noip模拟52[有一些不知]
noip模拟52 solutions
不写总结了，没有时间了，直接上正解
T1 异或
你发现\(+1\)之后有影响的只是第一个0后面的数，所以枚举这个0的位置，
0前面的就是当前的方案数，直接计算就好了
AC_code
#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define int long long
#define re register int
int n;
ull ans=0;
signed main(){
scanf("%llu",&n);n--;
for(re i=0;i<=60;i++){
int baf=((1ll<<i+1)-1);
int bag=((1ll<<i)-1),res=0;
if(bag<=n)res++;
if((baf&n)<bag)res--;
res+=(n>>i+1);
if(res>0)ans+=res*(i+1);
}
printf("%llu",ans);
}
T2 赌神
按照官方题解上的式子推
AC_code
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define re register int
const int N=1e6+5;
const int mod=998244353;
int n,num[N],ans,jc[N],sum;
int ksm(int x,int y){
int ret=1;
while(y){
if(y&1)ret=ret*x%mod;
x=x*x%mod;
y>>=1;
}
return ret;
}
int C(int x,int y){return jc[x]*ksm(jc[y]*jc[x-y]%mod,mod-2)%mod;}
signed main(){
scanf("%lld",&n);
for(re i=1;i<=n;i++)scanf("%lld",&num[i]),sum+=num[i];
ans=ksm(n,sum);
jc[0]=1;for(re i=1;i<=sum;i++)jc[i]=jc[i-1]*i%mod;
ans=ans*ksm(jc[sum],mod-2)%mod;
for(re i=1;i<=n;i++)ans=ans*jc[num[i]]%mod;
printf("%lld",ans);
}
T3 路径
这个我是FFT+点分治做的，直接优化一下找到距离为k的点对有多少
AC_code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
#define re register int
const int N=1e6+100000;
const int mod=998244353;
int n,m,mxd,md;
ll an[N],ans;
ll s[N],t[N],o[N];
struct FFT{
struct coex{
double r,i;
coex(){}
coex(double x,double y){r=x;i=y;}
coex operator + (coex x)const{return coex(r+x.r,i+x.i);}
coex operator - (coex x)const{return coex(r-x.r,i-x.i);}
coex operator * (coex x)const{return coex(r*x.r-i*x.i,r*x.i+i*x.r);}
}a[N],b[N],w[N];
const double pi=acos(-1.0);
int af[N],lim,len;
void fft(coex a[]){
for(re i=0;i<lim;i++)if(af[i]>i)swap(a[i],a[af[i]]);
for(re t=lim>>1,d=1;d<lim;d<<=1,t>>=1)
for(re i=0;i<lim;i+=(d<<1))
for(re j=0;j<d;j++){
coex tmp=w[t*j]*a[i+j+d];
a[i+j+d]=a[i+j]-tmp;
a[i+j]=a[i+j]+tmp;
}
}
void mul(int ln,int lm){
for(lim=1,len=0;lim<=ln+lm;lim<<=1,len++);
for(re i=0;i<lim;i++){
af[i]=(af[i>>1]>>1)|((i&1)<<(len-1));
w[i]=coex(cos(2.0*i*pi/lim),sin(2.0*i*pi/lim));
}
for(re i=0;i<=ln;i++)a[i].r=1.0*s[i];
for(re i=0;i<=lm;i++)b[i].r=1.0*t[i];
fft(a);fft(b);
for(re i=0;i<lim;i++)a[i]=a[i]*b[i],w[i].i=-w[i].i;
fft(a);
for(re i=0;i<lim;i++){
if(i<=ln+lm)o[i]=(int)(a[i].r/lim+0.5);
a[i]=b[i]=w[i]=coex(0,0);
}
}
}work;
ll ksm(ll x,ll y){
ll ret=1;
while(y){
if(y&1)ret=ret*x%mod;
x=x*x%mod;
y>>=1;
}
return ret;
}
int to[N*2],nxt[N*2],head[N],rp;
void add_edg(int x,int y){
to[++rp]=y;
nxt[rp]=head[x];
head[x]=rp;
}
int siz[N],rt,ms[N],mx;
bool vis[N];
void get_rt(int x,int f,int sz){
ms[x]=0;siz[x]=1;
for(re i=head[x];i;i=nxt[i]){
int y=to[i];
if(y==f||vis[y])continue;
get_rt(y,x,sz);
siz[x]+=siz[y];
ms[x]=max(ms[x],siz[y]);
}
ms[x]=max(ms[x],sz-siz[x]);
if(ms[x]<mx)mx=ms[x],rt=x;
}
void dfs(int x,int f,int dep){
t[dep]++;md=max(md,dep);
for(re i=head[x];i;i=nxt[i]){
int y=to[i];
if(y==f||vis[y])continue;
dfs(y,x,dep+1);
}
}
void sol(int x){
s[0]=1;mxd=0;
for(re i=head[x];i;i=nxt[i]){
int y=to[i];md=0;
if(vis[y])continue;
dfs(y,x,1);
work.mul(mxd,md);
for(re j=0;j<=md;j++)s[j]=s[j]+t[j]%mod,t[j]=0;
for(re j=0;j<=mxd+md;j++)an[j]=(an[j]+o[j])%mod;
mxd=max(mxd,md);
}
for(re i=1;i<=mxd;i++)s[i]=0;
}
void divd(int x,int sz){
vis[x]=true;sol(x);
for(re i=head[x];i;i=nxt[i]){
int y=to[i];
if(vis[y])continue;
