题意：给定一个$R imes C$的草坪，草坪上有一些羊和一些狼，你可以在羊的周围放一些狗，有狗的地方狼就过不去了，问是否存在一种方案使得所有的羊都能不被吃掉，并输出任意一组解。
题解：随便输出一组解的话，我们可以贪心地思考，把每一块空地都放一只狗，然后以每一只狼为起点进行一遍DFS，如果能碰到羊就无解，如果到最后都没有羊被吃掉，那么就输出方案。
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
char a[510][510];
bool vis[510][510];
int r,c;
const int dx[]={-1,1,0,0},dy[]={0,0,1,-1};
void dfs(int x,int y)
{
if(a[x][y]=='S')
{
puts("No");
exit(0);
}
int xx,yy;
for(int i=0;i<4;i++)
{
xx=x+dx[i],yy=y+dy[i];
if(xx>=0&&yy>=0&&xx<r&&yy<c&&a[xx][yy]!='D')
{
if(!vis[xx][yy])
{
vis[xx][yy]=1;
dfs(xx,yy);
}
}
}
}
int main()
{
scanf("%d%d",&r,&c);
for(int i=0;i<r;i++)
scanf("%s",a[i]);
for(int i=0;i<r;i++)
for(int j=0;j<c;j++)
if(a[i][j]=='.')
a[i][j]='D';
for(int i=0;i<r;i++)
for(int j=0;j<c;j++)
if(a[i][j]=='W')
dfs(i,j);
puts("Yes");
for(int i=0;i<r;i++)
printf("%s
",a[i]);
return 0;
}
这么水的题竟然写了接近50行qwq。。。
题意：两个人玩游戏，先指定一个$X_0$，第$i$轮的时候他会选择一个满足$ple X_i$的质数，然后找到一个最小的数满足$X_{i+1}>X_i,X_{i+1}mid p$，如此进行若干轮。现在给定你$X_2$，你要求出最小的$X_0$。
题解：令$P(N)$为$N$的最大质因数，显然我们可以通过选择$P(N)$作为每次操作的质数获得区间$[N-P(N)+1,N]$中的任意数，并且我们不能从其他任何数获得$N$。通过分解$X_2$，我们可以找到$X_1$的区间，通过分解$X_1$区间中的所有数，我们可以找到$X_0$的区间，答案是$X_1$区间交集的最小元素。
题意：有$N$堆雪，每堆雪都有一定体积，第$i$天每堆雪的体积都会缩小$T[i]$，求$1$到$N$天融化的雪的体积。
题解：由于$T[i]$是固定的，我们只需要确定每堆雪融化完的时间即可。对每堆雪的融化量做一个前缀和，然后就可以维护每天化了多少雪了。对于每堆雪的融化完成时间，直接在做前缀和的过程中就能处理完。
Alice has a very important message M consisting of some non-negative integers that she wants to keep secret from Eve. Alice knows that the only theoretically secure cipher is one-time pad. Alice generates a random key K of the length equal to the message's length. Alice computes the bitwise xor of each element of the message and the key (
, where 
denotes the bitwise XOR operation) and stores this encrypted message A. Alice is smart. Be like Alice.
For example, Alice may have wanted to store a message M = (0, 15, 9, 18). She generated a key K = (16, 7, 6, 3). The encrypted message is thus A = (16, 8, 15, 17).
Alice realised that she cannot store the key with the encrypted message. Alice sent her key K to Bob and deleted her own copy. Alice is smart. Really, be like Alice.
Bob realised that the encrypted message is only secure as long as the key is secret. Bob thus randomly permuted the key before storing it. Bob thinks that this way, even if Eve gets both the encrypted message and the key, she will not be able to read the message. Bob is not smart. Don't be like Bob.
In the above example, Bob may have, for instance, selected a permutation (3, 4, 1, 2) and stored the permuted key P = (6, 3, 16, 7).
One year has passed and Alice wants to decrypt her message. Only now Bob has realised that this is impossible. As he has permuted the key randomly, the message is lost forever. Did we mention that Bob isn't smart?
Bob wants to salvage at least some information from the message. Since he is not so smart, he asks for your help. You know the encrypted message A and the permuted key P. What is the lexicographically smallest message that could have resulted in the given encrypted text?
More precisely, for given A and P, find the lexicographically smallest message O, for which there exists a permutation π such that 
for every i.
