牛客国庆集训派对Day4 Solution

A    深度学习

puts(n)

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    double n;
    while (scanf("%lf", &n) != EOF)
        printf("%.10f
", n);
    return 0;
}

B    异或求和

思路：  有一个$O(nlogn^2)$ 的做法，但是T了。

设$bit(x, i)$为 x在二进制下第i位为1还是0

考虑 $sum_{1 <= i < j < k <= n}(a_i oplus a_j)(a_j oplus a_k)(a_i oplus a_k)$

将最后一部分的 $(a_i oplus a_k)$ 拆分成 $sum_{l = 0}^{l = 29}(bit(a_i, l))(bit(a_k, l))$

那么我们再枚举j  就可以

T了的代码：

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

#define N 100010
#define ll long long

ll MOD = (ll)998244353;

int n;
int arr[N], prefix[30][2][30][2], suffix[30][2][30][2]; 
 
void Run()
{
    while (scanf("%d", &n) != EOF) 
    {
        for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
        bitset <30> b; 
        for (int i = 2; i <= n; ++i) 
        {
            b = arr[i];
            for (int j = 29; j >= 0; --j)
                for (int k = 29; k >= 0; --k)
                    ++suffix[j][b[j]][k][b[k]]; 
        }
        b = arr[1]; 
        for (int j = 29; j >= 0; --j)
            for (int k = 29; k >= 0; --k)
                ++prefix[j][b[j]][k][b[k]];
        ll ans = 0;
        for (int i = 2; i < n; ++i)
        {
            b = arr[i];
            for (int j = 29; j >= 0; --j)
                for (int k = 29; k >= 0; --k)
                    --suffix[j][b[j]][k][b[k]]; 
            for (int j = 29; j >= 0; --j) 
            {
                ll l = 0, r = 0;
                // enum 0 1
                for (int k = 29; k >= 0; --k)
                {
                    
                    l = (l + ((1ll << k) * prefix[j][0][k][b[k] ^ 1]) % MOD) % MOD;
                    r = (r + ((1ll << k) * suffix[j][1][k][b[k] ^ 1]) % MOD) % MOD;
                }
                ans = (ans + (1ll << j) * l % MOD * r % MOD) % MOD;
                l = 0, r = 0;
                // enum 1 0 
                for (int k = 29; k >= 0; --k)
                {
                    l = (l + ((1ll << k) * prefix[j][1][k][b[k] ^ 1]) % MOD) % MOD;
                    r = (r + ((1ll << k) * suffix[j][0][k][b[k] ^ 1]) % MOD) % MOD;
                }
                ans = (ans + (1ll << j) * l % MOD * r % MOD) % MOD; 
            }
            for (int j = 29; j >= 0; --j) 
                for (int k = 29; k >= 0; --k) 
                    ++prefix[j][b[j]][k][b[k]]; 
        }
        printf("%lld
", ans);
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif

    Run();
    return 0;
}

C    异或计数

留坑。

D    最小生成树

思路：用最小的点权去连边

#include <bits/stdc++.h>
using namespace std;
 
#define N 100010
#define ll long long
 
int n;
 
int main()
{
    while (scanf("%d", &n) != EOF)
    {
        ll res = 0, num, Min = 0x3f3f3f3f3f3f3f3f;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%lld
", &num);
            Min = min(Min, num);
            res += num;
        }
        printf("%lld
", res + Min * (n - 2));
    }
    return 0;
}

E    乒乓球

留坑。

F    导数卷积

留坑。

G    区间权值

显然可以发现长度为l和n-l的权值相同，然后枚举每个长度l，发现出现次数为两端向中间增加，然后中间不变，最后O(n)扫一遍即可

#include<bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
 
const int maxn = (int)3e5 + 10;
const ll MOD = 1e9 + 7;
int n;
ll ans;
ll w[maxn];
ll val[maxn];
 
int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; ++i)
        {
            scanf("%lld", val + i);
        }
        for(int i = 1; i <= n; ++i)
        {
            scanf("%lld", w + i);
        }
        for(int i = 1; i <= n; ++i)
        {
            val[i] += val[i - 1];
            val[i] %= MOD;
        }
        ans = 0;
        ll tmp = 0;
        for(int i = 1; i <= n / 2; ++i)
        {
            tmp += (val[n - i + 1] - val[i - 1] + MOD) % MOD;
            tmp %= MOD;
            ans += (tmp * w[i] % MOD + tmp * w[n - i + 1] % MOD) % MOD;
            ans %= MOD;
        }
        if(n & 1)
        {
            tmp += val[(n + 1) / 2] - val[n / 2];
            tmp %= MOD;
            ans += tmp * w[(n + 1) / 2] % MOD;
            ans %= MOD;
        }
        printf("%lld
", ans);
    }
    return 0;
}

H    树链博弈

思路：每层上都有偶数个黑点，后手必胜，否则，先手必胜

假设所有点都是白点，是先手必败态

如果先手必胜，那么走一步，走到先手必败态

如果是先手必败态，走一步，可以走到先手必胜态

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

#define N 1010

int n;
int color[N], deep[N];
vector <int> G[N];

void DFS(int u, int fa, int deepth)
{
    deep[deepth] += color[u]; 
    for (auto v : G[u])
    {
        if (v == fa) continue;
        DFS(v, u, deepth + 1);
    }
}

void Run()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%d", color + i);
        for (int i = 1, u, v; i < n; ++i)
        {
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        DFS(1, 1, 0);
        int res = 0;
        for (int i = 0; i < n; ++i) if (deep[i] & 1) res = 1;
        puts(res ? "First" : "Second");
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif

    Run();
    return 0;
}

I    连通块计数

分两种情况考虑，一种是含根，一种是不含根，分别计数

#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
    long long ans, t ,x, i, n, MOD;
    scanf("%lld
",&n);
    MOD = 998244353;
    ans = 0; t = 1;
    for (i = 1; i <= n; ++i) {
        scanf("%lld
",&x);
        ans = (ans + (x * x + x) / 2) % MOD;
        t = (t * (x + 1)) % MOD;
    }
    ans = (ans + t) % MOD;
    printf("%lld
",ans);
    return 0;
}

J    寻找复读机

按题意模拟即可

#include <bits/stdc++.h>
using namespace std;
 
#define N 1010
 
int n, m;
int used[N], vis[N];
vector <int> ans;
char s[N][110];
 
int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        memset(vis, 0, sizeof vis);
        memset(used, 0, sizeof used);
        ans.clear();
        for (int i = 1, x; i <= m; ++i)
        {
            scanf("%d %s", &x, s[i]);
            if (i == 1) vis[x] = 1;
            else if (strcmp(s[i], s[i - 1]) != 0) vis[x] = 1;
            used[x] = 1;   
        }
        for (int i = 1; i <= n; ++i) if (!vis[i])
            ans.push_back(i);
        if (!ans.size()) puts("");
        for (int i = 0, len = ans.size(); i < len; ++i) printf("%d%c", ans[i], " 
"[i == len - 1]);
    }
    return 0;
}