mx=0x3f3f3f3f;
if(siz[y]<siz[x])get_rt(y,x,siz[y]),divd(rt,siz[y]);
else get_rt(y,x,sz-siz[x]),divd(rt,sz-siz[x]);
}
}
signed main(){
scanf("%lld%lld",&n,&m);
for(re i=1;i<n;i++){
int x,y;scanf("%lld%lld",&x,&y);
add_edg(x,y);add_edg(y,x);
}
mx=0x3f3f3f3f;
get_rt(1,0,n);
divd(rt,n);
for(re i=1;i<=n;i++)ans=(ans+an[i]*ksm(i,m))%mod;
printf("%lld",ans);
}
T4 树
直接根号分治，小于的用一种，大于的用另外一种
这样的做法很常见吧
放上我被zero4338卡掉的代码
60pts
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define re int
const int N=3e6+5;
const int B=550;
int n,q,ans[N];
int to[N*4],nxt[N*4],head[N],rp;
void add_edg(int x,int y){
to[++rp]=y;
nxt[rp]=head[x];
head[x]=rp;
}
int dfn[N],dfm[N],cnt,idf[N],dep[N],mxd;
void dfs_fi(int x,int f){
dfn[x]=++cnt;idf[cnt]=x;
mxd=max(mxd,dep[x]);
for(re i=head[x];i;i=nxt[i]){
int y=to[i];
if(y==f)continue;
dep[y]=dep[x]+1;
dfs_fi(y,x);
}
dfm[x]=cnt;
}
int typ[N],v[N],x[N],y[N],yy[N],z[N];
int bas,bel[N],l[B],r[B];
vector<int> one[B][B],two[N];
int so1[N],sol1[N],so2[N],sol2[N];
signed main(){
scanf("%lld%lld",&n,&q);
for(re i=1;i<n;i++){
int u,v;scanf("%lld%lld",&u,&v);
add_edg(u,v);add_edg(v,u);
}
dep[1]=1;dfs_fi(1,0);bas=sqrt(n);
memset(l,0x3f,sizeof(l));
for(re i=1;i<=n;i++){
bel[i]=(i-1)/bas+1;
l[bel[i]]=min(l[bel[i]],i);
r[bel[i]]=max(r[bel[i]],i);
}
for(re i=1;i<=q;i++){
scanf("%lld",&typ[i]);
if(typ[i]==1){
scanf("%lld%lld%lld%lld",&v[i],&x[i],&y[i],&z[i]);
yy[i]=y[i];y[i]=(y[i]+dep[v[i]])%x[i];
if(x[i]<=bas){
if(!one[x[i]][y[i]].size())one[x[i]][y[i]].push_back(0);
one[x[i]][y[i]].push_back(i);
}
else if(dep[v[i]]+yy[i]<=mxd)two[dep[v[i]]+yy[i]].push_back(i);
}
else {
scanf("%lld",&v[i]);
for(re j=1;j<=bas;j++)if(one[j][dep[v[i]]%j].size())one[j][dep[v[i]]%j].push_back(i);
if(two[dep[v[i]]].size())two[dep[v[i]]].push_back(i);
}
}
for(re i=1;i<=bas+1;i++){
for(re j=0;j<i;j++){
for(re k=1;k<one[i][j].size();k++){
int now=one[i][j][k];
if(typ[now]==1){
for(re o=max(dfn[v[now]],l[bel[dfn[v[now]]]]);o<=r[bel[dfn[v[now]]]];o++)so1[o]+=z[now];
for(re o=l[bel[dfm[v[now]]]];o<=min(dfm[v[now]],r[bel[dfm[v[now]]]]);o++)so1[o]+=z[now];
for(re o=bel[dfn[v[now]]]+1;o<bel[dfm[v[now]]];o++)sol1[o]+=z[now];
}
else ans[now]+=so1[dfn[v[now]]]+sol1[bel[dfn[v[now]]]];
}
for(re k=1;k<one[i][j].size();k++){
int now=one[i][j][k];
if(typ[now]==1){
for(re o=max(dfn[v[now]],l[bel[dfn[v[now]]]]);o<=r[bel[dfn[v[now]]]];o++)so1[o]-=z[now];
for(re o=l[bel[dfm[v[now]]]];o<=min(dfm[v[now]],r[bel[dfm[v[now]]]]);o++)so1[o]-=z[now];
for(re o=bel[dfn[v[now]]]+1;o<bel[dfm[v[now]]];o++)sol1[o]-=z[now];
}
}
}
}
for(re i=1;i<=mxd;i++){
if(!two[i].size())continue;
sort(two[i].begin(),two[i].end());
for(re j=0;j<two[i].size();j++){
int now=two[i][j];
if(typ[now]==1){
so2[dfn[v[now]]]+=z[now];
sol2[bel[dfn[v[now]]]]+=z[now];
so2[dfm[v[now]]+1]-=z[now];
sol2[bel[dfm[v[now]]+1]]-=z[now];
if(i+x[now]<=mxd)two[i+x[now]].push_back(now);
}
else {
for(re o=l[bel[dfn[v[now]]]];o<=min(r[bel[dfn[v[now]]]],dfn[v[now]]);o++)ans[now]+=so2[o];
for(re o=1;o<bel[dfn[v[now]]];o++)ans[now]+=sol2[o];
}
}
for(re j=0;j<two[i].size();j++){
int now=two[i][j];
if(typ[now]==1){
so2[dfn[v[now]]]-=z[now];
sol2[bel[dfn[v[now]]]]-=z[now];
so2[dfm[v[now]]+1]+=z[now];
sol2[bel[dfm[v[now]]+1]]+=z[now];
}
}
}
for(re i=1;i<=q;i++)if(typ[i]==2)printf("%lld\n",ans[i]);
}
QQ:2953174821
