ICPC南京补题

由于缺的题目比较多，竟然高达3题，所以再写一篇补题的博客

Lpl and Energy-saving Lamps

During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secret. After a pause, Dragon added:

— We have a contract. A rental agreement. He always works all day long. He likes silence. Besides that, there are many more advantages of living here in the Castle. Say, it is easy to justify a missed call: a phone ring can't reach the other side of the Castle from where the phone has been left. So, the imprisonment is just a tale. Actually, he thinks about everything. He is smart. For instance, he started replacing incandescent lamps with energy-saving lamps in the whole Castle...

Lpl chose a model of energy-saving lamps and started the replacement as described below. He numbered all rooms in the Castle and counted how many lamps in each room he needs to replace.

At the beginning of each month, Lpl buys

As soon as all the work is done, he ceases buying new lamps. They are very high quality and have a very long-life cycle.

Your task is for a given number of month and descriptions of rooms to compute in how many rooms the old lamps will be replaced with energy-saving ones and how many energy-saving lamps will remain by the end of each month.

Input

Each input will consist of a single test case.

The first line contains integers $\le n \le 100000, 1 \le m \le 100)m(1≤n≤100000,1≤m≤100)— the number of rooms in the Castle and the number of energy-saving lamps, which Lpl buys monthly.$

The second line contains $k_1, k_2, ..., k_nk1,k2,...,kn(1 \le k_j \le 10000, j = 1, 2, ..., n)(1≤kj≤10000,j=1,2,...,n) — the number of lamps in the rooms of the Castle. The number in position jj is the number of lamps in jj-th room. Room numbers are given in accordance with Lpl's list.$

The third line contains one integer $\le q \le 100000)q(1≤q≤100000) — the number of queries.$

The fourth line contains $d_1, d_2, ..., d_qd1,d2,...,dq(1 \le d_p \le 100000, p = 1, 2, ..., q)(1≤dp≤100000,p=1,2,...,q)— numbers of months, in which queries are formed.$

Months are numbered starting with

Output

Line $d_pdp month.$

Hint

Explanation for the sample:

In the first month, he bought

样例输入

5 4
3 10 5 2 7
10
5 1 4 8 7 2 3 6 4 7

样例输出

题目来源

ACM-ICPC 2018 南京赛区网络预赛

从队友扒到的题意

城堡里有n个房间每个房间里有ki盏旧灯，Lpl每天购买m盏灯，他会按顺序，如果他的新灯>=房间的旧灯，他会拿新灯替换所有旧灯，直到不能替换，q个询问，每个询问输出两个数字分别为当前月有几个房间已经替换过了，还剩下几盏灯

要想到预处理这个题就解决一大半了，所以直接线段树啊，换完了就是变为inf就好

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pq priority_queue
const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
int a[N<<2];
void build(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&a[rt]);
        return;
    }
    int mid=(l+r)>>1;
    build(lson),build(rson);
    a[rt]=min(a[rt<<1],a[rt<<1|1]);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
        return a[rt];
    int mid=(l+r)>>1,ans=1e9;
    if(L<=mid)ans=min(ans,query(L,R,lson));
    if(R>mid)ans=min(ans,query(L,R,rson));
    return ans;
}
int f,La,Lb;
void update(int L,int R,int l,int r,int rt)
{
    if(L>r||R<l)return;
    if(L<=l&&r<=R&&a[rt]>La)return;
    if(l==r)
    {
        ++f,La-=a[rt],Lb=l,a[rt]=INF;
        return;
    }
    int mid=(l+r)>>1;
    if(L<=mid)update(L,R,lson);
    if(R>mid)update(L,R,rson);
    a[rt]=min(a[rt<<1],a[rt<<1|1]);
}
int n,m,q;
int ans1[N],ans2[N];
int main()
{
    scanf("%d%d",&n,&m);
    build(1,n,1);
    for(int i=1; i<=100000; i++)
    {
        if(f==n)
        {
            ans1[i]=f,ans2[i]=La;
            continue;
        }
        La+=m,Lb=1;
        while(query(Lb,n,1,n,1)<=La)update(Lb,n,1,n,1);
        ans1[i]=f,ans2[i]=La;
    }
    scanf("%d",&q);
    for(int i=0,x; i<q; i++)scanf("%d",&x),printf("%d %d\n",ans1[x],ans2[x]);
    return 0;
}

Skr

256000K

A number is skr, if and only if it's unchanged after being reversed. For example, "12321", "11" and "1" are skr numbers, but "123", "221" are not. FYW has a string of numbers, each substring can present a number, he wants to know the sum of distinct skr number in the string. FYW are not good at math, so he asks you for help.

