10.2 (晚上) 考试

T1

就是个签到题

第一个可以选m

第二个选m-1

第三个 m-2

之后都是m-2

特判n=1/2，m一开始就要mod，不然炸long long

(就这么道不是题的题，我特么考试的时候竟然觉得他是 矩阵乘优化dp....，100分啊啊啊啊啊啊啊啊啊)

 

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
using namespace std;
const int mod=1000000007;

ll n,m;
ll ans;
int T;

ll mi(ll a,ll ci,int mod)
{
    ll ans=1;
    a%=mod;
    while(ci)
    {
        if(ci&1)
            ans=ans*a%mod;
        a=a*a%mod;
        ci>>=1;
    }
    return ans;
}

int main(){
    
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&m);

        m%=mod;

        ans=m%mod;
        if(n>1)
            ans=ans*(m-1)%mod;
        if(n>2)
            ans=ans*mi(m-2,n-2,mod)%mod;
        printf("%lld
",ans);
    }
}

 

 

 

T2

两遍dfs

分每个点做root和不做root 两种情况

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define dd double
using namespace std;
const dd esp=0.000000001;
inline dd abss(dd x){return x<0?-x:x;}
inline int read()
{
    char q=getchar();int ans=0;
    while(q<'0'||q>'9')q=getchar();
    while(q>='0'&&q<='9'){ans=ans*10+q-'0';q=getchar();}
    return ans;
}
const int N=1000006;
struct son
{
    int v,next;
}a1[N*3];
int first[N*3],e;
void addbian(int u,int v)
{
    a1[e].v=v;
    a1[e].next=first[u];
    first[u]=e++;
}

int n;
int A[N];

dd deg[N];

dd sh[N],xi[N],shr[N],xir[N];

int fa[N];
void dfs1(int x)
{
    xi[x]=(dd)A[x];
    xir[x]=(dd)A[x];
    int temp;
    for(int i=first[x];i!=-1;i=a1[i].next)
    {
        temp=a1[i].v;
        if(temp==fa[x])
            continue;
        fa[temp]=x;
        dfs1(temp);
        xir[x]+=(xi[temp]*(1.0/deg[x]));
        if(deg[x]>1)
            xi[x]+=(xi[temp]*(1.0/(deg[x]-1.0)));
    }
}

void dfs2(int x)
{
    int temp;
    dd hh;
    for(int i=first[x];i!=-1;i=a1[i].next)
    {
        temp=a1[i].v;
        if(temp==fa[x])
            continue;
        hh=0;
        if(deg[x]>1)
            hh=(xi[temp]*(1.0/(deg[x]-1.0)));
        shr[temp]=( sh[x]+( xi[x]-hh ) )*(1.0/deg[temp]);
        if(deg[temp]>1)
            sh[temp]=( ( xi[x]-hh )+sh[x] )*(1.0/(deg[temp]-1.0));
        dfs2(temp);
    }
}

int main(){

    //freopen("T2.in","r",stdin);
    //freopen("T2.out","w",stdout);

    mem(first,-1);

    n=read();
    for(int i=1;i<=n;++i)
        A[i]=read();
    int tin1,tin2;
    for(int i=1;i<n;++i)
    {
        tin1=read();tin2=read();
        deg[tin1]+=1.0;
        deg[tin2]+=1.0;
        addbian(tin1,tin2);
        addbian(tin2,tin1);
    }

    dfs1(1);
    dfs2(1);
    dd mn=shr[1]+xir[1];
    int order=1;
    for(int i=2;i<=n;++i)
        if(mn>shr[i]+xir[i]-esp)
        {
            mn=shr[i]+xir[i];
            order=i;
        }

    /*printf("
");
    for(int i=1;i<=n;++i)
        printf("i=%d %lf
",i,shr[i]+xir[i]);
    printf("
");
    for(int i=1;i<=n;++i)
        printf("i=%d %lf
",i,shr[i]);
    printf("
");
    for(int i=1;i<=n;++i)
        printf("i=%d %lf
",i,xir[i]);
    printf("
");*/

    cout<<order;
}

 

 

 

T3

splay区间翻转

然后并查集维护相同的

在用类似于26进制的方法解(luan)出(gao)

 

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ls(x) ((x)->ch[0])
#define p(x) ((x)->p)
#define v(x) ((x)->v)
#define rs(x) ((x)->ch[1])
#define flag(x) ((x)->flag)
#define s(x) ((x)->s)
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
inline int read()
{
    char q=getchar();int ans=0;
    while(q<'0'||q>'9')q=getchar();
    while(q>='0'&&q<='9'){ans=ans*10+q-'0';q=getchar();}
    return ans;
}
inline char readchar()
{
    char q=getchar();
    while( (q<'a'||q>'z')&&q!='?' )q=getchar();
    return q;
}
const int N=600006;
const int INF=0x7fffffff;

int n,m;
ll K;
char s[N];
int ji[N],ccc;

struct sp
{
    int s,flag,v;
    sp *ch[2],*p;
    sp(int size,int val,sp *fa)
    {
        p=fa;s=size;v=val;
        flag=0;
    }
}*null;

sp* newsp(int size,int val,sp *fa)
{
    sp *temp=new sp(size,val,fa);
    ls(temp)=rs(temp)=null;
    return temp;
}

