广大校赛

A

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//先考虑3
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const int N=5e2+10;
int n,m;
char a[N][N];
int res1=0,res2=0,res3=0;
void check1(int x,int y){
    int f=1;
    for(int i=0;i<5;i++){
        if(a[x+i][y]!='#') f=0;
    }
    for(int j=1;j<=2;j++){
        if(a[x][y+j]!='#') f=0;
        if(a[x+2][y+j]!='#') f=0;
    }
    if(a[x+1][y+2]!='#') f=0;
    if(f){
        res1++;
        for(int i=0;i<5;i++){
            a[x+i][y]='.';
        }
        for(int j=1;j<=2;j++){
            a[x][y+j]='.';
            a[x+2][y+j]='.';
        }
        a[x+1][y+2]='.';
    }
}
void check2(int x,int y){
    int f=1;
    for(int i=0;i<5;i++){
        if(a[x+i][y+2]!='#') f=0;
    }
    if(a[x][y]!='#') f=0;
    if(a[x][y+1]!='#') f=0;
    if(a[x+2][y+1]!='#') f=0;
    if(f){
        res2++;
        for(int i=0;i<5;i++){
            a[x+i][y+2]='.';
        }
        a[x][y]='.';
        a[x][y+1]='.';
        a[x+2][y+1]='.';
    }
}
void check3(int x,int y){
    int f=1;
    for(int i=0;i<5;i++){
        if(a[x+i][y]!='#') f=0;
        if(a[x+i][y+2]!='#') f=0;
    } 
    if(a[x][y+1]!='#') f=0;
    if(a[x+2][y+1]!='#') f=0;
    if(a[x+4][y+1]!='#') f=0;
    if(f){
        res3++;
        for(int i=0;i<5;i++){
            a[x+i][y]='.';
            a[x+i][y+2]='.';
        } 
        a[x][y+1]='.' ;
        a[x+2][y+1]='.' ;
        a[x+4][y+1]='.' ;
    }
}
 
signed main()
{
    IOS; 
    cin>>n>>m;
     for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)
            cin>>a[i][j];
}
    for(int j=1;j<=m;j++)
    {
        for(int i=1;i<=n;i++)
        {
            if(a[i][j]=='#')
            {
                check3(i,j);//优先考虑，因为符三包含符二和符三 
            }
            if(a[i][j]=='#')
            {
                check1(i,j);
            }
            if(a[i][j]=='#')
            {
                check2(i,j);
            }
            //每次都判断一下该点 
        }
    }
 
    cout<<res1<<" "<<res2<<" "<<res3<<endl;
     return 0;
}

写到这里舍友在打atcoder，于是写了一道B

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//中途打atcoder的舍友说这题有趣，他wa了两个点
//于是来看看
#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
using namespace std;
const int N=1e6+100;
int sum1[N];
int sum2[N];
int main()
{
    string s="wbwbwwbwbwbw";
    string a=" ";
    int x,y;
    cin>>x>>y;
    rep(i,1,100)
    {
        a+=s;
    }
    int t=s.length()-1;
    rep(i,1,t*100)
{
if(a[i]=='w')
{
    sum1[i]=sum1[i-1]+1;
    sum2[i]=sum2[i-1];
}
else
{
    sum1[i]=sum1[i-1];
    sum2[i]=sum2[i-1]+1;
}
}
rep(i,x+y,t*100-1)
{
if(sum1[i]-sum1[i-x-y]==x&&sum2[i]-sum2[i-x-y]==y)
{cout<<"Yes";return 0;
}
}
cout<<"No";

}

G

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//一眼二分
//没想到敲一遍就过了
#include<iostream>
#include<vector>
using namespace std;
signed main(){

 ios::sync_with_stdio(false);

 cin.tie(0);

 
 int n;

 cin >> n;

 vector<long long> a(n + 2);

 
 for(int i = 1; i <= n; i++){

 cin >> a[i];

 }

 auto check = [&](long long x){

 auto b = a; 
 for(int i = 1; i <= n; i++){

 if(x <= b[i]){

 return false;