Note that the sequence S is lexicographically smaller than the sequence T, if there is an index i such that Si < Ti and for all j < i the condition Sj = Tj holds.
The first line contains a single integer N (1 ≤ N ≤ 300000), the length of the message.
The second line contains N integers A1, A2, ..., AN (0 ≤ Ai < 230) representing the encrypted message.
The third line contains N integers P1, P2, ..., PN (0 ≤ Pi < 230) representing the permuted encryption key.
Output a single line with N integers, the lexicographically smallest possible message O. Note that all its elements should be non-negative.
3
8 4 13
17 2 7
10 3 28
5
12 7 87 22 11
18 39 9 12 16
0 14 69 6 44
10
331415699 278745619 998190004 423175621 42983144 166555524 843586353 802130100 337889448 685310951
226011312 266003835 342809544 504667531 529814910 684873393 817026985 844010788 993949858 1031395667
128965467 243912600 4281110 112029883 223689619 76924724 429589 119397893 613490433 362863284
In the first case, the solution is (10, 3, 28), since 
, 
and 
. Other possible permutations of key yield messages (25, 6, 10), (25, 3, 15), (10, 21, 10), (15, 21, 15) and (15, 6, 28), which are all lexicographically larger than the solution.
题意：给定两组数，求这两组数两两异或后连接起来字典序最小的一组。
题解：我们贪心的解密每一对数，考虑要维护一些数的异或和，我们直接上Trie树把数的每一位分解，记录个数，然后在Trie树上寻找最优解即可。
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
- A
BC - B AC
- C AB
- AAA empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
10011
In the first query we can achieve the result, for instance, by using transitions 
.
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
题意：给两个只有$A,B,C$的字符串$S$和$T$，有$Q$次询问，每次询问能否将$S$的一段区间$[a_i,b_i]$变换为$T$的一段区间$[c_i,d_i]$。
变换方式如下：
egin{equation*}
egin{aligned}
A&
ightarrow BC\
B&
ightarrow AC\
C&
ightarrow AB\
AAA&
ightarrowvarnothing
end{aligned}
end{equation*}
题解：
根据所给条件，我们对字符串进行一些变换，容易发现一些单个字符变换的规律：
egin{equation*}
egin{aligned}
&B
ightarrow AC
ightarrow AAB
ightarrowmathbf{AAA}C
ightarrow C\
&C
ightarrow AB
ightarrow AAC
ightarrowmathbf{AAA}B
ightarrow B\
&B
ightarrow AC
ightarrow AAB
ightarrowmathbf{AAA}C
ightarrowmathbf{AAA}AB
ightarrow AB\
&C
ightarrow AB
ightarrow AAC
ightarrowmathbf{AAA}B
ightarrowmathbf{AAA}AC
ightarrow AC\
&A
ightarrow BC
ightarrow BAB
ightarrow BAAC
ightarrow Bmathbf{AAA}B
ightarrow BB\
end{aligned}
end{equation*}
由上述一些规律我们还容易得到：
egin{equation*}
egin{aligned}
&B
ightarrow AC
ightarrow AB
ightarrow BBB\
&AAB
ightarrow mathbf{AAA}C
ightarrow C
ightarrow B\
&AB
ightarrow ACB
ightarrow ABB\
end{aligned}
end{equation*}
综上所述，我们得到以下一些结论：
egin{align}
&BLeftrightarrow C\
&B每次可以增加偶数个\
&在B之前出现的A都可以删除\
end{align}
于是我们得到了原字符串和目标字符串之间的关系：
- 目标字符串中的$B$与$C$的数量和必须大于等于原字符串中的$B$与$C$的数量和
- 目标字符串中的$B$与$C$的数量和的奇偶性必须与原字符串中的$B$与$C$的数量和相同
最后，如果原字符串中的$B$和$C$的数量与目标字符串中的$B$和$C$的数量相等，那么没有操作$1$和操作$2$，只能将后面的$A$减少$3n,nin N$个。由于所有操作都无法增加原字符串末尾的$A$的个数，如果原字符串末尾$A$的个数少于目标字符串就无解，否则对3取个模然后与目标字符串末尾$A$的个数对3取模的结果判一下是否相等就行了。
至于怎么维护两个字符串中$B$和$C$的数量，直接上前缀和预处理一下就行了。