Problem - D - Codeforces

思路

只选两个小数计算贡献，可对序列排序，每对$(i,j)$ 其中$(i<j)$的的贡献是$gcd(i,j)\cdot (n-j)$
通过预处理因数+欧拉反演可将复杂度降为$O(nV)$：
$$\begin{array}{rl}n& ={\Sigma }_{d|n}\phi (d)\\ 令n& =gcd(i,j)\\ 得到gcd(i,j)& ={\Sigma }_{d|(i,j)}\phi (d)\\ & ={\Sigma }_{d|i}{\Sigma }_{d|j}\phi (d).\\ & \\ 可以求{\Sigma }_{i=1}^{n}gcd(i,n)& ={\Sigma }_{i=1}^n{\Sigma }_{d|i}{\Sigma }_{d|n}\phi (d)\\ & ={\Sigma }_{d|n}{\Sigma }_{i=1}^n{\Sigma }_{d|i}\phi (d)\\ & ={\Sigma }_{d|n}\frac{n}{d}\phi (d).\\ & \\ 题目要求& {\Sigma }_{i=1}{\Sigma }_{j=i+1}gcd(a[i],a[j])\cdot (n-j)\\ & ={\Sigma }_{j=1}^n{\Sigma }_{i=1}^{j}gcd(a[i],a[j])\cdot (n-j)\\ & ={\Sigma }_{j=1}^n{\Sigma }_{d|a[j]}\frac{a[j]}{d}\phi (d)\cdot (n-j)\\ & ={\Sigma }_{j=1}^n{\Sigma }_{d|a[j]}cnt[d]\phi (d)\cdot (n-j).\end{array}$$

参考博客：欧拉函数|(扩展)欧拉定理|欧拉反演 - Morning_Glory - 博客园 (cnblogs.com)

另：

一个序列的两两$gcd$和 $=\phi (d)\cdot C(cnt[d],2)$，仿照莫队可以推得，当$cnt[d]$增加1时，对当前位置答案的影响为$\phi (d)\cdot cnt[d]$，对每一位计算贡献后乘位权即可。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
typedef long long ll;
//#define int ll
#define pb push_back
#define eb emplace_back
#define m_p make_pair
const int mod = 998244353;
#define mem(a,b) memset(a,b,sizeof a)
#define pii pair<int,int>
#define fi first
#define se second
const int inf = 0x3f3f3f3f;
const int N = 3e5 + 50;
//__builtin_ctzll(x);后导0的个数
//__builtin_popcount计算二进制中1的个数
int a[N];
vector<int> fac[N];
bool is[N]; //筛数组
int pri[N]; //素数数组
int phi[N]; //欧拉函数
int tot=0;  //素数个数
int n;
ll cnt[N];void init(){for(int i=1;i<N;++i){is[i]=1;}is[0]=is[1]=0;phi[1]=1;for(int i=2;i<N;++i){if(is[i]){pri[++tot]=i;phi[i]=i-1;}for(int j=1;j<=tot && pri[j]*i<N;++j){is[pri[j]*i]=0;if(i%pri[j]==0){phi[i*pri[j]]=phi[i]*pri[j];break;}else{phi[i*pri[j]]=phi[i]*(pri[j]-1);}}}
}void work() {int n;cin >> n;for (int i = 1; i < N; ++i) {cnt[i] = 0;}int maxx = 0;for (int i = 1; i <= n; ++i) {cin >> a[i];}sort(a + 1, a + 1 + n);ll ans = 0;for (int i = 1; i <= n; ++i) {ll now = 0;for (auto x:fac[a[i]]) {now += phi[x] * cnt[x];cnt[x]++;}ans += now * (n - i);}cout << ans << '\n';
}signed main() {io;int t = 1;cin >> t;for (int i = 1; i < N; ++i) {for (int j = 1; j * i < N; ++j) {fac[i * j].pb(i);}}init();while (t--) {work();}return 0;
}

