csp模拟-3

没有题目，没有题解，甚至没有正解，自然不是外漏。

T1 . span

很有趣的一道题，关键是要注意到条件一其实就是棍木，然后排序 + 维护区间最大/小值（我这里用的 st 表）就结束了。

#include <bits/extc++.h>
using namespace std;
using ling = long long;
#define en_ putchar_unlocked('\n')
#define e_ putchar_unlocked(' ')
inline int in() {
    int n = 0; char p = getchar_unlocked();
    while(p < '-') p = getchar_unlocked();
    bool f = p == '-' ? p = getchar_unlocked() : 0;
    do n = n*10 + (p^48), p = getchar_unlocked();
    while(isdigit(p));
    return f ? -n : n;
}
inline void in(int &n) { n = in();}
const int N = 2e5 + 10;
int n;
struct P{
    int val, id;
} p[N];
int stk[N], k, f[N][30], lgn[N], f2[N][30];
inline void pre_st() {
    for(int i = 2; i <= n; i++) lgn[i] = lgn[i >> 1] + 1;
    for(int i = 1; i <= 29; i++) 
        for(int j = 1; j <= n - (1 << i) + 1; j++) {
            f[j][i] = max(f[j][i-1], f[(1<<(i-1))+ j][i-1]);
        }
    for(int i = 1; i <= 29; i++) 
        for(int j = 1; j <= n - (1 << i) + 1; j++) {
            f2[j][i] = min(f2[j][i-1], f2[(1<<(i-1))+ j][i-1]);
        }
}
inline int rmq(int l, int r) {
    int t = lgn[r - l + 1];
    return max(f[l][t], f[r - (1 << t) + 1][t]);
}
inline int rxq(int l, int r) {
    int t = lgn[r - l + 1];
    return min(f2[l][t], f2[r - (1 << t) + 1][t]);
}
signed main() {
    freopen("span.in","r",stdin);
    freopen("span.out","w",stdout);
    int l = 0;
    in(n), in(k);
    for(int i = 1; i <= n; i++) in(p[i].val), p[i].id = i;
    sort(p+1, p+1+n, [](P a, P b){return a.val < b.val;});
    int xiao = 292929292;
    for(int i = 1; i <= n; i++) f2[i][0] = f[i][0] = p[i].id;
    pre_st();
    p[0].val = - 12321321;
    for(int i = 1; i <= n; i++) {
        if(p[i].val != p[i-1].val + 1) {
            l = 0;
        }
        l++;
        if(l >= k) {
            xiao = min(xiao, rmq(i-k+1,i) - rxq(i-k+1,i));
        }
    }
    cout << xiao;
}

T2 . graph

正解是根号分治，看起来也很有意思。不过，还是再说罢，这里是赛时代码（ \(60\) pts）。

#include <bits/extc++.h>
using namespace std;
using ling = long long;
#define en_ putchar_unlocked('\n')
#define e_ putchar_unlocked(' ')
inline int in() {
    int n = 0; char p = getchar_unlocked();
    while(p < '-') p = getchar_unlocked();
    bool f = p == '-' ? p = getchar_unlocked() : 0;
    do n = n*10 + (p^48), p = getchar_unlocked();
    while(isdigit(p));
    return f ? -n : n;
}
inline void in(int &n) { n = in();}
const int N = 2e5 + 10;
int to[N*2], nxt[N*2], hd[N], c[N];
int q, n ,m, tot;
signed main() {
    freopen("graph.in","r",stdin);
    freopen("graph.out","w",stdout);
    in(n), in(m), in(q);
    for(int i = 1; i <= n; i++) c[i] = i;
    for(int i = 1, u, v; i <= m; i++) {
        in(u), in(v);
        to[++tot] = v;
        nxt[tot] = hd[u];
        hd[u]=tot;
        to[++tot] = u;
        nxt[tot] = hd[v];
        hd[v]=tot;
    }
    for(int i = 1, x, lx; i <= q; i++) {
        in(x);
        if(lx == x) continue;
        for(int v = hd[x];v;v=nxt[v]) c[to[v]] = c[x];
        lx = x;
    }
    for(int i = 1; i <= n; i++) {
        printf("%d ", c[i]);
    }
}

T3 . mod

据说是数学 + 随机化（高概率）。
但是我又不会数学也没看出随机化。
通过观察数据得出模数是个质数，然后就暴力 + 卡时，本以为 \(40\) 结果 \(70\)？

