脚本宝典收集整理的这篇文章主要介绍了noip模拟56，脚本宝典觉得挺不错的，现在分享给大家，也给大家做个参考。
原题，但是我的复杂度是假的.
#include<bITs/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long int
#define lf double
#define ull unsigned ll
#define mp make_pair
#define lb lower_bound
#define ub upPEr_bound
#define Fill(x,y) memset(x,y,sizeof x)
#define Copy(x,y) memcpy(x,y,sizeof x)
#define File(x,y) freopen(#x,"r",stdin),freopen(#y,"w",stdout)
#define FILE(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
inline ll read(){
ll ss=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) ss=(ss<<3)+(ss<<1)+(ch^48),ch=getchar();
return cit?ss:-ss;
}
} using namespace BSS;
const ll N=1e6+21;
ll m,n,ts,ans,tail,tot;
ll head[N],fa[N],dep[N],stk[N],rk[N];
vector<ll> vec[N];
struct I { ll u,v,nxt; } e[N<<1];
inline void add(ll u,ll v){
e[++ts].u=u,e[ts].v=v,e[ts].nxt=head[u],
head[u]=ts;
}
inline bool comp(ll i,ll j){ return i > j ; }
void DFs(ll u,ll depth){
dep[u]=depth,rk[u]=(tail ? stk[tail--] : ++tot );
ll x=rk[u],cnt=0;
if(!head[u]){
ans+=dep[u];
vec[x].push_back(dep[u]);
return ;
}
for(int i=head[u];i;i=e[i].nxt){
dfs(e[i].v,depth+1);
for(int j : vec[rk[e[i].v]]){
vec[x].push_back(j);
}
vec[rk[e[i].v]].clear(),stk[++tail]=rk[e[i].v],rk[e[i].v]=0;
}
sort(vec[x].begin(),vec[x].end(),comp);
for(int i=vec[x].size()-1;i>=1;i--){
if(vec[x][i]-depth<depth){
cnt++,ans+=vec[x][i]-depth*2;
}
}
while(cnt--) vec[x].pop_back();
}
signed main(){
FILE(a);
n=read(); ll u,v;
for(int i=2;i<=n;i++){
u=read(),add(u,i),fa[i]=u;
}
dfs(1,0);
PRintf("%lldn",ans),exit(0);
}
首先需要了解一下珂朵莉树，而且这题卡不掉珂朵莉，因为题目性质.
按照右端点排序后，我们选择移动一个指针指向右端点.
然后分别根据 (l) 的位置进行加减即可.
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long int
#define ull unsigned ll
#define lf double
#define lBT(x) (x&(-x))
#define mp(x,y) make_pair(x,y)
#define lb lower_bound
#define ub upper_bound
#define Fill(x,y) memset(x,y,sizeof x)
#define Copy(x,y) memcpy(x,y,sizeof x)
#define File(x,y) freopen(#x,"r",stdin),freopen(#y,"w",stdout)
#define FILE(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
inline ll read() {
int ss=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) ss=(ss<<3)+(ss<<1)+(ch^48),ch=getchar();
return cit?ss:-ss;
}
} using namespace BSS;
const ll N=1e6+21,mod=1e9+7;
ll n,m;
ll L[N],R[N],ans[N];
struct I {
ll l,r,w;
I(){}
I(ll x,ll y,ll z){ l=x,r=y,w=z; }
I(ll x){ l=x; }
inline bool operator <(const I &x)const{
return l<x.l;
}
};
#define sit set<I>::iterator
set<I> s;
struct BIT{
ll res; ll c[N<<1];
inline void update(ll x,ll w){ for(;x;x&=x-1) c[x]=(c[x]+w)%mod; }
inline ll query(ll x){ res=0; for(;x<=n;x+=lbt(x)) res=(res+c[x])%mod; return res; }
}b,bl,br,blr;
inline ll ksm(ll a,ll b,ll c){
ll res=1; a%=c;
for(;b;b>>=1,a=(a*a)%c) if(b&1) res=(res*a)%c;
return res%c;
}
inline sit split(ll pos){
sit it=s.lb(I(pos));
if(it!=s.end() and it->l==pos) return it;
--it;
