Codeforces Round 911 (Div. 2) D、E

CF1900D

从小到大排列，第 $i$ 个元素作为最大值对答案的贡献为 $\sum_{x=1}^{i-1} \sum_{y=x+1}^{i-1} gcd(a_x,a_y)$ ，即区间 $[1,i-1]$ 两两元素 $gcd$ 和，记作 $val_{i-1}$ 。然后考虑用第 $i$ 个元素更新 $val_{i-1}$ ，即 $val_i=val_{i-1}+\sum_{x=1}^{i-1}gcd(a_x,a_i)$ 。处理这玩意有两种做法。

方法一

容斥。首先维护出 $cnt[j](j\le1e5)$ 表示第 $i$ 位之前 $j$ 的倍数有多少，从大到小枚举 $a_i$ 因子 $d$ ， $cnt[d]$ 就是 $gcd(a_x,a_i)=d$ 的个数，然后把 $d$ 的因子的 $cnt$ 全都减 $cnt[d]$ 就筛完了，这个时间复杂度不太会证，初始化的时间复杂度肯定是调和级数，近似 $O(nlogn)$ ，但是嵌套两个for枚举因子的就不太懂了（ $\sum_{i}^{n}a_i所有因子的因子数量和$ ？）。

方法二

欧拉函数反演。

n = \sum_{d | n} ϕ (d) \sum_{x = 1}^{i - 1} g c d (a_{x}, a_{i}) = \sum_{x = 1}^{i - 1} \sum_{d | g c d (a_{x}, a_{i})} ϕ (d) = \sum_{x = 1}^{i - 1} \sum_{d | a_{i}} ϕ (d) [d | a_{x}] = \sum_{d | a_{i}} ϕ (d) \sum_{x = 1}^{i - 1} [d | a_{x}]

$n=\sum_{d|n}\phi (d) \\ \sum_{x=1}^{i-1}gcd(a_x,a_i)=\sum_{x=1}^{i-1}\sum_{d|gcd(a_x,a_i)}\phi (d)=\sum_{x=1}^{i-1}\sum_{d|a_i}\phi (d)[d|a_x]=\sum_{d|a_i}\phi (d)\sum_{x=1}^{i-1}[d|a_x]$

方法一

 #include<bits/stdc++.h>
using namespace std;
 
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0);
 
typedef long long LL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
 
const int INF = 0x3f3f3f3f, N = 1e5 + 10, MOD = 998244353;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7, pi = acos(-1);
 
int cnt[N], s[N];
 
vector<int> t[N];
 
void init() {
    rep (i, 1, N - 1) {
        for (int j = i; j < N; j += i) {
            t[j].push_back(i);
        }
    }
}
 
void work() {
	int n;
	cin >> n;
	rep (i, 1, N - 1) cnt[i] = 0;
	vector<int> a(n + 1);
	rep (i, 1, n) cin >> a[i];
	sort(a.begin() + 1, a.end());
	LL res = 0, sum = 0;
	rep (i, 1, n) {
		res += sum;
		rrep (j, t[a[i]].size() - 1, 0) {
			int u = t[a[i]][j];
			sum += 1ll * u * cnt[u];
			s[j] = cnt[u];
			for (auto &x: t[u]) {
				cnt[x] -= s[j];
			}
		}
		rrep (j, t[a[i]].size() - 1, 0) {
			int u = t[a[i]][j];
			int v = s[j];
			for (auto &x: t[u]) {
				cnt[x] += s[j];
			}
		}
		for (auto &x: t[a[i]]) {
			cnt[x]++;
		}
	}
	cout << res << "\n";
}
 
/*
 
 
*/
 
signed main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    IO
    init();
    
    int test = 1;
    cin >> test;
    while (test--) work();
    
    return 0;
}

方法二

 #include<bits/stdc++.h>
using namespace std;
 
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0);
 
typedef long long LL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
 
const int INF = 0x3f3f3f3f, N = 1e5 + 10, MOD = 998244353;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7, pi = acos(-1);
 
int phi[N];
int primes[N], st[N];
int cnt[N];
vector<int> t[N];
 
void init() {
    rep (i, 1, N - 1) {
        for (int j = i; j < N; j += i) {
            t[j].push_back(i);
        }
    }
    int tot = 0;
	phi[1] = 1; 
	for (int i = 2; i <= N - 1; i++) {
		if (!st[i]) {
			primes[tot++] = i;
			phi[i] = i - 1;
		}
		for (int j = 0; primes[j] * i <= N - 1; j++) {
			st[primes[j] * i] = 1;
			if (i % primes[j] == 0) {
				phi[i * primes[j]] = phi[i] * primes[j];
				break;
			} else phi[i * primes[j]] = phi[i] * (primes[j] - 1);
		}
	}
}
 
void work() {
	int n;
	cin >> n;
	rep (i, 1, N - 1) cnt[i] = 0;
	vector<int> a(n + 1);
	rep (i, 1, n) cin >> a[i];
    
	sort(a.begin() + 1, a.end());
	LL res = 0, sum = 0;
    
	rep (i, 1, n) {
		res += sum;
		for (auto &x: t[a[i]]) {
            sum += 1ll * phi[x] * cnt[x];
            cnt[x]++;
        }
	}
	cout << res << "\n";
}
 
