状态很不好，恼了。虚拟机太卡了，根本交不上去。
flandre
发现选取的肯定是从大到小排序后的一个后缀，然后就做完了，时间复杂度\(O(n\log n)\)。
点此查看代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t;i += p)
#define drep(i,s,t,p) for(int i = s;i >= t;i -= p)
#ifdef LOCALFILE *I = fopen("in.in","r"),*O = fopen("out.out","w");
#else// FILE *I = stdin,*O = stdout;FILE *I = fopen("flandre.in","r"),*O = fopen("flandre.out","w");
#endif
using ll = long long;using ull = unsigned long long;
using db = double;using ldb = long double;
#define int long long
const int N = 1e6 + 10;
#define pii pair<int,int>
#define mk make_pair
pii a[N];int n,k;
inline void solve(){cin>>n>>k;rep(i,1,n,1) cin>>a[i].first,a[i].second = i;sort(a+1,a+1+n);int pos = n+1,ans = 0,sum = 0;map<int,int> mp;drep(i,n,1,1){sum += a[i].first;mp[a[i].first]++;sum += (n-i+1-mp[a[i].first])*k;if(sum > ans){ans = sum;pos = i;}}cout<<ans<<' '<<n-pos+1<<'\n';rep(i,pos,n,1) cout<<a[i].second<<' ';
}
signed main(){cin.tie(nullptr)->sync_with_stdio(false);solve();
}
meirin
因为有且仅有对\(b\)的操作，考虑将\(b\)提出来。
考虑什么时候\(b_i,a_j\)有贡献，当且仅当区间\([l,r]\)满足\(l\le\min\{i,j\}\le\max\{i,j\}\le r\le n\)。
假设\(b_i,a_j\)的贡献为\(b_i\times a_j\times p_{i,j}\)，那么有\(p_{i,j}=\begin{cases}i\times(n-j+1)&i\le j\\j\times(n-i+1)&j<i\end{cases}\)。
将\(a_j\)乘进去，再令\(p_i=\sum\limits_{j=1}^np_{i,j}\)，有\(p_i=\sum\limits_{j=1}^{i-1}a_j\times j\times (n-i+1)+\sum\limits_{j=i}^na_j\times (n-j+1)\times i\)。
发现如果\(i\)恒定，那么就是求\(a_i\times i\)的前缀和和\(a_i\times (n-i+1)\)的后缀和，预处理即可。
对于区间加，增加的贡献就是\(k\times (\sum\limits_{i=l}^rp_i)\)，预处理前缀和即可。
时间复杂度\(O(n+m)\)。
点此查看代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t;i += p)
#define drep(i,s,t,p) for(int i = s;i >= t;i -= p)
using ll = long long;using ull = unsigned long long;
using db = double;using ldb = long double;
#ifdef LOCALFILE *I = freopen("in.in","r",stdin),*O = freopen("out.out","w",stdout);
#else// FILE *I = stdin,*O = stdout;FILE *I = freopen("meirin.in","r",stdin),*O = freopen("meirin.out","w",stdout);
#endif
#define int long long
const int N = 5e5 + 10,mod = 1e9 + 7;
int n,m,a[N],b[N],s1[N],s2[N],p[N],sum[N],ans;
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cin>>n>>m;rep(i,1,n,1) cin>>a[i];rep(i,1,n,1) cin>>b[i];rep(i,1,n,1) s1[i] = (s1[i-1] + i*a[i]%mod)%mod;drep(i,n,1,1) s2[i] = (s2[i+1] + (n-i+1)*a[i]%mod)%mod;rep(i,1,n,1) p[i] = ((n-i+1)*s1[i-1]%mod + (s2[i])%mod*i%mod)%mod;rep(i,1,n,1) sum[i] = (sum[i-1] + p[i])%mod,ans = (ans + b[i]*p[i]%mod)%mod;rep(test,1,m,1){int l,r,k;cin>>l>>r>>k;ans = (ans + (sum[r]-sum[l-1]+mod)%mod*k%mod)%mod;cout<<(ans+mod)%mod<<'\n';}
}
sakuya