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <bits/extc++.h>
using namespace std;
// #define int long long
#define en_ putchar_unlocked('\n')
#define e_ putchar_unlocked(' ')
inline int in() {
    int n = 0; char p = getchar_unlocked();
    while(p < '-') p = getchar_unlocked();
    bool f = p == '-' ? p = getchar_unlocked() : 0;
    do n = n*10 + (p^48), p = getchar_unlocked();
    while(isdigit(p));
    return f ? -n : n;
}
inline void in(int &n) { n = in();}
const int N = 2e5 + 10, V = 1e7 + 10;
inline bool isprime(int n) {
    const int l = sqrt(n);
    for(int i = 2; i <= l; i++) {
        if(n % i == 0) return 0;
    }
    return 1;
}
// bitset<V> vis;
bool vis[V];
int prime[1000000];
int s[N], v, n, p, tot;
inline void line() {
    for (int i = 2; i <= v; i++) {
        if (!vis[i])
            prime[++tot] = i;
        for (int j = 1; j <= tot && i * prime[j] <= v; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}
// __gnu_pbds:: cc_hash_table<int,int> mp;
// unordered_map<int,int> mp;
int mp[V];
signed main() {
    freopen("mod.in","r",stdin);
    freopen("mod.out","w",stdout);
    int cnt2 = 0;
    in(n), in(v);
    v = min(v, V - 10);
    for(int i = 1; i <= n; i++) {
        in(s[i]);
    }
    line();
    int ma = 0, MA = 0, I, tim = 0;
    for(int i = 1, j, o; i <= tot; i++) {
        // mp.clear();
        // for(int k = 0; k <= prime[i]; k++) mp[k] = 0;
        memset(mp, 0, 4 * (prime[i] + 1));
        ma = 0;
        if(tim>257222222) break;
        for(j = 1; j <= n; j++) {
            mp[o = s[j] % prime[i]]++;
            if(ma < mp[o]) ma = mp[o];
        }
        if(ma > MA) MA = ma, I = prime[i];
        tim += n;
    }
    cout << MA << endl << I;
}

T4 . loquat

直说不会，\(1.5h\) 码了个暴力，还行吧，至少不罚坐。

#include <bits/extc++.h>
using namespace std;
#define int long long
#define en_ putchar_unlocked('\n')
#define e_ putchar_unlocked(' ')
inline int in() {
    int n = 0; char p = getchar_unlocked();
    while(p < '-') p = getchar_unlocked();
    bool f = p == '-' ? p = getchar_unlocked() : 0;
    do n = n*10 + (p^48), p = getchar_unlocked();
    while(isdigit(p));
    return f ? -n : n;
}
inline void in(int &n) { n = in();}
const int N = 13000, mod = 1e9 + 7;
struct node {
    int v, w;
};
inline int qpow(int i, int k) {
    int res = 1;
    for(; k; i = i * i % mod, k >>= 1) if(k & 1) res = res * i % mod;
    return res;
}
struct T{
    vector<node>e[N];
    int sz;
    int g[15000], dis[15000], siz[15000];
    void dfs(int u, int fa) {
        siz[u] = 1;
        for(node v : e[u]) {
            if(v.v != fa) {
                dis[v.v] = v.w + dis[u] % mod;
                dfs(v.v, u);
                siz[u] += siz[v.v];
            }
        }
    }
    void dps(int u, int fa) {
        for(node op : e[u]) {
            int v = op.v, w = op.w;
            if(v == fa) continue;
            g[v] = g[u] - (siz[v] * w % mod) + ((siz[0] - siz[v]) * w % mod) % mod;
            dps(v,u);
        }
    }
    int getdis() {
        dfs(0,-1);
        for(int i = 0; i < sz; i++) {
            g[0] += dis[i];
            g[0] %= mod;
        }
        dps(0, -1);
        int res = 0;
        // for(int i = 0; i < sz; i++) {
        //     cout << g[i] << endl;
        // }
        for(int i = 0; i < sz; i++) res += g[i], res %= mod;
        return (res * qpow(2, mod-2)) % mod;
    }

}t[301], p;
signed main() {
    freopen("loquat.in","r",stdin);
    freopen("loquat.out","w",stdout);
    int m = in();
    t[0].sz = 1;
    for(int i = 1, x, y, u, v, w; i <= m; i++) {
        in(x), in(y), in(u), in(v), in(w);
        t[i] = t[x];
        for(int j = 0; j < t[x].sz ; j++) t[i].e[j] = t[x].e[j];
        for(int j = 0; j < t[y].sz ; j++) {
            t[i].e[j+t[x].sz] = t[y].e[j];
            for(uint k = 0; k < t[i].e[j+t[x].sz].size(); k++) t[i].e[j+t[x].sz][k].v += t[x].sz;
        }
        t[i].e[u].push_back({v+t[x].sz, w});
        t[i].e[v + t[x].sz].push_back({u, w});
        t[i].sz = t[x].sz + t[y].sz;
    }    
    // t[3].getdis();
    for(int i = 1; i <= m; i++) {
        cout << t[i].getdis() << '\n';
    }
}