2021-4-27 CSUST 银川选拔赛补题

题目传送门

A:查询区间众数出现次数

莫队板子题，太菜了看了好久ww

#include 
using namespace std;
typedef long long LL;
const int N = 1e5 + 7, M = 207;
int a[N], ans[N], group[N], n, m;
struct node
{
     
    int l, r, id;
} q[N];
int res = 0, block;
int numcnt[N]; 
int cntcnt[N];
void add(int pos)
{
     
    int t = a[pos];
    numcnt[t]++;        
    cntcnt[numcnt[t]]++; 
    res = max(res, numcnt[a[pos]]);
}
void del(int pos)
{
     
    int t = a[pos];
    cntcnt[numcnt[t]]--; 
    if (res == numcnt[t] && cntcnt[numcnt[t]] == 0)
    {
     
        res--; 
    }
    numcnt[t]--;
}
bool cmp(node a, node b)
{
     
    if (a.l / block == b.l / block)
    {
     
        return a.r < b.r;
    }
    return a.l < b.l;
}
int main()
{
     
    cin >> n >> m;
    block = sqrt(n); //块的长度
    for (int i = 1; i <= n; i++)
    {
     
        cin >> a[i];
    }
    for (int i = 1; i <= m; i++)
    {
     
        cin >> q[i].l >> q[i].r;
        q[i].id = i;
    }
    sort(q + 1, q + 1 + m, cmp);
    int l = 1, r = 0;
    for (int i = 1; i <= m; i++)
    {
     
        while (q[i].l < l)
        {
     
            add(--l);
        }
        while (q[i].r > r)
        {
     
            add(++r);
        }
        while (q[i].l > l)
        {
     
            del(l++);
        }
        while (q[i].r < r)
        {
     
            del(r--);
        }
        ans[q[i].id] = res;
    }
    for (int i = 1; i <= m; i++)
    {
     
        cout << ans[i] << endl;
    }
    return 0;
}

B:PC玩游戏

二分

#include 
using namespace std;
typedef long long lli;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
const int maxm = 200005;
typedef long long ll;
#define gcd __gcd
int pos[200005];
int vis[200005];
int n, k, m, length;
bool check(int key)
{
     
    int cnt = 0, ma = 0;
    for (int i = 1; i <= key; i++)
    {
     
        vis[pos[i]] = 1;
    }
    int last = 0;
    vis[n + 1] = 1;
    for (int i = 1; i <= n + 1; i++)
    {
     
        if (vis[i])
        {
     
            int d = i - last - 1;
            if (d >= length)
            {
     
                cnt++;
                d -= length;
            }
            cnt += (d / (length + 1));
            last = i;
        }
    }
    memset(vis, 0, sizeof vis);
    return cnt < k;
}
int main()
{
     
    int ans = inf;
    cin >> n >> k >> length >> m;
    for (int i = 1; i <= m; i++)
    {
     
        scanf("%d", &pos[i]);
    }
    if (k * length + k - 1 > n)
    {
     
        cout << 0;
        return 0;
    }
    int l = 1, r = m;
    while (l <= r)
    {
     
        int mid = (l + r) >> 1;
        if (check(mid))
        {
     
            r = mid - 1;
            ans = mid;
        }
        else
        {
     
            l = mid + 1;
        }
    }
    if (ans == inf)
    {
     
        cout << -1;
        return 0;
    }
    cout << ans;
    return 0;
}

C:PC买礼物

DAG上dp

#include 
using namespace std;
typedef long long lli;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
const int maxm = 200005;
typedef long long ll;
int w[2005], dp[2005][2005];
vector<int> edge[2005];
void addedge(int u, int v)
{
     
    edge[v].push_back(u); //到达v点的上一个点
}
int main()
{
     
    //通过dp计算有多少种方法  晚训好像写过一次（太菜了没写出来）
    //(xqqdltxdy代码一看就懂%%%%%%%)
    ll ans = 0;
    int n, m, k, u, v;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
     
        cin >> w[i];
    }
    for (int i = 1; i <= m; i++)
    {
     
        cin >> u >> v;
        addedge(u, v);
    }
    //dp[i][j]: 到第i间商店花费j元的方案数
    dp[1][0] = 1;
    dp[1][w[1]] = 1;
    for (int i = 1; i <= n; i++)
    {
     
        //不同方法指 不同路径或购买的不同礼物
        //所以需要枚举路径 与 是否购买当前礼物
        for (int j = 0; j < edge[i].size(); j++)
        {
     
            int p = edge[i][j]; //枚举到达i点 上一个点有多少点（路径枚举）
            for (int h = 0; h <= k; h++)
            {
     
                dp[i][h] = (dp[i][h] + dp[p][h]) % mod;
                if (h - w[i] >= 0)
                    dp[i][h] = (dp[i][h] + dp[p][h - w[i]]) % mod;
            }
        }
    }
    for (int i = 0; i <= k; i++)
    {
     
        ans += dp[n][i];
        ans %= mod;
    }
    cout << ans;
    return 0;
}

D:game

#include 
using namespace std;
typedef long long lli;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
lli a[1000005],vis[10000005];
int main()
{
     
    int n;
    while(~scanf("%d",&n)){
     
    if(n==0) break;
      if(n&1) cout<<"Bob"<<'\n';
      else cout<<"Alice"<<'\n';
    }
    return 0;
}

