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  • HourRank 20

    第一题略

    第二题组合数学

    s1 = min(cnt['a'],cnt['b']),

    s2 = min(cnt['c'],cnt['d']),

    b1 = max(cnt['a'],cnt['b']),

    b2 = max(cnt['c'],cnt['d']);

    简单分开枚举'a','b'一组和'c','d'一组然后乘起来就好了

    ans = sigma(C(s1,i) * C(b1,i))(0<=i<=s1) * sigma(C(s2,i) * C(b2,i))(0<=i<=s2) - 1;

    #include<bits/stdc++.h>
    #define MP make_pair
    using namespace std;
    typedef long long LL;
    const int N = 5e5 + 5;
    const LL mod = 1e9 + 7;
    int Tm[333];
    char s[N];
    LL fac[N];
    LL inv[N];
    LL qpow(LL x,LL b){
        LL res = 1;
        while(b){
            if(b&1)res = res * x % mod;
            x = x * x % mod;
            b >>= 1;
        }
        return res;
    }
    void init(){
        fac[0] = 1;
        for(int i = 1; i < N; i++)fac[i] = fac[i-1] * i % mod;
        inv[N - 1] = qpow(fac[N - 1],mod - 2);
        for(int i = N - 2; i >= 0; i--){
            inv[i] = inv[i + 1] * (i + 1) % mod;
        }
    }
    inline LL C(int a,int b){
        return fac[a] * inv[b] % mod * inv[a - b] % mod;
    }
    int s1,b1,s2,b2;
    LL solve(){
        LL res1 = 0,res2 = 0;
        for(int i = 0; i <= s1; i++){
            res1 = (res1 + C(s1,i) * C(b1,i) % mod) % mod;
        }
        for(int i = 0; i <= s2; i++){
            res2 = (res2 + C(s2,i) * C(b2,i) % mod) % mod;
        }
    //    cout<<s1<<" "<<s2<<endl;
    //    cout<<res1<<" "<<res2<<endl;
        return (res1 * res2 % mod - 1 + mod)%mod;
    }
    int main() {
       // freopen("input", "r", stdin);
        // freopen("input", "w", stdout);
        ios::sync_with_stdio(false);
        cin.tie(0);
        init();
        int q;
        cin>>q;
        while(q--){
            memset(Tm,0,sizeof Tm);
            cin>>s;
            int len = strlen(s);
            for(int i = 0; i < len; i++)Tm[(int)s[i]]++;
            s1 = min(Tm['a'],Tm['b']);
            b1 = max(Tm['a'],Tm['b']);
            s2 = min(Tm['c'],Tm['d']);
            b2 = max(Tm['c'],Tm['d']);
            cout<<solve()<<endl;
        }
        return 0;
    
    }
    View Code

    第三题对于每个询问建一颗虚树,根节点要放在虚树里,如果u , fa,fa是u的祖先,并且u到fa这条路径上只有这两个点,那这条路径的点子树黑点都等于u上的子树黑点。

    #include<bits/stdc++.h>
    #define MP make_pair
    #define PB push_back
    using namespace std;
    typedef long long LL;
    const int N = 6e5 + 5;
    const LL mod = 1e9 + 7;
    int L[N];
    int n;
    int R[N];
    int ans[N];
    int Fa[N][20];
    int deep[N];
    int tin = 0;
    vector<int>Tree[N];
    vector<int>Mtree[N];
    int rev[N];
    bool cmp(int a, int b) {
        return L[a] < L[b];
    }
    int lca(int a, int b) {
        if(deep[a] < deep[b])swap(a, b);
        int d = deep[a] - deep[b];
        for(int i = 0; i < 20; i++) {
            if(d & (1 << i))a = Fa[a][i];
        }
        if(a == b)return a;
        for(int i = 19; i >= 0; i--) {
            if(Fa[a][i] != Fa[b][i])a = Fa[a][i], b = Fa[b][i];
        }
        return Fa[a][0];
    }
    int v[N];
    int St[N];
    int top = 0;
    inline bool isFa(int fa,int u){
        return L[fa] <= L[u] && R[fa] >= R[u];
    }
    inline void dfs(int u,int fa){
        for(int to : Mtree[u]){
            dfs(to,u);
            v[u] += v[to];
        }
        ans[v[u]] += deep[u] - deep[fa];
    }
    inline void solve() {
        int k, x;
        cin >> k;
        int tol = 0;
        for(int i = 0; i < k; i++) {
            cin >> x;
            rev[tol++] = x;
            v[x] = 1;
        }
        sort(rev, rev + tol, cmp);
        for(int i = 1; i < k; i++) {
            rev[tol++] = lca(rev[i - 1], rev[i]);
        }
        rev[tol++] = 1;
        sort(rev, rev + tol, cmp);
        tol = unique(rev , rev + tol) - rev;
        top = 0;
        St[++top] = 1;
        for(int i = 1; i < tol; i++){
            int to = rev[i];
            while(top > 0 && !isFa(St[top],to))top--;
            int nt = St[top];
            Mtree[nt].PB(to);
            St[++top] = to;
        }
        dfs(1,0);
        ans[0] = n;
        for(int i = 1; i <= k; i++)ans[0] -= ans[i];
        for(int i = 0; i <= k; i++)cout<<ans[i]<<" ",ans[i] = 0;
        for(int i = 0; i < tol; i++){
            Mtree[rev[i]].clear();v[rev[i]] = 0;
        }
        cout<<endl;
    }
    void pre(int u, int fa) {
        deep[u] = deep[fa] + 1;
        L[u] = ++tin;
        Fa[u][0] = fa;
        for(int to : Tree[u]) {
            if(to == fa)continue;
            pre(to, u);
        }
        R[u] = tin;
    }
    void init() {
        cin>>n;
        for(int i = 1; i < n; i++) {
            int u, to;
            cin >> u >> to;
            Tree[u].PB(to), Tree[to].PB(u);
        }
        pre(1, 0);
        for(int i = 1; (1 << i) < n; i++) {
            for(int j = 1; j <= n; j++) {
                Fa[j][i] = Fa[Fa[j][i - 1]][i - 1];
            }
        }
        int q;
        cin >> q;
        while(q--) {
            solve();
        }
    }
    int main() {
       // freopen("input", "r", stdin);
        // freopen("input", "w", stdout);
        ios::sync_with_stdio(false);
        cin.tie(0);
        init();
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/scau-zk/p/6804123.html
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