NOIP2024加赛8
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 LOCAL
FILE *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 LOCAL
FILE *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 LOCAL
FILE *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 LOCAL
FILE *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_unlocked
template<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
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本文来自博客园，作者：CuFeO4，转载请注明原文链接：https://www.cnblogs.com/hzoi-Cu/p/18572926
