B-序列的与和_2023河南萌新联赛第（四）场：河南大学 (nowcoder.com)（dfs）
cpp
#include<iostream>
#include<string>
using namespace std;
#define ull unsigned long long
int n,k;
ull a[21];
ull ans=0;
int check(ull sum){
int q=0;
while(sum){
q++;
sum&=(sum-1);
}
return q;
}
void dfs(int x,ull sum){
if(check(sum)==k)ans++;
for(int i=x+1;i<=n;i++){
dfs(i,sum&a[i]);
}
}
int main(){
cin>>n>>k;
for(int i=1;i<=n;i++){
cin>>a[i];
}
for(int i=1;i<=n;i++){
dfs(i,a[i]);
}
cout<<ans<<endl;
}D-幂运算_2023河南萌新联赛第（四）场：河南大学 (nowcoder.com)(快速幂类似)
cpp
#include<iostream>
#include<string>
using namespace std;
long long qowe(long long a,long long n,long long p){//指数幂
long long ans=2;
while(n){
ans=(long long)ans*ans%p;
a=(long long)a*a%p;
n=n-1;
}
return ans;
}
int main(){
int n,k;
cin>>n>>k;
cout<<qowe(2,n,k);
}P4554 小明的游戏 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)(双端队列bfs--模板)
cpp
#include<iostream>
#include<string>
#include<vector>
#include<deque>
#include<cstring>
using namespace std;
int n,m;
struct node{
int xx,yy,step;
};
string s[505];
int x1,y1,a,b;
bool vis[505][505];
int cnt[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int in(int x,int y){
if(x<0||y<0||x>=n||y>=m)return 0;
return 1;
}
int bfs(int x,int y){
node c;
c.xx=x,c.yy=y,c.step=0;
deque<node>q;
q.push_front(c);
while(!q.empty()){
node temp=q.front();
q.pop_front();
node now;
vis[temp.xx][temp.yy]=true;
if(temp.xx==a&&temp.yy==b)return temp.step;
for(int i=0;i<4;i++){
now.xx=temp.xx+cnt[i][0];
now.yy=temp.yy+cnt[i][1];
if(!vis[now.xx][now.yy]&&in(now.xx,now.yy)&&s[temp.xx][temp.yy]==s[now.xx][now.yy]){
now.step=temp.step;
q.push_front(now);
}
else if(!vis[now.xx][now.yy]&&in(now.xx,now.yy)&&s[temp.xx][temp.yy]!=s[now.xx][now.yy]){
now.step=temp.step+1;
q.push_back(now);
}
}
}
return 0;
}
int main(){
while(cin>>n>>m&&(n||m)){
for(int i=0;i<n;i++){
cin>>s[i];
}
cin>>x1>>y1>>a>>b;
memset(vis,0,sizeof(vis));
cout<<bfs(x1,y1)<<"\n";
}
}Labyrinth - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)(双端队列bfs)
cpp
#include<iostream>
#include<string>
#include<vector>
#include<deque>
#include<cstring>
#include<queue>
using namespace std;
const int N=2005;
int n,m;
struct node{
int xx,yy,step1,step2;
};
string s[N];
int x1,y1,a,b;
bool vis[N][N];
int cnt[4][2]={{1,0},{-1,0},{0,-1},{0,1}};
int ans=1;
int in(int x,int y){
if(x<0||y<0||x>=n||y>=m)return 0;
return 1;
}
int bfs(int x,int y){
node c;
c.xx=x,c.yy=y,c.step1=0,c.step2=0;
deque<node>q;
q.push_front(c);
vis[x][y]=true;
while(!q.empty()){
node temp=q.front();
q.pop_front();
node now;
for(int i=0;i<4;i++){
now.xx=temp.xx+cnt[i][0];
now.yy=temp.yy+cnt[i][1];
if(!in(now.xx,now.yy)||vis[now.xx][now.yy]||s[now.xx][now.yy]=='*')continue;
if(i==0||i==1){
now.step1=temp.step1;
now.step2=temp.step2;
q.push_front(now);
vis[now.xx][now.yy]=true;
ans++;
}
else if(i==2){
if(temp.step1>=a)continue;
now.step1=temp.step1+1;
now.step2=temp.step2;
q.push_back(now);
vis[now.xx][now.yy]=true;
ans++;
}
else if(i==3){
if(temp.step2>=b)continue;
now.step1=temp.step1;
now.step2=temp.step2+1;
q.push_back(now);
vis[now.xx][now.yy]=true;
ans++;
}
}
}
return 0;
}
int main(){
cin>>n>>m;
cin>>x1>>y1>>a>>b;
for(int i=0;i<n;i++){
cin>>s[i];
}
bfs(x1-1,y1-1);
cout<<ans<<"\n";
}175. 电路维修 - AcWing题库(双端队列bfs--不是很理解待重新补)
cpp
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <deque>
using namespace std;
typedef pair<int, int > PII;
int n, m;
const int N = 510;
char g[N][N];
int d[N][N];
inline bool check(int x, int y)
{
if(x >= 0 && x <= n && y >= 0 && y <= m) return true;
return false;
}
inline int bfs()
{
memset(d, 0x3f, sizeof d);
deque<PII> dq;
dq.push_front({0, 0});
d[0][0] = 0;
int moved[4][2] = {{-1, -1},{-1, 1},{1, 1},{1, -1}};
int movei[4][2] = {
{-1, -1},
{-1, 0},
{0, 0},
{0, -1},
};
char cp[] = "\\/\\/";
while(dq.size())
{
auto t = dq.front();
dq.pop_front();
int x = t.first, y = t.second;
for(int i = 0; i < 4; i ++)
{
int a = x + moved[i][0], b = y + moved[i][1];
if(check(a, b))
{
int j = x + movei[i][0], k = y + movei[i][1];
int w;
if(g[j][k] != cp[i]) w = 1;
else w = 0;
if(d[a][b] > d[x][y] + w)
{
d[a][b] = d[x][y] + w;
if(w) dq.push_back({a, b});
else dq.push_front({a, b});
}
}
}
}
if(d[n][m] == 0x3f3f3f3f) return -1;
else return d[n][m];
}
inline void solve()
{
cin >> n >> m;
for(int i = 0; i < n; i ++) cin >> g[i];
int res = bfs();
if(res == -1) puts("NO SOLUTION");
else cout << res << endl;
}
int main()
{
int t;
cin >> t;
while( t -- ) solve();
return 0;
}