/*
 
 
*/
 
signed main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    IO
    init();
    
    int test = 1;
    cin >> test;
    while (test--) work();
    
    return 0;
}

CF1900E

操作完成后，每个强连通分量都是完全图，强连通分量外部的的”新边“对答案没有影响。

缩完点后， $SCC$ 的”长度“为内部点数，权值为内部点权和，在这个 $DAG$ 上 $dp$ 即可。

自环和重边无影响，不用处理。

 const int INF = 0X3f3f3f3f, N = 2e5 + 10, MOD = 998244353;
const double eps = 1e-8, pi = acos(-1);
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 2e5 + 10;
 
int n;
LL a[N], len[N], val[N];
vector<int> g[N], G[N];
 
int dfn[N], low[N], timestamp;
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt, Size[N];
int dout[N], din[N];
 
void tarjan(int u) {
	dfn[u] = low[u] = ++timestamp;
	stk[++top] = u, in_stk[u] = true;
	for (auto &v: g[u]) {
		if (!dfn[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if (in_stk[v]) low[u] = min(low[u], dfn[v]);
	}
 
	if (dfn[u] == low[u]) {
		++scc_cnt;
		int y;
		do {
			y = stk[top--];
			in_stk[y] = false;
			id[y] = scc_cnt;
			Size[scc_cnt]++;
            val[scc_cnt] += a[y];
            len[scc_cnt]++;
		} while (y != u);
	}
}
 
void Count_D() {
	rep (u, 1, n) {
        for (auto &v: g[u]) {
            if (id[u] != id[v]) {
                dout[id[u]]++, din[id[v]]++;
                G[id[u]].push_back(id[v]);
            } 
        }
    }
}
 
void init() {
    timestamp = top = scc_cnt = 0;
    rep (i, 1, n) {
        g[i].clear();
        G[i].clear();
        dfn[i] = in_stk[i] = id[i] = Size[i] = din[i] = dout[i] = len[i] = val[i] = 0;
    }
}
 
void topo() {
    queue<int> q; 
    vector<LL> f(scc_cnt + 1), g(scc_cnt + 1);
    rep (i, 1, scc_cnt) {
        if (din[i] == 0) {
            f[i] = len[i];
            g[i] = val[i];
            q.push(i);
        } 
    } 
    while (q.size()) {
        int u = q.front();
        q.pop();
        
        for (auto &v: G[u]) {
            din[v]--;
            if (f[v] < f[u] + len[v]) {
                f[v] = f[u] + len[v];
                g[v] = g[u] + val[v];
            }
            else if (f[v] == f[u] + len[v]) {
                g[v] = min(g[v], g[u] + val[v]);
            }
            if (din[v] == 0) {
                q.push(v);
            }
        }
    }
    
    LL res_len = 0, res_val = 0;
    rep (i, 1, scc_cnt) {
        if (res_len < f[i]) {
            res_len = f[i];
            res_val = g[i];
        }
        else if (res_len == f[i]) res_val = min(res_val, g[i]);
    }
    cout << res_len << " " << res_val << "\n";
}
 
void work() {
    int m;
    cin >> n >> m;
    init();
    rep (i, 1, n) cin >> a[i];
    rep (i, 1, m) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
    }
    rep (i, 1, n) if (!id[i]) tarjan(i);
    Count_D();
    topo();
}
/*
*/

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本文作者：xhy666

本文链接：https://www.cnblogs.com/xhy666/p/17860930.html

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Codeforces Round 911 (Div. 2) D、E

