SDUT 2022 Winter Team Contest - 1

Posted 2023-02-18 佐鼬Jun

题目链接：link

A - Carryless Square Root

思路：

两个数字进行题目说的运算，的出来的长度一定是奇数，因为$a\cdot a=n$ $a$的长度为$len$话，最终结果的长度为$2\ast len-1$，把运算展开就可以的到这个结论，然后从高位到低位，从小数开始试，$dfs$

#include <bits/stdc++.h>
using namespace std;
const int N = 666;
string s;
int a[N], res[N], sum[N];
int len;
bool flag;

bool check(int pos) 
    memset(sum, 0, sizeof(sum));
    for (int i = 0; i <= pos; i++) 
        for (int j = 0; j <= pos; j++) 
            sum[i + j] = sum[i + j] + (res[i] * res[j]) % 10;
            sum[i + j] %= 10;
        
    
    if (pos == len - 1) 
        pos = s.size() - 1;
    
    for (int i = 0; i <= pos; i++) 
        if (sum[i] != a[i]) 
            return 0;
        
    
    return 1;


void dfs(int pos) 
    if (pos == len) 
        for (int i = 0; i < pos; i++) 
            cout << res[i];
        
        cout << endl;
        flag = 1;
        exit(0);
    

    for (int i = 0; i < 10; i++) 
        res[pos] = i;
        if (check(pos)) 
            dfs(pos + 1);
        
    

int main() 
    cin >> s;
    len = s.size();
    if (len % 2 == 0) 
        cout << -1 << endl;
        return 0;
    
    for (int i = 0; i < len; i++) 
        a[i] = s[i] - '0';
    
    len = (len + 1) / 2;
    dfs(0);
    if (flag == 0) 
        cout << -1 << endl;
    

I - Maze Connect

思路：

有些空间，例如空白的空地，在原图中是没有坐标的，所以不好判断图的封闭性，但是把原图放大为原来的$2$倍后，再按照原图的相对位置，把图案还原，就能发现无论是空地还是墙壁都是有坐标的，然后直接$dfs$搜索，搜有多少个封闭空间

#include <bits/stdc++.h>
using namespace std;
const int N = 6666;
char s[N];
int n, m;
int g[N][N];
int dx[4] = -1, 1, 0, 0;
int dy[4] = 0, 0, -1, 1;

void dfs(int x, int y) 
    g[x][y] = 1;
    for (int i = 0; i < 4; i++) 
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (nx < 1 || nx > 2 * n || ny < 1 || ny > 2 * m) continue;
        if (g[nx][ny]) continue;
        dfs(nx, ny);
    

int main() 
    cin >> n >> m;
    for (int i = 1; i <= n; i++) 
        scanf("%s", s + 1);
        for (int j = 1; j <= m; j++) 
            if (s[j] == '/') 
                int x = 2 * i, y = 2 * j;
                g[x][y - 1] = 1;
                g[x - 1][y] = 1;
             else if (s[j] == '\\\\') 
                int x = 2 * i, y = 2 * j;
                g[x - 1][y - 1] = 1;
                g[x][y] = 1;
            
        
    

    for (int i = 1; i <= 2 * m; i++) 
        if (!g[1][i]) 
            dfs(1, i);
        
        if (!g[2 * n][i]) 
            dfs(2 * n, i);
        
    

    for (int i = 1; i <= 2 * n; i++) 
        if (!g[i][1]) 
            dfs(i, 1);
        
        if (!g[i][2 * m]) 
            dfs(i, 2 * m);
        
    
    int res = 0;

    for (int i = 1; i <= 2 * n; i++) 
        for (int j = 1; j <= 2 * m; j++) 
            if (g[i][j] == 0) 
                dfs(i, j);
                res++;
            
        
    
    cout << res << endl;

To be continued
如果你有任何建议或者批评和补充，请留言指出，不胜感激