CF 168(div2)

A.直接模拟即可，最后根据翻转次数奇偶判断。

B.类似连连看的判断，要求判断任意两点间是否能最多只转一次方向就相连。对每一点，先拓展出和它相同行，相同列能连到的点，再对那些点再做一次拓展，记录下总共能到达的点数目，判断是否为总点数即可。代码比较挫

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 50 + 5;
const int INF = 1000000000;
typedef long long LL;

char M[maxn][maxn];
int vis[maxn][maxn];
struct Node{
    int x,y;
};

int n,m;
queue<Node> Q;

int Bfs(int x,int y){
    memset(vis,0,sizeof(vis));
    while(!Q.empty()) Q.pop();
    int ret = 1;
    for(int i = y+1;i < m;i++){
        if(M[x][i] == 'B'){
            Q.push(Node{x,i});
            if(vis[x][i] == 0){
                ret++;
                vis[x][i] = 1;
            }
        }
        else break;
    }
    for(int i = y-1;i >= 0;i--){
        if(M[x][i] == 'B'){
            Q.push(Node{x,i});
            if(vis[x][i] == 0){
                vis[x][i] = 1;
                ret++;
            }
        }
        else break;
    }
    while(!Q.empty()){
        int X = Q.front().x;
        int Y = Q.front().y;
        Q.pop();
        for(int i = X-1;i >= 0;i--){
            if(M[i][Y] == 'B'){
                if(vis[i][Y] == 0){
                    ret++;
                    vis[i][Y] = 1;
                }
            }
            else break;
        }
        for(int i = X+1;i < n;i++){
            if(M[i][Y] == 'B'){
                if(vis[i][Y] == 0){
                    vis[i][Y] = 1;
                    ret++;
                }
            }
            else break;
        }
    }

    while(!Q.empty()) Q.pop();
    for(int i = x+1;i < n;i++){
        if(M[i][y] == 'B'){
            Q.push(Node{i,y});
            if(vis[i][y] == 0){
                vis[i][y] = 1;
                ret++;
            }
        }
        else break;
    }
    for(int i = x-1;i >= 0;i--){
        if(M[i][y] == 'B'){
            Q.push(Node{i,y});
            if(vis[i][y] == 0){
                vis[i][y] = 1;
                ret++;
            }
        }
        else break;
    }
    while(!Q.empty()){
        int X = Q.front().x;
        int Y = Q.front().y;
        Q.pop();
        for(int i = Y-1;i >= 0;i--){
            if(M[X][i] == 'B'){
                if(vis[X][i] == 0){
                    vis[X][i] = 1;
                    ret++;
                }
            }
            else break;
        }
        for(int i = Y+1;i < m;i++){
            if(M[X][i] == 'B'){
                if(vis[X][i] == 0){
                    vis[X][i] = 1;
                    ret++;
                }
            }
            else break;
        }
    }
    return ret;
}

int main(){
    while(cin >> n >> m){
        int total = 0;
        for(int i = 0;i < n;i++) scanf("%s",M[i]);
        for(int i = 0;i < n;i++){
            for(int j = 0;j < m;j++)
                if(M[i][j] == 'B') total++;
        }
        int ans = 1;
        for(int i = 0;i < n;i++){
            for(int j = 0;j < m;j++){
                if(M[i][j] == 'B' && Bfs(i,j) != total){
                    //printf("%d %d %d\n",i,j,Bfs(i,j));
                    ans = 0;
                    break;
                }
            }
            if(ans == 0) break;
        }
        if(ans == 0) cout << "NO\n";
        else cout << "YES\n";
    }
    return 0;
}

C.贪心的做法，从小到大排序后，优先选小的，然后把这个数a*k的值标记为不能选的，因为数的范围很大，用个map即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 100000 + 5;
const int INF = 1000000000;
typedef long long LL;

LL a[maxn];
map<LL,bool> M;

int main(){
    int n;
    LL k;
    while(cin >> n >> k){
        for(int i = 0;i < n;i++) cin >> a[i];
        sort(a,a+n);
        M.clear();
        int ans = 0;
        for(int i = 0;i < n;i++){
            if(M.count(a[i]) == 0){
                ans++;
                M[a[i]*k] = true;
            }
        }
        cout << ans << endl;
    }
    return 0;
}