ll l=it->l,r=it->r,w=it->w;
s.erase(it),s.insert(I(l,pos-1,w));
return s.insert(I(pos,r,w)).First;
}
inline void assign(ll l,ll r,ll x){
sit itr=split(r+1),itl=split(l);
for(sit it=itl;it!=itr;it++){
ll k=it->r-it->l+1,t=it->w;
bl.update(x,(x-1)*k%mod),bl.update(t,(1-x+mod)*k%mod);
br.update(x,(x+1)*k%mod),br.update(t,(mod-1-t)*k%mod);
blr.update(x,mod-k),blr.update(t,k);
b.update(x,(1-x*x%mod+mod)%mod*k%mod);
b.update(t,(x*t%mod+x-t-1+mod)%mod*k%mod);
}
s.erase(itl,itr),s.insert(I(l,r,x));
}
struct II { ll l,r,id; } q[N];
inline bool comp(II i,II j){ return i.r<j.r; }
signed main(){
FILE(b);
n=read(),m=read();
for(int i=1;i<=n;i++) L[i]=read(),R[i]=read()-1;
for(int i=1;i<=m;i++) q[i].l=read(),q[i].r=read(),q[i].id=i;
s.insert(I(1,1e9,0)),sort(q+1,q+1+m,comp);
ll p=0,resl,resr,reslr,resb,res;
for(int i=1;i<=m;i++){
while(p<q[i].r) p++,assign(L[p],R[p],p);
resl=q[i].l*bl.query(q[i].l)%mod;
resr=q[i].r*br.query(q[i].l)%mod;
reslr=q[i].l*q[i].r%mod*blr.query(q[i].l)%mod;
resb=b.query(q[i].l);
res=(resl+resr+reslr+resb)%mod;
ans[q[i].id]=res*ksm((q[i].r-q[i].l+1)*(q[i].r-q[i].l+2)>>1,mod-2,mod)%mod;
}
for(int i=1;i<=m;i++) printf("%lldn",ans[i]);
exit(0);
}
二项式反演，比较裸的题目了.
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long int
#define ull unsigned ll
#define lf double
#define lbt(x) (x&(-x))
#define mp(x,y) make_pair(x,y)
#define lb lower_bound
#define ub upper_bound
#define Fill(x,y) memset(x,y,sizeof x)
#define Copy(x,y) memset(x,y,sizeof x)
#define File(x,y) freopen(#x,"r",stdin),freopen(#y,"w",stdout)
#define FILE(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
inline int read() {
int ss=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) ss=(ss<<3)+(ss<<1)+(ch^48),ch=getchar();
return cit?ss:-ss;
}
} using namespace BSS;
const ll N=5e3+21,mod=1e9+7;
ll n,m,ans;
ll frc[(int)1e6+21],inv[(int)1e6+21];
ll f[N][N],g[N][N];
inline ll ksm(ll a,ll b,ll c){
a%=c; ll res=1;
while(b){
if(b&1) res=(res*a)%c;
a=(a*a)%c,b>>=1;
}
return res%c;
}
inline ll C(ll a,ll b){
if(a>b) return 0;
return frc[b]*inv[a]%mod*inv[b-a]%mod;
}
signed main(){
FILE(c);
n=read(),m=read(),frc[0]=1,inv[0]=1; ll res;
for(ll i=1;i<=1e5;i++){
frc[i]=frc[i-1]*i%mod,inv[i]=ksm(frc[i],mod-2,mod);
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m/i;j++){
for(int k=i;k<=m/j;k++){
res=C(i,k)*C(k,n)%mod*C(n-1,m+n-1-j*k)%mod;
if((k&1)==(i&1)) g[i][j]=(g[i][j]+res)%mod;
else g[i][j]=(g[i][j]-res+mod)%mod;
}
}
}
for(int i=n;i>=1;i--){
for(int j=1;j<=m;j++){
f[i][j]=(f[i+1][j]+g[i][j])%mod;
}
}
res=ksm(C(n-1,n+m-1),mod-2,mod);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
ans=(ans+f[i][j])%mod;
}
printf("%lldn",(ans*res%mod+i)%mod);
}
exit(0);
}
以上是脚本宝典为你收集整理的noip模拟56全部内容，希望文章能够帮你解决noip模拟56所遇到的问题。
本图文内容来源于网友网络收集整理提供，作为学习参考使用，版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ：384754419，请注明来意。