考虑如果一条边有贡献，那么就是它两端子树内的关键点的乘积乘上\(w\)。这个东西直接预处理。
考虑如果对一个点所连的边进行\(+k\)操作，那么其实就是所有与之相连的边的边的贡献变成\(k+w\times sth.\)，记\(f_x\)表示与\(x\)相连的边的贡献即可。
但是还没完，题目让我们求得是期望，不是贡献，只需要求个平均数就好了，注意要乘一个二。
点此查看代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t;i += p)
#define drep(i,s,t,p) for(int i = s;i >= t;i -= p)
#ifdef LOCALFILE *I = freopen("in.in","r",stdin),*O = freopen("out.out","w",stdout);
#else// FILE *I = stdin,*O = stdout;FILE *I = freopen("sakuya.in","r",stdin),*O = freopen("sakuya.out","w",stdout);
#endif
using ll = long long;using ull = unsigned long long;
using db = double;using ldb = long double;
#define pii pair<int,int>
#define eb emplace_back
const int N = 5e5 + 10,mod = 998244353;
vector<pii> e[N];
int n,m,f[N],p[N],dep[N],ans = 0;
bitset<N> pd;
void dfs1(int x,int fa){if(pd[x]) p[x]++;dep[x] = dep[fa] + 1;for(auto [y,w]:e[x]){if(y == fa) continue;dfs1(y,x);p[x] += p[y];}
}
inline int getval(int x,int y){if(dep[x] > dep[y]) return 1ll*p[x]*(m-p[x])%mod;else return 1ll*p[y]*(m-p[y])%mod;
}
void dfs2(int x,int fa){for(auto [y,w]:e[x]){f[x] = (f[x] + getval(x,y))%mod;if(y == fa) continue;dfs2(y,x);ans = (ans + 1ll*w*getval(x,y)%mod)%mod;}
}
inline int power(int a,int b,int mod){int res = 1;for(;b;b >>= 1,a = 1ll*a*a%mod)if(b&1) res = 1ll*res*a%mod;return res;
}
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cin>>n>>m;rep(i,2,n,1){int x,y,w;cin>>x>>y>>w;e[x].eb(y,w);e[y].eb(x,w);}rep(i,1,m,1){int x;cin>>x;pd.set(x);}dfs1(1,0);dfs2(1,0);int q;cin>>q;int more = power(m,mod-2,mod)*2ll%mod;rep(test,1,q,1){int x,k;cin>>x>>k;ans = (ans + 1ll*f[x]*k%mod)%mod;cout<<1ll*ans*more%mod<<'\n';}
}
红楼 ~ Eastern Dream
初始化的强化？
根号分治是显然的，对于\(x\le\sqrt n\)可以去看我的[Ynoi2011] 初始化题解。对于\(x>\sqrt n\)的，显然有一个线段树解法，具体的从\(1\)暴力跳，步长为\(x\)，将\(x\sim x+y-1\)所有的加上\(k\)。
但是这样是很不优秀的，修改的复杂度为\(O(\sqrt n\log n+\sqrt n)\)，查询的复杂度为\(O(\log n+\sqrt n)\)，考虑根号平衡，将区间加变为\(O(1)\)的，修改变成\(O(\sqrt n)\)的。
考虑差分和分块，具体的，设\(c_i\)表示序列\(a\)的差分数组。
对于一次查询，显然有\(ans_{l,r}=\sum\limits_{i=l}^r\sum\limits_{j=1}^ic_j\)，如果您知道树状数组区间修改怎么推的和如何操作那么您就过了。
考虑这个柿子怎么化简到方便维护的形式。
\[\begin{aligned}ans_{l,r}&=\sum\limits_{i=l}^r\sum\limits_{j=1}^ic_j\\&=\sum_{i=1}^{l-1}c_i\times (r-l+1)+\sum_{i=l}^rc_i(r-i+1)\\&=(r-l+1)\sum_{i=1}^{l-1}+(r+1)\sum_{i=l}^rc_i-\sum_{i=l}^rc_i\times i\end{aligned}
\]
然后后面这个东西就像树状数组区间修改一样维护就行了，具体实现看代码中的qry函数，时间复杂度\(O(n\sqrt n)\)。
点此查看代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t;i += p)
#define drep(i,s,t,p) for(int i = s;i >= t;i -= p)
#ifdef LOCALFILE *I = freopen("in.in","r",stdin),*O = freopen("out.out","w",stdout);
