A. 俄罗斯国旗

直接模拟就好了

#define MAX 55
int W[MAX], B[MAX], R[MAX];
string mp[MAX];
signed main() {
	IOS;
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> mp[i];
		mp[i] = ' ' + mp[i];
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			W[i] += mp[i][j] != 'W';
			B[i] += mp[i][j] != 'B';
			R[i] += mp[i][j] != 'R';
		}
		W[i] += W[i - 1];
		B[i] += B[i - 1];
		R[i] += R[i - 1];
	}
	int ans = 1e9;
	for (int i = 1; i <= n; i++) {
		for (int j = i + 1; j < n; j++) {
			ans = min(ans, W[i] + B[j] - B[i] + R[n] - R[j]);
		}
	}
	cout << ans << endl;

	return 0;
}

B. 被感染的二进制数

模拟一下二进制乘法就好了

signed main()
{
    IOS;
    string s;
    cin >> s;
    int n = s.size();
    s = ' ' + s;
    vector<int> a(11111, 0), b(11111, 0);
    for (int i = 1; i <= n; i++)
        b[i + 4] = a[i] = s[i] - '0';
    int op = 0;
    for (int i = n + 4, t = 0; i >= 0; i--)
    {
        a[i] += b[i] + op;
        op = a[i] / 2;
        a[i] %= 2;
    }
    for (int i = a[0] ? 0 : 1; i <= n + 4; i++)
    {
        cout << a[i];
    }

    return 0;
}

C. 造篱笆

因为h很小所以可以用DP的思路去实现一下

signed main()
{
    IOS;
    int n;
    cin >> n;
    vector<int> v(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> v[i];
    }
    vector<int> h(4500), k(4500);
    int cnt = 0, mx = 0;
    for (int i = 1; i <= n; i++)
    {
        h[v[i]]++;
    }
    for (int i = 1; i <= 2000; i++)
    {
        for (int j = i; j <= 2000; j++)
        {
            if (h[i] == 0 || h[j] == 0)
                continue;
            if (i == j)
            {
                k[i + j] += h[i] / 2;
                continue;
            }
            k[i + j] += min(h[i], h[j]);
        }
    }
    for (int i = 1; i <= 4200; i++)
    {
        // cout << k[i] << ' ';
        mx = max(mx, k[i]);
    }
    for (int i = 1; i <= 4200; i++)
    {
        if (mx == k[i])
        {
            cnt++;
        }
    }
    cout << mx << ' ' << cnt << endl;

    return 0;
}

D. 回文数字

还是去模拟的去做

const int N = 1e5 + 10;

string s;
int a[N];

void solve() {
    cin >> s;
    int n = s.size();
    s = "$" + s;
    
    for (int i = 1; i <= (n + 1) / 2; i ++ ){
        a[i] = a[n - i + 1] = s[i] - '0';
    }
    
    int ans = 0;
    for (int i = 1; i <= n; i ++ ) //大
        if (a[i] != s[i] - '0'){
            ans = a[i] > (s[i] - '0');
            break;
        }
    if (ans){
        for (int i = 1; i <= n; i ++ )
            cout << a[i] ;
        return ;
    }
    
    int j = (n + 1) / 2;
    while (j >= 1 && a[j] == 9){
        j --;
    }
    if (j == 0){
        //中间添加
        for (int i = 1; i <= n + 1; i ++ )
            cout << "01"[i == 1 || i == n + 1];
    }else{
        a[j] ++;
        for (int i = j + 1; i <= (n + 1) / 2; i ++ ) a[i] = 0;
        for (int i = 1; i <= n; i ++ )
            cout << (i <= (n + 1) / 2 ? a[i] : a[n - i + 1]);
    }
}

signed main() {
    IOS;
    int t = 1;
//    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

E. 礼物

双端搜索就好了

#define MAX 35
int a[MAX], b[MAX];
int l;
int n;
vector<int> ans[MAX];
int res = 1e18;
void dfs1(int idx, int sum, int num) {
    if (idx == l + 1) {
        ans[num].push_back(sum);
        return ;
    }
    dfs1(idx + 1, sum + a[idx], num + 1);
    dfs1(idx + 1, sum - b[idx], num);
}

void get(int num, int sum) {
    for (auto tmp : ans[num]) {
        //cout << tmp << ' ' << sum << endl;
        res = min(res, abs(tmp + sum));
    }
}
void dfs2(int idx, int sum, int num) {
    if (idx == n + 1) {
        get((n + 1) / 2 - num, sum);
        get(n / 2 - num, sum);
        return ;
    }
    dfs2(idx + 1, sum + a[idx], num + 1);
    dfs2(idx + 1, sum - b[idx], num);
}

signed main() {
    IOS;
    cin >> n;
    l = (n + 1) / 2;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
        cin >> b[i];
    }
    dfs1(1, 0, 0);
    for (int i = 0; i <= n; i ++ ){
        sort(ans[i].begin(), ans[i].end());
    }
    dfs2(l + 1, 0, 0);
    cout << res << endl;
    return 0;
}

F. 必经过点

缩点模版 + 数学

#define MAX 500005
vector<int> G[MAX];
int dfn[MAX], low[MAX], cnt, sz[MAX], n, m;
int ans[MAX];
void dfs(int u, int fa) {
    dfn[u] = low[u] = ++cnt;
    sz[u] = 1;
    int son = 0;
    for (auto v : G[u]) {
        if (v == fa) continue;
        if (dfn[v]) {
            low[u] = min(low[u], dfn[v]);
        } else {
            dfs(v, u);
            low[u] = min(low[u], low[v]);
            sz[u] += sz[v];
            if (dfn[u] <= low[v]) {
                ans[u] += son * (ll)sz[v];
                son += sz[v];
            }
        }
    }
    ans[u] += son * (ll)(n - 1 - son);
}


signed main() {
    IOS;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v); G[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
        cout << ((ans[i] -1  + n) << 1) << endl;
    }
    return 0;
}