10.18模拟赛

sol：不难吧，首先假设所有的狮子会一口气吃到底，（只剩一只），

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=100005,inf=2e9+7;
int n,a[N],rt,re[N],pp,pre[N];
struct segtree{int l,r,mi,ma,miid,maid;inline int mid(){return (l+r)>>1;}}Tree[N<<2];
#define c1 x<<1
#define c2 x<<1|1
inline void Ami(int x,int y){Tree[x].mi=Tree[y].mi;Tree[x].miid=Tree[y].miid;}
inline void Ama(int x,int y){Tree[x].ma=Tree[y].ma;Tree[x].maid=Tree[y].maid;}
inline void Up(int x){if(Tree[c1].mi!=Tree[c2].mi){if(Tree[c1].mi<Tree[c2].mi)Ami(x,c1); else Ami(x,c2);}else Ami(x,c1);if(Tree[c1].ma!=Tree[c2].ma){if(Tree[c1].ma>Tree[c2].ma)Ama(x,c1); else Ama(x,c2);}else Ama(x,c2);}
inline void build(int l,int r,int x){Tree[x].l=l;Tree[x].r=r;if(l==r){Tree[x].mi=Tree[x].ma=a[l];Tree[x].maid=Tree[x].miid=l;return;}int mid=(l+r)>>1;build(l,mid,c1);build(mid+1,r,c2); Up(x);}
inline void ins(int po,int x,int v){if(Tree[x].l==Tree[x].r){Tree[x].mi=Tree[x].ma=v;return;}int mid=Tree[x].mid();if(po<=mid)ins(po,c1,v);else ins(po,c2,v); Up(x);}
inline void del(int po,int x){if(Tree[x].l==Tree[x].r){Tree[x].mi=inf;Tree[x].ma=-1;return;}int mid=Tree[x].mid();if(po<=mid)del(po,c1);else del(po,c2); Up(x);}
int main()
{
    int i,p1,p2; scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%d",&a[i]); build(1,n,1); pp=n-1;
    for(i=1;i<n;i++)
    {
        p1=Tree[1].miid; p2=Tree[1].maid; pre[p2]=i; a[p2]-=a[p1]; del(p1,1); ins(p2,1,a[p2]); re[i]=p1;
    }
    for(i=n-1;i>=1;i=min(--i,pp))
    {
        if(pre[re[i]])pp=pre[re[i]]-1;
    }printf("%d
",pp); sort(re+1,re+pp+1); for(i=1;i<pp;i++)printf("%d ",re[i]);printf("%d
",re[pp]);
}

sol：线段树裸题，多打打胆子练大点，这种题应该很快就能过

维护8个标记，左边的高度，左边相同的点的个数，左边这段连续的是否合法（即左段最右边是否大于右边那个数），右边和左边维护一样的三个标记，再维护答案，lazy，这段数字这否全部相同

