After a single solution to a problem or to put it.
A.Divisors
Root sieve demand factors of all numbers, you can sweep it again to re-count.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
const int N=205;
int a[N],m,n;
map<int,int> bu;
vector<int> res,app;
int ans[N];
int main()
{
//freopen("dt.in","r",stdin);
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
scanf("%d",&a[i]);
for(int j=1;1LL*j*j<=a[i];j++)
{
if(a[i]%j)continue;
if(j*j==a[i])res.push_back(j);
else res.push_back(j),res.push_back(a[i]/j);
}
}
int sz=res.size();
for(int i=0;i<sz;i++)
{
if(res[i]>n)continue;
if(bu.find(res[i])==bu.end())app.push_back(res[i]);
bu[res[i]]++;
}
sz=app.size();
for(int i=0;i<sz;i++)
ans[bu[app[i]]]++;
ans[0]=n-sz;
for(int i=0;i<=m;i++)
printf("%d\n",ans[i]);
return 0;
}
B.Market
Offline asked. The inquiry and shops are in chronological order, all maintain a pointer to the current plan can be purchased are put to run a backpack.
He noted that the bill is very large and very small value, then the value of the array index as dp, min suffix taken to ensure that half of the monotonous find the optimal solution can be.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
const int inf=0x3f3f3f3f;
int n,m,f[100005],ans[100005];
struct shop
{
int c,v,t;
friend bool operator < (shop a,shop b)
{
return a.t<b.t;
}
}s[305];
struct plan
{
int t,num,id;
friend bool operator < (plan a,plan b)
{
return a.t<b.t;
}
}p[100005];
void show()
{
for(int i=1;i<=n*300;i++)
cout<<f[i]<<endl;
}
int main()
{
/*freopen("dt.in","r",stdin);
freopen("my.out","w",stdout);*/
n=read();m=read();
for(int i=1;i<=n;i++)
s[i].c=read(),s[i].v=read(),s[i].t=read();
for(int i=1;i<=m;i++)
p[i].t=read(),p[i].num=read(),p[i].id=i;
sort(s+1,s+n+1);sort(p+1,p+m+1);
memset(f,0x3f,sizeof(f));
f[0]=0;
int j=1;
for(int i=1;i<=m;i++)
{
while(j<=n&&s[j].t<=p[i].t)
{
for(int k=n*300;k>=s[j].v;k--)
f[k]=min(f[k],f[k-s[j].v]+s[j].c);
for(int k=n*300;k;k--)
f[k]=min(f[k],f[k+1]);
j++;
}
//ow();
//cout<<p[i].id<<' '<<p[i].num<<endl;
ans[p[i].id]=upper_bound(f+1,f+n*300+1,p[i].num)-f-1;
}
for(int i=1;i<=m;i++)
printf("%d\n",ans[i]);
return 0;
}
C.Dash Speed
Then a change of perspective from side to consider not seem to do, starting from the feasible speed, continue to add to the current collection side, the diameter is the answer to the ultimate edge set.
This approach and T2 examination of a very similar.
To establish a rate for the next segment tree standard, and comply with the interval limits insertion side (to the satellite, or directly before the row vector also manner similar).
After which you can by $ O (n \ log \ n) $ of a one-time segment tree divide and conquer all the answers. A segment tree arrival interval, there will be a set of points of the connected edges and herein. Specific merger is like that question before, to discuss six kinds of circumstances to determine the new China Unicom block endpoint.
Since the partition requires backtracking, backtracking need to reverse impact, so that the structure can not disjoint-set irreversibly changed halfway, i.e., the path can not be compressed. So here we need another way to merge - merge by rank.