Problem - E - Codeforces

思路

可以发现，通过传递拓展到的边一定是将原有路径变短，不符合我们需要的答案，所以拓展是无用的操作，只管原图即可。图中最难处理的是有向环，有向环可以跑到环上任意一点出去，自然联想$tarjan$缩点，缩点后在$DAG$图上跑$dp$就可以得到答案。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
typedef long long ll;
//#define int ll
#define pb push_back
#define eb emplace_back
#define m_p make_pair
const int mod = 998244353;
#define mem(a,b) memset(a,b,sizeof a)
#define pii pair<int,ll>
#define fi first
#define se second
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int N = 3e5 + 50;
//__builtin_ctzll(x);后导0的个数
//__builtin_popcount计算二进制中1的个数
int a[N];
int n, m, val[N], h[N], e[N], ne[N], idx;
int in[N];
int dfn[N], low[N], dfncnt, s[N], in_stack[N], tp;
int scc[N], sc;  // 结点 i 所在 SCC 的编号
int sz[N];       // 强连通 i 的大小
ll sum[N];      // 强连通 i 的权值
pii dp[N];void add(int u, int v) {e[++idx] = v, ne[idx] = h[u], h[u] = idx;
}void tarjan(int u) {low[u] = dfn[u] = ++dfncnt, s[++tp] = u, in_stack[u] = 1;for (int i = h[u]; i; i = ne[i]) {int v = e[i];if (!dfn[v]) {tarjan(v);low[u] = min(low[u], low[v]);} else if (in_stack[v]) {low[u] = min(low[u], dfn[v]);}}if (dfn[u] == low[u]) {++sc;while (s[tp] != u) {scc[s[tp]] = sc;sz[sc]++;sum[sc] += val[s[tp]];in_stack[s[tp]] = 0;--tp;}scc[s[tp]] = sc;sz[sc]++;sum[sc] += val[s[tp]];in_stack[s[tp]] = 0;--tp;}
}vector<int> ed[N];void work() {cin >> n >> m;dfncnt = tp = sc = idx = 0;for (int i = 1; i <= n; i++) {h[i] = 0;sum[i] = dfn[i] = low[i] = s[i] = in_stack[i] = sz[i] = scc[i] = 0;in[i] = 0;ed[i].clear();}for (int i = 1; i <= n; ++i) {cin >> val[i];dp[i].fi = 0;dp[i].se = inf;}for (int i = 1; i <= m; ++i) {int u, v;cin >> u >> v;add(u, v);}for (int i = 1; i <= n; ++i) {if (!dfn[i]) tarjan(i);}for (int i = 1; i <= n; ++i) {int u = scc[i];for (int j = h[i]; j ; j = ne[j]) {int v = scc[e[j]];if (u == v) continue;ed[v].pb(u);in[u]++;}}queue<int>q;for (int i = 1; i <= sc; ++i) {if (in[i] == 0) {q.push(i);if (dp[i].fi == sz[i]) dp[i].se = min(dp[i].se, sum[i]);else if (dp[i].fi < sz[i]) dp[i] = {sz[i], sum[i]};}}while (!q.empty()) {int now = q.front();q.pop();for (auto x: ed[now]) {in[x]--;if (dp[x].fi == sz[x] + dp[now].fi) dp[x].se = min(dp[x].se, sum[x] + dp[now].se);else if (dp[x].fi < sz[x] + dp[now].fi) dp[x] = {sz[x] + dp[now].fi, sum[x] + dp[now].se};if (in[x] == 0) {q.push(x);}}}ll len = 0, ans = inf;for (int i = 1; i <= sc; ++i) {if (dp[i].fi > len) len = dp[i].fi, ans = dp[i].se;else if (dp[i].fi == len) ans = min(ans, dp[i].se);}cout << len << " " << ans << '\n';
}signed main() {io;int t = 1;cin >> t;while (t--) {work();}return 0;
}