struct Splay
{
    sp *root,*temp;
    void sw(sp *x)
    {
        if(x==null)
            return ;
        flag(x)^=1;
        temp=ls(x);ls(x)=rs(x);rs(x)=temp;
    }
    void pushdown(sp *x)
    {
        if(x==null||!flag(x))
            return ;
        if(ls(x)!=null)sw(ls(x));
        if(rs(x)!=null)sw(rs(x));
        flag(x)=0;
    }
    void pushup(sp *x)
    {
        if(x==null)
            return ;
        s(x)=s(ls(x))+s(rs(x))+1;
    }
    inline int islc(sp *x){return ls(p(x))==x;}
    void rot(sp *x)
    {
        sp *y=p(x);int d=islc(x);
        pushdown(y);pushdown(x);
        if(p(y)==null)root=x;
        else p(y)->ch[islc(y)^1]=x;
        p(x)=p(y);
        y->ch[d^1]=x->ch[d];
        if(x->ch[d]!=null)p(x->ch[d])=y;
        x->ch[d]=y;p(y)=x;
        pushup(y);pushup(x);
    }
    void splay(sp *x,sp *t)
    {
        sp *y;
        while(p(x)!=t)
        {
            y=p(x);
            if(p(y)!=t){ islc(x)==islc(y)?rot(y):rot(x);}
            rot(x);//printf("aaa
");
        }
    }
    sp* kth(int k)
    {
        sp *x=root;
        while(x!=null)
        {
            //printf("s=%d k=%d
",s(ls(x))+1,k);
            pushdown(x);
            if( s(ls(x))+1==k )return x;
            if( s(ls(x))+1>k )
                x=ls(x);
            else
            {k-=(s(ls(x))+1);x=rs(x);}
        }
        return null;
    }
    void change(int l,int r)
    {
        
        sp *tt=kth(l);
        splay(tt,null);
        splay(kth(r+2),root);
        sw(ls(rs(root)));//printf("l=%d r=%d
",l,r);
    }
    sp* build(int l,int r,sp *fa)
    {
        if(l>r)
            return null;
        //printf("bl=%d br=%d
",l,r);
        int mid=(l+r)>>1;
        sp *x=newsp(1,mid,fa);
        ls(x)=build(l,mid-1,x);
        rs(x)=build(mid+1,r,x);
        pushup(x);
        return x;
    }
    void build()
    {
        root=newsp(1,-INF,null);
        rs(root)=newsp(1,INF,root);
        ls(rs(root))=build(1,n,rs(root));
        pushup(ls(rs(root)));
        pushup(rs(root));
        pushup(root);
        //printf("sizeisss %d
",s(root));
    }
    void dfs(sp *x)
    {
        if(x==null)
            return ;
        //printf("dsf
");
        pushdown(x);
        dfs(ls(x));
        //printf("v=%d
",v(x));
        ji[++ccc]=v(x);
        dfs(rs(x));
    }
    void dfs()
    {
        splay(kth(1),null);
        splay(kth(n+2),root);
        dfs(ls(rs(root)));
    }
}T;

int ff[N];
char vv[N];
int fin(int x)
{
    if(ff[x]==-1)
        return x;
    return ff[x]=fin(ff[x]);
}
inline void match(int u,int v)
{
    int x=fin(u),y=fin(v);
    if(x==y)
        return ;
    if(x<y)
    {
        ff[y]=x;
        if(vv[y]!='?')
            vv[x]=vv[y];
    }
    else
    {
        ff[x]=y;
        if(vv[x]!='?')
            vv[y]=vv[x];
    }
}

int q[N],con;
char qv[N];
char fina[N];

ll mi(ll a,int ci)
{
    ll ans=1;
    while(ci)
    {
        if(ci&1)
            ans=ans*a;
        printf("ci=%d
",ci);
        a=a*a;
        ci>>=1;
    }
    return ans;
}

ll tt[N];
void work()
{
    for(int i=1;i<=n;++i)
        vv[i]=s[i];

    /*printf("
");
    for(int i=1;i<=ccc;++i)
        printf("%d ",ji[i]);
    printf("
");*/

    for(int i=1;i<=n;++i)
    {
        match(i,ji[i]);
    }
    for(int i=1;i<=n;++i)
        if(fin(i)==i&&vv[i]=='?')
            q[++con]=i;

    tt[con+1]=1;
    int order=con;
    for(int i=con;i>=1;--i)
    {
        order=i;
        if(K/tt[i+1]<26)
            break;
        tt[i]=tt[i+1]*26;
    }

    for(int i=1;i<order;++i)
        qv[i]='a';
    ll temp;
    for(int i=order;i<=con;++i)
    {
        /*if(K<tt[i+1])
        {
            qv[i]='a';
            continue;
        }*/
        temp=K/tt[i+1];
        qv[i]='a'+temp;
        if(temp*tt[i+1]==K)
            --qv[i];
        K-=(temp*tt[i+1]);
    }
    for(int i=1;i<=con;++i)
        vv[q[i]]=qv[i];
    for(int i=1;i<=n;++i)
        fina[i]=vv[fin(i)];
}

int main(){

    //freopen("T3.in","r",stdin);

    mem(ff,-1);

    null=new sp(0,0,NULL);
    ls(null)=rs(null)=NULL;

    n=read();m=read();scanf("%lld",&K);
    for(int i=1;i<=n;++i)
        s[i]=readchar();

    T.build();

    int l0,r0;
    for(int i=1;i<=m;++i)
    {
        l0=read();r0=read();
        T.change(l0,r0);
    }
    T.dfs();
    work();
    for(int i=1;i<=n;++i)
        printf("%c",fina[i]);
}

 

 

总之，昨天晚上的考试 长了教训

还是要相信 T1是最简单的... (其实在打后面题的时候，就隐隐约约有一种感觉是T1很简单，但是也没在意这种感觉)

T1误导我的 其实是n，m的范围... 以后还是要花上10min好好审题啊啊啊

(这感觉就像 你好不容易把地狱领主砍死了，结果剩下一滴血，被小兵A死了)