2020CCPC河南省赛题解

A. 班委竞选

签到题，模拟。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long
//#define double long double

using namespace std;
typedef unsigned long long ULL ;
typedef pair<int,int> PII ;
typedef pair<double,double> PDD ;
const int N = 200010 , M = N * 2 , mod = 998244353 ;
int n , m ;
int a[20] , id[20] ;

void solve()
{
    cin >> n >> m ;
    for(int i = 1 ; i <= n ; i ++) {
        int c , x ;
        cin >> c >> x ;
        if(x > a[c]) {
            a[c] = x ;
            id[c] = i ;
        }
    }
    for(int i = 1 ; i <= m ; i ++)
        cout << id[i] << " " ;
    cout << "\n" ;
}
signed main()
{
    std::ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;
    int t = 1 ;
//    cin >> t ;
    while (t --) solve() ;
    return 0;
}

C. 我得重新集结部队

模拟。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long
//#define double long double

using namespace std;
typedef unsigned long long ULL ;
typedef pair<int,int> PII ;
typedef pair<double,double> PDD ;
const int N = 200010 , M = N * 2 , mod = 998244353 ;
int n ;
struct node{
    int id , x , y , h , c , r , f ;
}a[2010];

void solve()
{
    cin >> n ;
    for(int i = 1 ; i <= n ; i ++) {
        cin >> a[i].id ;
        if(a[i].id == 1) {
            cin >> a[i].x >> a[i].y >> a[i].h ;
            a[i].f = 1 ;
        }
        else {
            cin >> a[i].x >> a[i].y >> a[i].c >> a[i].r ;
            a[i].f = 1 ;
        }
    }
    for(int i = 1 ; i <= n ; i ++) {
        if(a[i].id == 2) {
            int uid = -1 , dis = 1e18 ;
            for(int j = 1 ; j < i ; j ++)
                if(a[j].id == 1 && a[j].f == 1) {
                    int s = (a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y) ;
                    if(s < dis) {
                        dis = s ;
                        uid = j ;
                    }
                }
            if(uid != -1) {
                int fx = a[uid].x , fy = a[uid].y , d = a[i].r * a[i].r , ct = a[i].c * 3 ;
                for(int j = 1 ; j < i ; j ++)
                    if(a[j].id == 1 && a[j].f == 1) {
                        int s = (fx - a[j].x) * (fx - a[j].x) + (fy - a[j].y) * (fy - a[j].y) ;
                        if(s <= d) {
                            if(a[j].h <= ct) {
                                a[j].f = 0 ;
                            }
                            else {
                                a[j].h -= ct ;
                                a[i].f = 0 ;
                            }
                        }
                    }
            }
        }
    }

    for(int i = 1 ; i <= n ; i ++)
        if(a[i].f) cout << "Yes\n" ;
        else cout << "No\n" ;
}
signed main()
{
    std::ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;
    int t = 1 ;
//    cin >> t ;
    while (t --) solve() ;
    return 0;
}

I. 太阳轰炸

这道题其实就是在问，投影的r+飞弹的r与轰炸范围的r的关系，如果前者大，所有飞弹都落在投影内，反之就需要去求落在投影内的概率，然后排列组合。用到了逆元，但不难。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 5000010, M = N * 2, mod = 1e9 + 7, md = 998244353;
int n, R1, R2, r, a, h;
int fa[N], infa[N], fu[N], fv[N];

int qmi(int a, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1)
            res = res * a % mod;
        a = a * a % mod;
        k >>= 1;
    }
    return res;
}
int C(int a, int b)
{
    if (b > a)
        return 0;
    return fa[a] * infa[b] % mod * infa[a - b] % mod;
}
void solve()
{
    cin >> n >> R1 >> R2 >> r >> a >> h;
    fa[0] = infa[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        fa[i] = i * fa[i - 1] % mod;
        infa[i] = infa[i - 1] * qmi(i, mod - 2) % mod;
    }
    int u = 1, v = 0;
    if (R2 > R1 + r)
    {
        u = ((R1 + r) * (R1 + r) % mod) * qmi((R2 * R2) % mod, mod - 2) % mod;
        v = (R2 * R2 - (R1 + r) * (R1 + r)) % mod * qmi((R2 * R2) % mod, mod - 2) % mod;
    }
    else
    {
        if (n * a >= h)
            cout << "1\n";
        else
            cout << "0\n";
        return;
    }
    int p = h / a;
    if (h % a)
        p++;

    fu[0] = fv[0] = 1;
    for (int i = 1; i <= n; i++)
        fu[i] = (fu[i - 1] * u) % mod;
    for (int i = 1; i <= n; i++)
        fv[i] = (fv[i - 1] * v) % mod;

    int res = 0;
    for (int i = p; i <= n; i++)
    {
        int s = C(n, i) % mod;
        res = (res + s * fu[i] % mod * fv[n - i] % mod) % mod;
    }

    cout << res << "\n";
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    //    cin >> t ;
    while (t--)
        solve();
    return 0;
}