After the information about the changes recorded by the stack, popping back when you can withdraw.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<cmath>
#define re register
using namespace std;
namespace IO{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
#define ll long long
//fread->read
bool IOerror=0;
inline char nc(){
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if (p1==pend){
p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
if (pend==p1){IOerror=1;return -1;}
//{printf("IO error!\n");system("pause");for (;;);exit(0);}
}
return *p1++;
}
inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
inline void read(int &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(ll &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(double &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (ch=='.'){
double tmp=1; ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
}
if (sign)x=-x;
}
inline void read(char *s){
char ch=nc();
for (;blank(ch);ch=nc());
if (IOerror)return;
for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
*s=0;
}
inline void read(char &c){
for (c=nc();blank(c);c=nc());
if (IOerror){c=-1;return;}
}
//fwrite->write
struct Ostream_fwrite{
char *buf,*p1,*pend;
Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
void out(char ch){
if (p1==pend){
fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
}
*p1++=ch;
}
void print(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(double x,int y){
static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
if (x<-1e-12)out('-'),x=-x;x*=mul[y];
ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
}
void println(double x,int y){print(x,y);out('\n');}
void print(char *s){while (*s)out(*s++);}
void println(char *s){while (*s)out(*s++);out('\n');}
void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
~Ostream_fwrite(){flush();}
}Ostream;
inline void print(int x){Ostream.print(x);}
inline void println(int x){Ostream.println(x);}
inline void print(char x){Ostream.out(x);}
inline void println(char x){Ostream.out(x);Ostream.out('\n');}
inline void print(ll x){Ostream.print(x);}
inline void println(ll x){Ostream.println(x);}
inline void print(double x,int y){Ostream.print(x,y);}
inline void println(double x,int y){Ostream.println(x,y);}
inline void print(char *s){Ostream.print(s);}
inline void println(char *s){Ostream.println(s);}
inline void println(){Ostream.out('\n');}
inline void flush(){Ostream.flush();}
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
const int N=1e5+5;
int n,m;
int to[N<<4],head[N<<4],nxt[N<<4],tot,fr[N<<4];
int fa[N],size[N],son[N],Top[N],dep[N],ans[N];
vector<int> g[N];
#define ls(k) (k)<<1
#define rs(k) (k)<<1|1
struct Stack
{
int x,y,op,nd1,nd2;
}s[N<<4];
int top;
void ins(int k,int l,int r,int L,int R,int x,int y)
{
if(L<=l&&R>=r)
{
to[++tot]=y;fr[tot]=x;
nxt[tot]=head[k];head[k]=tot;
return ;
}
int mid=l+r>>1;
if(L<=mid)ins(ls(k),l,mid,L,R,x,y);
if(R>mid)ins(rs(k),mid+1,r,L,R,x,y);
}
void dfs1(int x,int f)
{
size[x]=1;fa[x]=f;
int sz=g[x].size();
for(int i=0;i<sz;i++)
{
int y=g[x][i];
if(y==f)continue;
dep[y]=dep[x]+1;
dfs1(y,x);
size[x]+=size[y];
if(size[y]>size[son[x]])son[x]=y;
}
return ;
}
void dfs2(int x,int f)
{
Top[x]=f;
if(!son[x])return ;
dfs2(son[x],f);
int sz=g[x].size();
for(int i=0;i<sz;i++)
{
int y=g[x][i];
if(!Top[y])dfs2(y,y);
}
}
int lca(int x,int y)
{
while(Top[x]!=Top[y])
{
if(dep[Top[x]]<dep[Top[y]])swap(x,y);
x=fa[Top[x]];
//cout<<x<<' '<<y<<endl;
}
return dep[x]<dep[y]?x:y;
}
int dis(int x,int y)
{
return dep[x]+dep[y]-(dep[lca(x,y)]<<1);
}
namespace U
{
int fa[N],node[N][3],rk[N];
void ini()
{
for(int i=1;i<=n;i++)
fa[i]=node[i][0]=node[i][1]=i;
}
int findf(int x)
{
return fa[x]==x?x:findf(fa[x]);
}
void merge(int xx,int yy,int &d)
{
int x=findf(xx),y=findf(yy),nd1,nd2,maxd=-1,nowd=dis(node[x][0],node[x][1]);
if(nowd>maxd)maxd=nowd,nd1=node[x][0],nd2=node[x][1];
nowd=dis(node[y][0],node[y][1]);
if(nowd>maxd)maxd=nowd,nd1=node[y][0],nd2=node[y][1];
for(re int i=0;i<2;i++)
for(re int j=0;j<2;j++)
{
nowd=dis(node[x][i],node[y][j]);
if(nowd>maxd)maxd=nowd,nd1=node[x][i],nd2=node[y][j];
}
d=max(d,maxd);
if(rk[x]<rk[y])swap(x,y);
s[++top]=(Stack){x,y,0,node[x][0],node[x][1]};
if(rk[x]==rk[y])++rk[x],s[top].op=1;
fa[y]=x;node[x][0]=nd1;node[x][1]=nd2;
}
}
void cancel(int k)
{
while(top>k)
{
U::rk[s[top].x]-=s[top].op;
U::fa[s[top].y]=s[top].y;
U::node[s[top].x][0]=s[top].nd1;
U::node[s[top].x][1]=s[top].nd2;
top--;
}
}
void work(int k,int l,int r,int sum)
{
int pos=top;
for(re int i=head[k];i;i=nxt[i])
U::merge(fr[i],to[i],sum);
if(l==r)
{
ans[l]=sum;
cancel(pos);
return ;
}
int mid=l+r>>1;
work(ls(k),l,mid,sum);
work(rs(k),mid+1,r,sum);
cancel(pos);
}
int main()
{
//freopen("speed2.in","r",stdin);
IO::read(n);IO::read(m);
for(re int i=1;i<n;i++)
{
int x,y,l,r;
IO::read(x);IO::read(y);
IO::read(l);IO::read(r);
g[x].push_back(y);
g[y].push_back(x);
ins(1,1,n,l,r,x,y);
}
dep[1]=1;
dfs1(1,0);dfs2(1,1);
/*for(int i=1;i<=n;i++)
cout<<i<<' '<<size[i]<<' '<<son[i]<<' '<<Top[i]<<endl;*/
U::ini();
work(1,1,n,0);
while(m--)
{
int q;
IO::read(q);
IO::println(ans[q]);
}
return 0;
}