牛客小白月赛59

1|0我会开摆

枚举一下左上角即可

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

string g[5];


void solve(){
    for( int i = 0 ; i < 4 ; i ++ )
        cin >> g[i];
    for( int i = 0 ; i < 3 ; i ++ )
        for( int j = 0 ; j < 3 ; j ++ )
            if( g[i][j] == g[i][j+1] && g[i][j] == g[i+1][j] && g[i][j] == g[i+1][j+1] ){
                cout << "Yes\n";
                return;
            }
    cout << "No\n";
    return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

2|0走廊的灯

找一下联系非1 的长度和连续非 0 的长度，取较大值

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

string g[5];


void solve(){
    int n = read();
    string s;
    cin >> s;
    int cnt = 0 , res = 0;
    for( auto c : s ){
        if( c != '1' ) cnt ++ , res = max( res , cnt );
        else cnt = 0;
    }
    cnt = 0;
    for( auto c : s ){
        if( c != '0' ) cnt ++ , res = max( res , cnt );
        else cnt = 0;
    }
    cout << res << "\n";
    return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

3|0输出练习

特判 0 的所有次方有 0 和 1，1的所有次方只有1。

对于当前数字 $t$ 当 $t\times k \ge r$ 时可以结束循环，对该式子变换可得 $t\ge \frac{r}{k}$

#include<bits/stdc++.h>
#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

void solve(){
    int l = read() , r = read() , k = read() , f = 0;
    if( k == 0 && l <= 0 && 0 <= r ) std::cout << "0 " , f = 1;
    if( l <= 1 && 1 <= r ) std::cout<< "0\n";

    for( int t = 1 ; k > 1 && t <= r / k; t *= k ){
        t *= k;
        if( t >= l ) std::cout << t  , f = 1;

    }

    if( f ) printf("\n");
    else printf("None.\n");
    return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

4|0国际象棋

我的做法是用两个 set 分别存储棋子是否存在，对于每次下一个棋子，统计他所在的四条线上相邻的同色棋子数量。

#include<bits/stdc++.h>
#define int long long
using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int dx[] = {1,1,0,-1};
const int dy[] ={0,1,1,1};

const int N = 1e3+5;
int h[N];
set< pair<int,int> > g[2];


int32_t main() {
    int n = read() , m = read() , k = read() , t = read();
    for( int i = 1 , x , y , c = 1 ; i <= t ; i ++ , c ^= 1 ){
        x = read() , y = ++h[x] ;
        g[c].insert(make_pair( x , y ) );
        for( int d = 0 ; d < 4 ; d ++ ){
            int cnt = 1;
            for( int j = 1 ; ; j ++ ){
                if( g[c].find(make_pair( x + dx[d]*j , y + dy[d]*j ) ) != g[c].end() )
                    cnt ++;
                else
                    break;
            }
            for( int j = -1 ; ; j -- ){
                if( g[c].find(make_pair( x + dx[d]*j , y + dy[d]*j ) ) != g[c].end() )
                    cnt ++;
                else
                    break;
            }
            if( cnt >= k ){
                cout << i << "\n";
                return 0;
            }
        }
    }
    return 0;
}