E. 发通知

异或前缀和，想到了这一点这道题就很明了了，还需要一个离散化就ok了。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 500010, M = N * 2, mod = 1e9 + 7, md = 998244353;
int n, k, cnt;
int a[N], b[N], w[N], sw[N * 3], s[N * 3];

void solve()
{
    cin >> n >> k;
    vector<int> q;
    map<int, int> mp;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i] >> b[i] >> w[i];
        q.push_back(a[i]), q.push_back(b[i]), q.push_back(b[i] + 1);
    }
    sort(q.begin(), q.end());
    for (int i = 0; i < q.size(); i++)
        if (!mp.count(q[i]))
            mp[q[i]] = ++cnt;
    for (int i = 1; i <= n; i++)
    {
        int l = mp[a[i]], r = mp[b[i]] + 1;
        s[l]++, s[r]--;
        sw[l] ^= w[i], sw[r] ^= w[i];
    }
    for (int i = 2; i <= cnt; i++)
        s[i] += s[i - 1], sw[i] ^= sw[i - 1];
    int res = -1;
    for (int i = 1; i <= cnt; i++)
        if (s[i] >= k)
            res = max(res, sw[i]);
    cout << res << "\n";
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    //    cin >> t ;
    while (t--)
        solve();
    return 0;
}

B. 广告投放

首先要知道 $\times b)$ 。
这样我们就可以预处理出m所以取值，只有 $\sqrt m$ 种可能。
这样就可以用dp来解决了， $f [i] [j]$ 表示在第 $i$ 集 $j$ 个人观看的最大收益。
转移方程： $f [i + 1] [j / d [i]] = m a x (f [i + 1] [j / d [i]], f [i] [j] + c * p [i])$
$\leq 10^5$ ，我们可以优化掉第一维度,因为 $\leq j$ ，所以第二维度需要从大到小枚举，其实只要我们预处理m的时候，从大到小处理就可以了。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 100010, M = N * 2, mod = 1e9 + 7, md = 998244353;
int n, m, cnt;
int p[N], d[N], f[N], a[N];

void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> p[i];
    for (int i = 1; i <= n; i++)
        cin >> d[i];
    a[++cnt] = 0;
    set<int> st;
    for (int i = m; i >= 1; i--)
    {
        if (!st[m / i])
        {
            st.insert(m / i);
            a[++cnt] = m / i;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= cnt; j++)
            f[a[j] / d[i]] = max(f[a[j]] + a[j] * p[i], f[a[j] / d[i]]);
    }
    int res = 0;
    for (int i = 0; i <= m; i++)
        res = max(res, f[i]);
    cout << res << "\n";
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    //    cin >> t ;
    while (t--)
        solve();
    return 0;
}

J. 二进制与、平方和

在线段树上维护区间或和区间平方和即可。
如果某一段的区间或&x没有变化，则说明这一段区间不需要修改。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 300010, M = N * 2, mod = 998244353;
int n, q;
int a[N];

struct Node
{
    int s, s2;
} tr[N * 4];

void pushup(int u)
{
    tr[u].s2 = (tr[u << 1].s2 + tr[u << 1 | 1].s2) % mod;
    tr[u].s = tr[u << 1].s | tr[u << 1 | 1].s;
}
void build(int u, int l, int r)
{
    if (l == r)
    {
        tr[u].s = a[l];
        tr[u].s2 = a[l] * a[l] % mod;
        return;
    }
    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
    pushup(u);
}
int query(int u, int l, int r, int L, int R)
{
    if (l >= L && r <= R)
        return tr[u].s2;

    int mid = l + r >> 1;
    int v = 0;
    if (L <= mid)
        v = (v + query(u << 1, l, mid, L, R)) % mod;
    if (R > mid)
        v = (v + query(u << 1 | 1, mid + 1, r, L, R)) % mod;
    return v;
}
void modify(int u, int l, int r, int L, int R, int x)
{
    if ((tr[u].s & x) == tr[u].s)
        return;
    if (l == r)
    {
        tr[u].s = tr[u].s & x;
        tr[u].s2 = tr[u].s * tr[u].s % mod;
        return;
    }

    int mid = l + r >> 1;
    if (L <= mid)
        modify(u << 1, l, mid, L, R, x);
    if (R > mid)
        modify(u << 1 | 1, mid + 1, r, L, R, x);
    pushup(u);
}
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    build(1, 1, n);
    cin >> q;
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int l, r, x;
            cin >> l >> r >> x;
            modify(1, 1, n, l, r, x);
        }
        else
        {
            int l, r;
            cin >> l >> r;
            cout << query(1, 1, n, l, r) << "\n";
        }
    }
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    //    cin >> t ;
    while (t--)
        solve();
    return 0;
}

