lkbsbird的博客

DescriptionThe little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the sa

DescriptionGiven two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponenti

DescriptionConsider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a syste

DescriptionA prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty str

DescriptionAndy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the le

DescriptionGiven a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers:Emergency 911

题意：L，T，OL代表长度，T代表颜色的数量，O代表操作的数量，C Ａ　B C 从A到B全染成CP A B 查询 A到B颜色的种类#include#include#includeusing namespace std;const int N=100010;#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)struct Tr

DescriptionYou have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

题意：给一个数N，下面给N行，每行给一个X，Y，表示坐标，求这些坐标能组成正方形最大个数解析：先枚举两个点，通过数学公式得到另外2个点，使得这四个点能够成正方形。然后检查散点集中是否存在计算出来的那两个点，若存在，说明有一个正方形。但这种做法会使同一个正方形按照不同的顺序被枚举了四次，因此最后的结果要除以4.数学公式：(x1,y1)  (x2,y2)则：   x3=x1+(y1

题意：有N件商品，知道了商品的价值和销售的最后期限，只要在最后日期之前销售处，就能得到相应的利润，并且销售该商品需要1天时间，求出最大利润。 解析：先进行利润从大到小的排序，然后利用vis数组进行标记来记录当前的时间是否满足，主要利用的是贪心算法#include<cstdio>#include<cstring>#include<algorithm>using namespace std;c

题意：有N件商品，知道了商品的价值和销售的最后期限，只要在最后日期之前销售处，就能得到相应的利润，并且销售该商品需要1天时间，求出最大利润。 解析：先进行利润从高到低的排序，利用并查集按利润排序，建立一个关于时间的并查集：每次插入一个物品时，若该物品时间为 i，找出find(i)，记为t，若t不为0，则将该物品安排到t这个时间完成，并使f[t]=t-1 亦即对于每个物品尽量#include<cst

题意：So easy的树状数组 解析：使用lowbit即可#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;int n;long long lowbit(long long a){ return a&(-a);}int main(){ s

题意：eee 解析：fsss#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;const int maxn=100009;int set[maxn];int top[maxn];int cnt[maxn];int n;int find(int x){

题意：简单并查集 解析：统计不同信仰的学生，标记即可#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=50009;int student[maxn];int n,m;int find(int x){ if(student[x]!=x) retu

DescriptionBeside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Inter