出来,看了一下榜,发现还是算了吧,直接去睡觉了.第二天一起床把代码一交,居然A了,发现交的话rating还能涨一点,囧. B:其实就是求一个最长不下降子序列的长度.注意到数据范围,使用二分的方式求解. #include <set> #include <map> #include <list> #include <cmat…

昨天想了一下D题,有点思路不过感觉很麻烦,就懒得去敲了: 今天上午也想了一下,还是没有结果,看了一下官方题解,证明得很精彩: 这道题目其实就是一道裸地最大上升子序列的题: 看到这里,直接怒码···· #include<cstdio> #include<algorithm> using namespace std; ]; int main() { ; scanf("%d",&n); ;i<n;i++) { scanf("%d",&…

C题很容易看懂题目,不过两个循环肯定会TLE,所以得用点小聪明: 首先排好序,因为是全排列,乱序和顺序的结果是一样的: 然后呢···· 如果是数列 1 2 3 4 5 元素1 被 2 3 4 5每个减了2次,它自己减0一次:相抵后为-7: 元素2 被 3 5 4 每个减了2次,它减1两次,减0一次:相抵后为 -3: 元素3 相抵后为1: 可以发现他们的数量相差4:这样就好办了,一个循环就搞定了: 代码: #include <iostream> #include <cstdio> #…

B题是一个计算几何的题,虽然以前看过计算几何的ppt,但一直都没有写过: 昨晚比赛的时候本来想写的,但是怕不熟练浪费时间,太可惜了! 其实没必要选出一个最大的矩形: 以矩形的一条对角线为轴,向上或者向下找到最大的三角形的面积就行了, 可以看看官方的题解,讲的挺不错的! 代码: #include<cstdio> #define eps 0.00000001 using namespace std; ][]; double ccw(int x,int y,int z) { ]-a[x][])*(a…

最水的题,可惜当时赶时间没有注意数据范围:暴力超时了! 其实应该用x,y的最大公约数来判断: 代码: #include<iostream> using namespace std; int gcd(int a,int b) { ?a:gcd(b,a%b); } int main() { int x,y,a,b; cin>>x>>y>>a>>b; if(x<y) { x=x^y; y=x^y; x=x^y; } int k=x/gcd(x,y…

D. Bubble Sort Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a1, a2, ..., an …

C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a…

B. Maximal Area Quadrilateral time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub has drawn a set of n points in the cartesian plane which he calls "special points". A quadrilateral…

C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a…

C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a…

E. Iahub and Permutations Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to h…

接着是C,D的题解 C. Tourist Problem Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destina…

A. The Wall 求下gcd即可. B. Maximal Area Quadrilateral 枚举对角线,根据叉积判断顺.逆时针方向构成的最大面积. 由于点坐标绝对值不超过1000,用int比较快. C. Tourist Problem 假设序列为\(p_1,p_2,...,p_n\),则距离总和为\(,,,p_1,|p_2-p_1|,...,|p_n-p_{n-1}|\). 第1个点\(p_1\)的贡献为\(\sum a_i(n-1)!\) \(|p_i-p_{i-1}|\)的贡献为\…

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…