Vegetable Chicken Contest Round 1

分数排名

题目-分数排名 (51nod.com)

把所有的数按照降序排序,每个人的分数在排序后第一次出现的位置就是这个人的排名,顺便说一句哈希的过程和这个很相似,只不过哈希需要把排序后的数组去重

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

const int N = 10005;
int n , a[N] , b[N];
string s[N];

int main()
{
    cin >> n;
    for( int i = 1 ; i <= n ; i ++ )
    {
        cin >> s[i] >> a[i];
        b[i] = a[i];
    }
    sort( b+1 , b+1+n , greater<int>() );
    for( int i = 1 , t ; i <= n ; i ++ )
    {
        t = lower_bound( b+1 , b+1+n , a[i] , greater<int>() ) - b;
        cout << s[i] << ' ' << t << endl;
    }
    return 0;

}

二维数组回形遍历

题目-二维数组回形遍历 (51nod.com)

首先把边界标记一下,如果遇到被标记为的点就转一下方向,如果被转过一次方向后依然是被标记的点就要退出迭代,这题也可以用循环来写

#include <bits/stdc++.h>
#define tx ( x + dx[d] )
#define ty ( y + dy[d] )
using namespace std;

const int N = 105;
const int dx[] = {0,1,0,-1} , dy[] ={1,0,-1,0};
int n , m , a[N][N] ;
bool vis[N][N];

void output( int x , int y , int d )
{
    vis[x][y] = 1;
    cout << a[x][y] << endl;
    if( vis[ tx ][ ty ] ) d = ( d + 1 ) % 4;
    if( vis[tx][ty] ) return;
    output( tx , ty , d );
    return ;
}

int main(){
    cin >> n >> m;
    for( int i = 1 ; i <= n ; i ++ )
        for( int j = 1 ; j <= m ; j ++ )
            cin >> a[i][j];
    for( int i = 1 ; i <= m ; i ++ )
        vis[0][i] = vis[n+1][i] = 1;
    for( int i = 1 ; i <= n ; i ++ )
        vis[i][0] = vis[i][m+1] = 1;
    output(1,1,0);
    return 0;
}

循环版

#include <bits/stdc++.h>
#define tx ( x + dx[d] )
#define ty ( y + dy[d] )
using namespace std;

const int N = 105;
const int dx[] = {0,1,0,-1} , dy[] ={1,0,-1,0};
int n , m , a[N][N] ;
bool vis[N][N];

int main(){
    cin >> n >> m;
    for( int i = 1 ; i <= n ; i ++ )
        for( int j = 1 ; j <= m ; j ++ )
            cin >> a[i][j];
    for( int i = 1 ; i <= m ; i ++ )
        vis[0][i] = vis[n+1][i] = 1;
    for( int i = 1 ; i <= n ; i ++ )
        vis[i][0] = vis[i][m+1] = 1;

    for( int x = 1 , y = 1 , d = 0 ; 1 ; )
    {
        vis[x][y] = 1;
        cout << a[x][y] << endl;
        if( vis[tx][ty] ) d = ( d + 1 ) % 4;
        if( vis[tx][ty] ) break;
        x = tx , y = ty;
    }
    return 0;
}

JJOOII 2

#3253. 「JOI 2020 Final」JJOOII 2 - 题目 - LibreOJ (loj.ac)

首先记录下每种字母出现的位置,然后枚举第一个\(J\),在数组中找到最近的第\(k\)个\(J\),然后通过二分找到后面最近的\(O\)以此类推,找到完整的序列后计算一下操作三的次数即可

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

const int N = 2e5+5;
int n , m , res = N;
vector<int> J , O , I;
string s;

int main(){
    cin >> n >> m >> s;
    for( int i = 0 ; i <= s.size() ; i ++ )
    {
        if( s[i] == 'J' ) J.push_back(i);
        else if( s[i] == 'O' ) O.push_back(i);
        else I.push_back(i);
    }
    for( int i = 0 , r , k ; i + m <= J.size() ; i ++ )
    {
        k = J[i+m-1] ;
        r = upper_bound(O.begin() , O.end() , k ) - O.begin();
        if( r+m-1 >= O.size() ) break;
        k = O[r+m-1];
        r = upper_bound( I.begin() , I.end() , k ) - I.begin();
        if( r + m >= I.size() ) break;
        r += m-1 , res = min( res , I[r]-J[i]+1 - 3*m );
    }
    if( res == N ) res = -1;
    cout << res << endl;
}