 }

 x += a[i] / 2;

 b[i + 1] += (a[i] + 1) / 2;

 }

 return true;

 };

 
 long long low = 1 , high = 1E9 + 10;

 while(low+1 < high){

 long long mid = (low + high) / 2;

 if(check(mid)){

 high = mid;

 }else{

 low = mid ;

 }

 }
 cout << high << "\n";
}

H

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#include <iostream>
using namespace std;
int main()
{
    char str[100];
    cin.get(str,100);
    for (int i = 0;i < 100;i++)
    {
        if (str[i] == '4')str[i] = 'a';
        if (str[i] == '6')str[i] = 'b';
        if (str[i] == '3')str[i] = 'e';
        if (str[i] == '9')str[i] = 'g';
        if (str[i] == '1')str[i] = 'l';
        if (str[i] == '0')str[i] = 'o';
        if (str[i] == '5')str[i] = 's';
        if (str[i] == '7')str[i] = 't';
        if (str[i] == '2')str[i] = 'z';
    }
    cout << str;
}

J

按位搜索，最精华的一部分

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//这题赛时想了2小时（）
//这下必须得补一下按位搜索了
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int  i=a;i>=n;i--)
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
int n,m;
cin>>n>>m;
vector<vector<int> >a(n+1,vector<int>(m+1));
//习惯用这个把
vector<vector<int> >vis(n+1,vector<int>(m+1));
rep(i,1,n)
{
    rep(j,1,m)
    {
        cin>>a[i][j];
    }
}
int ans=0;
frep(i,29,0)
{
rep(i,1,n)
{
    rep(j,1,m)
    {
        vis[i][j]=0;
    }
}
int tag=ans+(1<<i);
queue<array<int,2>>que;
if((a[1][1]&tag)==tag)
{
que.push({1,1});
vis[1][1]=1;
}
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
while(que.size())
{
    array<int,2>t=que.front();
    int x=t[0];
    int y=t[1];
    que.pop();
    rep(i,0,3)
    {
        int tx=x+dx[i];
        int ty=y+dy[i];
        bool flag=(tx>=1)&&(tx<=n)&&(ty<=m)&&(ty>=1)&&(vis[tx][ty]==0);
        if(flag)
{
    if((a[tx][ty]&tag)==tag)
    {
        vis[tx][ty]=1;
        que.push({tx,ty});
    }
    else
    {
        vis[tx][ty]=2;
    }
}
    }
    if(vis[n][m]==1)
{
    ans|=(1<<i);
}
}
}
cout<<ans<<'\n';



}

K

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//思路很容易，直接进行模拟即可
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
map<char,int> mp;
string s;
int main(){
    int n,ans=0,cur=0;
    cin>>n;
    cin>>s;
    for(int i=0;i<n;i++){
       if(cur>0){
           cur--;
           continue;
       } 
        else{
            ans++;
            mp[s[i]]++;
            if(mp['g']>0&&mp['h']>0&&mp['z']>0&&mp['u']>0){
                cur+=mp['g']+mp['h']+mp['z']+mp['u'];
                mp['g']=0;
                mp['h']=0;
                mp['z']=0;
                mp['u']=0;           
            } 
        }
    }
    cout<<ans;
     
     
    return 0;
}

M

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//赛时这道没出我未免也太胆小了
//明明那么多人过
#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
using namespace std;
const int N=1e6+100;
struct node
{
    int x,y,cnt;
}a[N];
bool cmp(node a,node b)
{
    if(a.y==b.y)
    {
        return a.x<b.x;
    }
    return a.y<b.y;
}
int main()
{
    int n;
    cin>>n;
rep(i,1,n)
{
    cin>>a[i].x>>a[i].y;
    a[i].cnt=i;
}
sort(a+1,a+1+n,cmp);
cout<<n/2<<'\n';
for(int i=1;i<=n;i+=2)
{
    if(i+1>n)
    {
break;
    }
    else
    {
        cout<<a[i].cnt<<" "<<a[i+1].cnt<<'\n';
    }
}

    
}