CF1900D

方法一

方法二

CF1900E

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

	#define fr first
	#define se second
	#define et0 exit(0);
	#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
	#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
	#define IO ios::sync_with_stdio(false),cin.tie(0);

	typedef long long LL;
	typedef pair<int, int> PII;
	typedef unsigned long long ULL;

	const int INF = 0x3f3f3f3f, N = 1e5 + 10, MOD = 998244353;
	const long long LLINF = 0x3f3f3f3f3f3f3f3f;
	const double eps = 1e-7, pi = acos(-1);

	int cnt[N], s[N];

	vector<int> t[N];

	void init() {
	rep (i, 1, N - 1) {
	for (int j = i; j < N; j += i) {
	t[j].push_back(i);
	}
	}
	}

	void work() {
	int n;
	cin >> n;
	rep (i, 1, N - 1) cnt[i] = 0;
	vector<int> a(n + 1);
	rep (i, 1, n) cin >> a[i];
	sort(a.begin() + 1, a.end());
	LL res = 0, sum = 0;
	rep (i, 1, n) {
	res += sum;
	rrep (j, t[a[i]].size() - 1, 0) {
	int u = t[a[i]][j];
	sum += 1ll * u * cnt[u];
	s[j] = cnt[u];
	for (auto &x: t[u]) {
	cnt[x] -= s[j];
	}
	}
	rrep (j, t[a[i]].size() - 1, 0) {
	int u = t[a[i]][j];
	int v = s[j];
	for (auto &x: t[u]) {
	cnt[x] += s[j];
	}
	}
	for (auto &x: t[a[i]]) {
	cnt[x]++;
	}
	}
	cout << res << "\n";
	}

	/*


	*/

	signed main() {
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
	IO
	init();

	int test = 1;
	cin >> test;
	while (test--) work();

	return 0;
	}

	const int INF = 0X3f3f3f3f, N = 2e5 + 10, MOD = 998244353;
	const double eps = 1e-8, pi = acos(-1);
	const long long LLINF = 0x3f3f3f3f3f3f3f3f;
	const int MAXN = 2e5 + 10;

	int n;
	LL a[N], len[N], val[N];
	vector<int> g[N], G[N];

	int dfn[N], low[N], timestamp;
	int stk[N], top;
	bool in_stk[N];
	int id[N], scc_cnt, Size[N];
	int dout[N], din[N];

	void tarjan(int u) {
	dfn[u] = low[u] = ++timestamp;
	stk[++top] = u, in_stk[u] = true;
	for (auto &v: g[u]) {
	if (!dfn[v]) {
	tarjan(v);
	low[u] = min(low[u], low[v]);
	} else if (in_stk[v]) low[u] = min(low[u], dfn[v]);
	}

	if (dfn[u] == low[u]) {
	++scc_cnt;
	int y;
	do {
	y = stk[top--];
	in_stk[y] = false;
	id[y] = scc_cnt;
	Size[scc_cnt]++;
	val[scc_cnt] += a[y];
	len[scc_cnt]++;
	} while (y != u);
	}
	}

	void Count_D() {
	rep (u, 1, n) {
	for (auto &v: g[u]) {
	if (id[u] != id[v]) {
	dout[id[u]]++, din[id[v]]++;
	G[id[u]].push_back(id[v]);
	}
	}
	}
	}

	void init() {
	timestamp = top = scc_cnt = 0;
	rep (i, 1, n) {
	g[i].clear();
	G[i].clear();
	dfn[i] = in_stk[i] = id[i] = Size[i] = din[i] = dout[i] = len[i] = val[i] = 0;
	}
	}

	void topo() {
	queue<int> q;
	vector<LL> f(scc_cnt + 1), g(scc_cnt + 1);
	rep (i, 1, scc_cnt) {
	if (din[i] == 0) {
	f[i] = len[i];
	g[i] = val[i];
	q.push(i);
	}
	}
	while (q.size()) {
	int u = q.front();
	q.pop();

	for (auto &v: G[u]) {
	din[v]--;
	if (f[v] < f[u] + len[v]) {
	f[v] = f[u] + len[v];
	g[v] = g[u] + val[v];
	}
	else if (f[v] == f[u] + len[v]) {
	g[v] = min(g[v], g[u] + val[v]);
	}
	if (din[v] == 0) {
	q.push(v);
	}
	}
	}

	LL res_len = 0, res_val = 0;
	rep (i, 1, scc_cnt) {
	if (res_len < f[i]) {
	res_len = f[i];
	res_val = g[i];
	}
	else if (res_len == f[i]) res_val = min(res_val, g[i]);
	}
	cout << res_len << " " << res_val << "\n";
	}

	void work() {
	int m;
	cin >> n >> m;
	init();
	rep (i, 1, n) cin >> a[i];
	rep (i, 1, m) {
	int u, v;
	cin >> u >> v;
	g[u].push_back(v);
	}
	rep (i, 1, n) if (!id[i]) tarjan(i);
	Count_D();
	topo();
	}
	/*
	*/