OIFC未来共同体20241030noip模拟四

T1

我们发现 \(1\) 其实根本没有用，只和一个连通块里的 \(0\) 的个数有关，直接 \(dfs\)，判断即可。

#include<iostream>
#include<cstring>

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}

const int N = 110;
int T, n, m, sum, ans;
int a[N][N], dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};
bool vis[N][N];
bool check(int x, int y){
    if (x < 1 || x > n || y < 1 || y > m || a[x][y] == -1) return 0;
    return 1;
}
void dfs(int x, int y){
    if (vis[x][y]) return;
    vis[x][y] = 1;
    if (a[x][y] == 0) sum++;
    for (int i = 0; i < 4; i++){
        int nx = x + dx[i], ny = y + dy[i];
        if (!check(nx, ny)) continue;
        dfs(nx, ny);
    }
}
void solve(){
    n = read(), m = read();
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++){
            char c; cin >> c;
            if (c == 'X') a[i][j] = -1;
            else a[i][j] = c - '0';
        }
    bool flag = 0;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            if (a[i][j] != -1 && !vis[i][j] && !flag){
                flag = 1, sum = 0;
                dfs(i, j);
                if (sum & 1) ans++;
            }else if (a[i][j] != -1 && !vis[i][j] && flag){
                cout << "No\n";
                ans = 0;
                memset(vis, 0, sizeof vis);
                memset(a, 0, sizeof a);
                return;
            }
        }
    }
    cout << (ans == 1 ? "Yes\n" : "No\n");
    ans = 0;
    memset(vis, 0, sizeof vis);
    memset(a, 0, sizeof a);
}

int main(){
    freopen("chess.in", "r", stdin);
    freopen("chess.out", "w", stdout);
    T = read();
    while (T--) solve();
    return 0;
}

T2

结论题，不为 \(1\) 的数的个数为奇数个就为 No，否则就为 Yes，其实也挺好证明的，这里略。

#include<iostream>
#include<algorithm>

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}

const int N = 5e5 + 10;
int T, n;
int a[N];
void solve(){
    n = read();
    int cnt = 0, el;
    for (int i = 1; i <= n; i++){
        a[i] = read();
        if (a[i] == 1) cnt++;
    }
    el = n - cnt;
    if (el & 1) cout << "Yes\n";
    else cout << "No\n";
}

int main(){
    freopen("delete.in", "r", stdin);
    freopen("delete.out", "w", stdout);
    T = read();
    while (T--) solve();
    return 0;
}

T3

将题目意思转化一下，就是求子序列和为 \(0\) 所在区间的个数，直接背包然后拆分贡献计算。

#include<iostream>
#define int long long

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}

const int N = 1e3 + 10, M = 2e5 + 10, K = 1e5, mod = 998244353;
int n, ans;
int a[N], dp[M];

signed main(){
    freopen("seq.in", "r", stdin);
    freopen("seq.out", "w", stdout);
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 2; i <= n - 1; i++){
        if (a[i] >= 0){
            for (int j = K; j >= -K; j--)
                if (j - a[i] >= -K && j - a[i] <= K)
                    dp[j + K] = (dp[j + K] + dp[j + K - a[i]]) % mod;
        }else{
            for (int j = -K; j <= K; j++)
                if (j - a[i] >= -K && j - a[i] <= K)
                    dp[j + K] = (dp[j + K] + dp[j + K - a[i]]) % mod;
        }
        dp[a[i] + K] = (dp[a[i] + K] + i - 1) % mod;
        ans = (ans + dp[K]) % mod;
    }
    cout << (ans + n * (n + 1) / 2 % mod) % mod;
    return 0;
}

T4

不会。