K. 子串翻转回文串

哈希，先找到最开始不同的位置l和最后不同的位置r，翻转一定是(l,i)或者(i,r)。
对正串和反串做哈希，枚举翻转位置，用哈希判断是否对称。（需要双哈希…

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 500010, M = N * 2, mod = 1e9 + 7, P = 131;
int n;
string s, u;
ULL h1[N], p[N], h2[N];

ULL get1(int l, int r)
{
    return (h1[r] - h1[l - 1] * p[r - l + 1] % mod + mod) % mod;
}
ULL get2(int l, int r)
{
    return (h2[r] - h2[l - 1] * p[r - l + 1] % mod + mod) % mod;
}
void solve()
{
    cin >> s;
    n = s.size();
    u = s;
    reverse(u.begin(), u.end());
    s = " " + s, u = " " + u;
    int l = 1, r = n;
    while (l <= n && s[l] == u[l])
        l++;
    while (r >= 1 && s[r] == u[r])
        r--;
    if (l >= r)
    {
        cout << "Yes\n";
        return;
    }

    p[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        p[i] = p[i - 1] * P % mod;
        h1[i] = (h1[i - 1] * P % mod + s[i]) % mod;
        h2[i] = (h2[i - 1] * P % mod + u[i]) % mod;
    }
    ULL f1 = h1[n], f2 = h2[n];
    for (int i = l; i <= r; i++)
    {
        ULL suma = ((h1[n] - get1(l, i) * p[n - i] % mod + mod) % mod + get2(n - i + 1, n - l + 1) * p[n - i] % mod) % mod, suma1 = ((h2[n] - get2(n - i + 1, n - l + 1) * p[l - 1] % mod + mod) % mod + get1(l, i) * p[l - 1] % mod) % mod;
        ULL sumb = ((h1[n] - get1(i, r) * p[n - r] % mod + mod) % mod + get2(n - r + 1, n - i + 1) * p[n - r] % mod) % mod, sumb1 = ((h2[n] - get2(n - r + 1, n - i + 1) * p[i - 1] % mod + mod) % mod + get1(i, r) * p[i - 1] % mod) % mod;
        if (suma == suma1 || sumb == sumb1)
        {
            cout << "Yes\n";
            return;
        }
    }
    cout << "No\n";
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

F. 旅游胜地

二分距离+2-set。

#include <bits/stdc++.h>
#define x first
#define y second
#define int long long

using namespace std;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int N = 1000010, M = N * 2, mod = 1e9 + 7, P = 131;
int n, m;
int a[N], b[N], u[N], v[N];
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], tmp, stk[N], top, in_stk[N], id[N], s_cnt;

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
void tarjan(int u)
{
    low[u] = dfn[u] = ++tmp;
    stk[++top] = u;
    in_stk[u] = 1;

    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if (in_stk[j])
            low[u] = min(low[u], dfn[j]);
    }

    if (dfn[u] == low[u])
    {
        ++s_cnt;
        int y;
        do
        {
            y = stk[top--];
            in_stk[y] = 0;
            id[y] = s_cnt;
        } while (y != u);
    }
}
bool check(int mid)
{
    for (int i = 0; i <= n * 2; i++)
        h[i] = -1, dfn[i] = low[i] = stk[i] = in_stk[i] = id[i] = 0;
    idx = top = s_cnt = tmp = 0;

    for (int i = 0; i < m; i++)
    {
        int fu = u[i], fv = v[i];
        if (abs(a[fu] - a[fv]) > mid)
        {
            add(fu, fv + n), add(fv, fu + n);
        }
        if (abs(a[fu] - b[fv]) > mid)
        {
            add(fu, fv), add(fv + n, fu + n);
        }
        if (abs(b[fu] - a[fv]) > mid)
        {
            add(fu + n, fv + n), add(fv, fu);
        }
        if (abs(b[fu] - b[fv]) > mid)
        {
            add(fu + n, fv), add(fv + n, fu);
        }
    }
    for (int i = 1; i <= n * 2; i++)
    {
        if (!dfn[i])
            tarjan(i);
    }
    for (int i = 1; i <= n; i++)
        if (id[i] == id[i + n])
            return false;
    return true;
}
void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        cin >> b[i];
    for (int i = 0; i < m; i++)
    {
        cin >> u[i] >> v[i];
    }

    int l = 0, r = 1e10;
    while (l < r)
    {
        int mid = (l + r) / 2;
        if (check(mid))
            r = mid;
        else
            l = mid + 1;
    }
    cout << l << "\n";
}

signed main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    // cin >> t;
    while (t--)
        solve();
    return 0;
}