#else// FILE *I = stdin,*O = stdout;FILE *I = freopen("scarlet.in","r",stdin),*O = freopen("scarlet.out","w",stdout);
#endif
using ll = long long;using ull = unsigned long long;
using db = double;using ldb = long double;
namespace IO{#define gc getchar_unlocked#define pc putchar_unlockedtemplate<class T>inline void read(T &x){x = 0;char s = gc();for(;s < '0' || '9' < s;s = gc());for(;'0' <= s && s <= '9';s = gc())x = (x<<1) + (x<<3) + (s^48);}template<class T,class... Args>inline void read(T &x,Args&... argc){read(x);read(argc...);}template<class T>inline void write(T x){static int sta[40],top = 0;do sta[++top] = x%10;while(x /= 10);while(top) pc(sta[top--]+'0');}inline void write(char x){pc(x);}template<class T,class... Args>inline void write(T x,Args... argc){write(x);write(argc...);}#undef gc#undef pc
}using IO::read;using IO::write;
const int N = 2e5 + 1,M = 450;
int n,m,a[N],pos[N],L[M],R[M],siz,len;
ll s[M],si[M],pre[M][M],suf[M][M],sum[M][M],num[M],c1[N],c2[N],qz[N];
signed main(){read(n,m);rep(i,1,n,1) read(a[i]),qz[i] = qz[i-1] + a[i];len = 450;siz = n/len;rep(i,1,siz,1) L[i] = R[i-1]+1,R[i] = i*len;if(R[siz] < n) siz++,L[siz] = R[siz-1] + 1,R[siz] = n;rep(i,1,siz,1) rep(j,L[i],R[i],1) pos[j] = i;int spl = sqrt(n/5.0);auto qry1 = [&](int l,int r){if(r == 0) return 0ll;int p = pos[l],q = pos[r];ll res = 0;if(p == q){rep(i,l,r,1) res += c1[i];return res;}rep(i,l,R[p],1) res += c1[i];rep(i,L[q],r,1) res += c1[i];rep(i,p+1,q-1,1) res += s[i];return res;};auto qry2 = [&](int l,int r){if(r == 0) return 0ll;int p = pos[l],q = pos[r];ll res = 0;if(p == q){rep(i,l,r,1) res += c2[i];return res;}rep(i,l,R[p],1) res += c2[i];rep(i,L[q],r,1) res += c2[i];rep(i,p+1,q-1,1) res += si[i];return res;};auto qry = [&](int l,int r){return 1ll*(r-l+1)*qry1(1,l-1)+1ll*(r+1)*(qry1(1,r)-qry1(1,l-1))-qry2(1,r)+qry2(1,l-1);};int tot = 0;rep(test,1,m,1){int op,x,y,k = 0;read(op,x,y);if(op == 1){read(k);y = min(y,x-1);if(x <= spl){rep(i,0,y,1) sum[x][i] += k;num[x] += 1ll*k*(y+1);pre[x][0] = sum[x][0];rep(i,1,x-1,1) pre[x][i] = pre[x][i-1] + sum[x][i];drep(i,x-1,0,1) suf[x][i] = suf[x][i+1] + sum[x][i];}else{y++;rep(i,1,n,x){int p = pos[i],iy = i+y;ll ik = 1ll*i*k;s[p] += k,si[p] += ik,c1[i] += k,c2[i] += ik;if(iy <= n) s[pos[iy]] -= k,si[pos[iy]] -= 1ll*iy*k,c1[iy] -= k,c2[iy] -= 1ll*iy*k;}}}else{ll ans = 0;rep(now,1,spl,1){if(y-x < now) rep(i,x,y,1) ans += sum[now][(i-1)%now];else{x--,y--;ans += suf[now][x%now];ans += pre[now][y%now];x++,y++;ans += ((y-((y-1)%now)-1)-(x+(now-(x-1)%now))+1)/now*num[now];}}write(qry(x,y)+ans+qz[y]-qz[x-1],'\n');}}
}
p