#include <cstdio>
#include <algorithm>
using namespace std;
const int N=200005;
int n,Q,h,cnt=0,has[N<<1];
struct node{int l,r,v;}as[N];
struct segtree{int l,r,s,lh,ln,lc,rh,rn,rc,ok,laz;inline int mid(){return (l+r)>>1;}inline int len(){return (r-l+1);}}Tree[N<<2];
#define c1 x<<1
#define c2 x<<1|1
inline void Up(int x)
{
    Tree[x].s=Tree[c1].s+Tree[c2].s; Tree[x].lh=Tree[c1].lh; Tree[x].rh=Tree[c2].rh;
    Tree[x].lc=Tree[c1].lc;Tree[x].rc=Tree[c2].rc;
    if(Tree[c1].ok){if(Tree[c1].rh>Tree[c2].lh)Tree[x].lc=1;else if(Tree[c1].rh==Tree[c2].lh)Tree[x].lc=Tree[c2].lc;}
    if(Tree[c2].ok){if(Tree[c2].lh>Tree[c1].rh)Tree[x].rc=1;else if(Tree[c2].lh==Tree[c1].rh)Tree[x].rc=Tree[c1].rc;}
    if(Tree[c1].ok){if(Tree[c1].rh==Tree[c2].lh)Tree[x].ln=Tree[c1].len()+Tree[c2].ln;else Tree[x].ln=Tree[c1].len();}else Tree[x].ln=Tree[c1].ln;
    if(Tree[c2].ok){if(Tree[c2].lh==Tree[c1].rh)Tree[x].rn=Tree[c2].len()+Tree[c1].rn;else Tree[x].rn=Tree[c2].len();}else Tree[x].rn=Tree[c2].rn;
    if(Tree[c1].ok&&Tree[c2].ok&&Tree[c1].rh==Tree[c2].lh)Tree[x].ok=1;else Tree[x].ok=0;
    if(Tree[c1].rh!=Tree[c2].lh)
    {
        if(Tree[c1].rh>Tree[c2].lh)
        {
            if(Tree[c1].rc) Tree[x].s+=has[Tree[c1].r]-has[Tree[c1].r-Tree[c1].rn+1]+1;
        }else if(Tree[c2].lc) Tree[x].s+=has[Tree[c2].l+Tree[c2].ln-1]-has[Tree[c2].l]+1;
    }else if(Tree[c1].rc&&Tree[c2].lc)Tree[x].s+=has[Tree[c2].l+Tree[c2].ln-1]-has[Tree[c1].r-Tree[c1].rn+1]+1;
}
inline void Ad(int x,int v){Tree[x].laz+=v; Tree[x].lh+=v; Tree[x].rh+=v;}
inline void Down(int x){if(Tree[x].laz){Ad(c1,Tree[x].laz);Ad(c2,Tree[x].laz);Tree[x].laz=0;}}
inline void build(int l,int r,int x)
{
    Tree[x].l=l; Tree[x].r=r;
    if(l==r)
    {
        Tree[x].lh=Tree[x].rh=h; Tree[x].ln=Tree[x].rn=Tree[x].ok=1; return;
    }int mid=(l+r)>>1; build(l,mid,c1); build(mid+1,r,c2); Up(x);
}
inline void ins(int l,int r,int x,int v)
{
    if(Tree[x].l==l&&Tree[x].r==r){Ad(x,v);return;} Down(x); int mid=Tree[x].mid();
    if(r<=mid)ins(l,r,c1,v);else if(l>mid)ins(l,r,c2,v);else ins(l,mid,c1,v),ins(mid+1,r,c2,v); Up(x);
}
int main()
{
    int i,x,y,v; scanf("%d%d%d",&n,&Q,&h); has[++cnt]=1; has[++cnt]=n;
    for(i=1;i<=Q;i++)
    {
        scanf("%d%d%d",&x,&y,&v); has[++cnt]=x; has[++cnt]=y; as[i]=(node){x,y,v}; if(x>1)has[++cnt]=x-1; if(y<n)has[++cnt]=y+1;
    }sort(has+1,has+cnt+1); cnt=unique(has+1,has+cnt+1)-has-1;
    for(i=1;i<=Q;i++)
    {
        as[i].l=lower_bound(has+1,has+cnt+1,as[i].l)-has; as[i].r=lower_bound(has+1,has+cnt+1,as[i].r)-has;
    }build(1,cnt,1);
    for(i=1;i<=Q;i++)
    {
        ins(as[i].l,as[i].r,1,as[i].v);    printf("%d
",Tree[1].s);
    }
}

#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;
inline int read(){int s=0,f=0; char ch; while(!isdigit(ch)){f|=(ch=='-');ch=getchar();} while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();} return f?-s:s;}
#define r(x) x=read()
const int N=1200005;
int n,md[10],fa[N],sz[N],tong[N];
inline void solve(int x)
{
    int i,j,s=0;
    for(i=2;i<=n;i++)
    {
        fa[i]=(fa[i]+x)%(i-1); if(fa[i]<0)fa[i]+=i;else fa[i]++;
    }memset(sz,0,sizeof sz); memset(tong,0,sizeof tong);
    for(i=n;i>=1;i--){sz[i]++;sz[fa[i]]+=sz[i];} for(i=1;i<=n;i++)tong[sz[i]]++;
    for(i=1;i<=n;i++) if(n%i==0)
    {
        s=0; for(j=i;j<=n;j+=i)s+=tong[j]; if(s>=n/i)printf("%d
",i);
    }
}
int main()
{
    int i,x; r(n); for(i=1;i<=9;i++)r(md[i]); md[0]=-1;
    for(i=2;i<=n;i++){r(x);fa[i]=x;}
    for(i=0;i<=9;i++)
    {
        printf("Case #%d:
",i+1); solve(md[i]);
    }return 0;
}