E:median

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define LL long long
const int inf = 0x3f3f3f3f;
const int mod = 1e5;
int a[1000005], n1[1000005], n2[1000005];
int main()
{
     
    int n, v, pos;
    cin >> n >> v;
    for (int i = 1; i <= n; i++)
    {
     
        scanf("%d", &a[i]);
        if (a[i] > v)
        {
     
            a[i] = -1;
        }
        else if (a[i] < v)
        {
     
            a[i] = 1;
        }
        else
        {
     
            a[i] = 0;
            pos = i;
        }
    }
    int sum = 0, ans = 0, cnt = 0;
    for (int i = pos; i >= 1; i--)
    {
     
        sum += a[i]; //zuo
        if (!sum)
        {
     
            ans++;
            cnt++;
        }
        if ((pos - i) % 2 == 0)
        {
     
            n1[mod + sum]++;
        }
        else
        {
     
            n2[mod + sum]++;
        }
    }
    sum = 0; //you
    for (int i = pos + 1; i <= n; i++)
    {
     
        sum += a[i];
        if (!sum)
        {
     
            ans += cnt;
        }
        else if ((i - pos) % 2 == 0)
        {
     
            ans += n1[mod - sum];
        }
        else
        {
     
            ans += n2[mod - sum];
        }
    }
    cout << ans;
    return 0;
}

F:重建网络

最大生成树

#include 
using namespace std;
typedef long long lli;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
const int maxm = 200005;
int f[maxn];
lli num[maxm];
int n, m, k, co;
struct node
{
     
    int u, v;
    lli w;
    
} edge[maxm];
int tot = 0;
void addedge(int u, int v, int w)
{
     
    edge[tot].u = u;
    edge[tot].v = v;
    edge[tot++].w = w;
}
bool cmp(node a, node b)
{
     
    return a.w > b.w;
}
int find(int x)
{
     
    if (f[x] == -1)
        return x;
    else
        return f[x] = find(f[x]);
}
int main()
{
     
    cin >> n >> m >> k;
    int x, y, z;
    for (int i = 1; i <= m; i++)
    {
     
        scanf("%d%d%d", &x, &y, &z);
        addedge(x, y, z);
    }
    memset(f, -1, sizeof f);
    sort(edge, edge + tot, cmp);
    int cnt = 0;
    int ans = 0;
    vector<lli> mii;
    vector<lli> maa;
    for (int i = 0; i < tot; i++)
    {
     
        int u = edge[i].u;
        int v = edge[i].v;
        int w = edge[i].w;
        int t1 = find(u);
        int t2 = find(v);
        if (t1 != t2)
        {
     
            if (w > k)
            {
     
                maa.push_back(w);
            }
            else if (w < k)
            {
     
                mii.push_back(w);
            }
            f[t1] = t2;
            //cout << "%" << t1 << ' ' << t2 << " cost " << w << endl;
            cnt++;
        }
        if (cnt == n - 1)
            break;
    }
    lli res = 0;
    if (mii.size())
    {
     
        for (auto i : mii)
        {
     
            res += k - i;
        }
        cout << res;
        return 0;
    }
    else
    {
     
        res = inf;
        for (int i = 0; i < tot; i++)
        {
     
            res = min(res, abs(edge[i].w - k));
        }
        cout << res;
        return 0;
    }
    return 0;
}

G:最大得分

三维dp，现在都还不太懂））
dp[i][j][k]表示在第i个数，第k个块gcd为j

#include 
using namespace std;
typedef long long LL;
const int N = 1e4 + 7;
int a[N], dp[N][105][55];
int main()
{
     
    //memset(dp, -1, sizeof(dp));
    int n, k;
    int ans = -1;
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
     
        cin >> a[i];
    }
    dp[1][a[1]][1] = a[1];
    for (int i = 2; i <= n; i++)
    {
     
        for (int j = 1; j <= 100; j++)
        {
     
            int g = __gcd(j, a[i]);
            for (int h = 1; h <= min(k, i); h++)
            {
     
                //放到新一组
                if (dp[i - 1][j][h - 1])
                    dp[i][a[i]][h] = max(dp[i][a[i]][h], dp[i - 1][j][h - 1] + a[i]);
                //合并
                if (dp[i - 1][j][h])
                    dp[i][g][h] = max(dp[i][g][h], dp[i - 1][j][h] - j + g);
            }
        }
    }
    for (int i = 1; i <= 100; i++)
    {
     
        ans = max(ans, dp[n][i][k]);
    }
    cout << ans;
    return 0;
}