Input

The only line contains the string of numbers

It is guaranteed that $\le S[i] \le 91≤S[i]≤9, the length of SS is less than 20000002000000.$

Output

Print the answer modulo

样例输入1

样例输出1

样例输入2

样例输出2

题目来源

ACM-ICPC 2018 南京赛区网络预赛

比赛的时候特别想把这个题搞出来，奈何自己菜，回文树是第一次接触

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pq priority_queue
const int N=3000005,MD=1e9+7,INF=0x3f3f3f3f;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
char s[N];
ll fac[N],inv[N],slen;
ll ans=0;
struct Palindrome_Automaton//回文自动机
{
    int i,len[N],nxt[N][10],fail[N],cnt[N],last,cur,S[N],p,n;
    int newnode(int l)//新建节点
    {
        for(int i=0; i<10; i++)nxt[p][i]=0; //新建的节点为p，先消除它的子节点
        cnt[p]=0;
        len[p]=l;
        return p++;//勿打成++p，因为此节点为p，我们应返回p
    }
    inline void init()//初始化
    {
        p=n=last=0;
        newnode(0);
        newnode(-1);
        S[0]=-1;
        fail[0]=1;
    }
    int get_fail(int x)
    {
        while(S[n-len[x]-1]!=S[n])x=fail[x];
        return x;
    }
    inline void add(int c,int pos)//插字符
    {
        c-='0';
        S[++n]=c;
        int cur=get_fail(last);
        if(!nxt[cur][c])
        {
            int now=newnode(len[cur]+2);
            fail[now]=nxt[get_fail(fail[cur])][c];
            nxt[cur][c]=now;
            ans=(ans+(fac[pos-len[now]+1]-fac[pos+1])*inv[slen-pos-1]%MD+MD)%MD;
        }
        last=nxt[cur][c];
        cnt[last]++;
    }
    void count()//统计本质相同的回文串的出现次数
    {
        //逆序累加，保证每个点都会比它的父亲节点先算完，于是父亲节点能加到所有子孙
        for(int i=p-1; i>=0; i--)
            cnt[fail[i]]+=cnt[i];
    }
} run;
ll POW(ll x,ll y)
{
    ll ans=1;
    for(; y; x=x*x%MD,y>>=1)if(y&1)ans=ans*x%MD;
    return ans;
}
int main()
{
    scanf("%s",&s);
    slen=strlen(s);//千万要先把这个记录下来，因为求长度的时间复杂度是O(n)——直接扫一遍，碰到结束符才停止，我一开始把它直接塞进下方循环的nn里，就T了一遍
    inv[0]=1,fac[slen]=0;
    ll t=POW(10,MD-2),now=1;
    for(int i=1; i<=slen; i++)inv[i]=t*inv[i-1]%MD;
    for(int i=slen-1; i>=0; i--)fac[i]=(now*(s[i]-'0')+fac[i+1])%MD,now=now*10%MD;
    run.init(),ans=0;
    for(int i=0; i<slen; i++)run.add(s[i],i);
    cout<<ans<<"\n";
}

biu哥这个hash很强大啊%%%留做模板

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pq priority_queue
const int N=2000005,MD=1e9+7,INF=0x3f3f3f3f,Mod=2000007;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
ull P=13331;
ull sqr[N],Hash_[N];
char str[N];
ull sumHash[N];
struct StringHash
{
    int first[Mod+2],num;
    ull EdgeNum[N];
    int nxt[N],close[N];
    void init ()
    {
        num=0,memset(first,0,sizeof first);
    }
    int insert (unsigned long long val,int id)
    {
        int u=val%Mod;
        for(int i=first[u],t;i;i=nxt[i])
        {
            if(val==EdgeNum[i])
            {
                t=close[i],close[i]=id;
                return t;
            }
        }
        ++num;
        EdgeNum[num]=val,close[num]=id,nxt[num]=first[u],first[u]=num;
        return 0;
    }
} H;
int r[N];
ll sum[N],ans=0,xp[N];
void init()
{
    xp[0]=1;
    for(int i=1;i<N;i++)xp[i]=(xp[i-1]*10)%MD;
    return ;
}
void make_hash(char str[])//处理出str的hash值
{

    int len=strlen(str+1);
    sum[len+1]=0;
    for(int i=len;i>=1;i--)
    {
        sum[i]=(sum[i+1]*10+str[i]-'0')%MD;
    }
    return ;
}
ll Get_hash(int i,int L)
{
    return (sum[i]%MD-sum[i+L]*xp[L]%MD+MD)%MD;//注意这里要对被减的先取模
}
void insert_(int i,int j)
{
    ull now=Hash_[j]-Hash_[i-1]*sqr[j-i+1];
    if(!H.insert(now,1))//先判重，再插入
        ans=(ans+Get_hash(i,j-i+1))%MD;
}
void solve(int n)
{
    sqr[0]=1;
    for(int i=1;i<=n;++i)sqr[i]=sqr[i-1]*P,Hash_[i]=Hash_[i-1]*P+str[i];
}
void manacher(int n)
{
    int x=0,pos=0;
    for(int i=1;i<=n;i++)
    {
        int j=0;
        insert_(i,i);
        if(pos>i) j=min(r[2*x-i],pos-i);
        while(i+j+1<=n&&str[i+j+1]==str[i-j-1])
        {
            insert_(i-j-1,i+j+1);
            j++;
        }
        r[i]=j;
        if(i+j>pos)
        {
            pos=i+j;
            x=i;
        }
    }
    x=0,pos=0;
    for(int i=2;i<=n;i++)
    {
        int j=0;
        if(pos>i)j=min(r[2*x-i],pos-i+1);
        while(i+j<=n&&str[i+j]==str[i-j-1])
        {
            insert_(i-j-1,i+j);
            ++j;
        }
        r[i]=j;
        if(i+j-1>pos)
        {
            pos=i+j-1;
            x=i;
        }
    }
}
int main()
{
    scanf("%s",str+1);
    int n=strlen(str+1);
    init();//hash每一位的贡献预处理
    make_hash(str);//hash贡献值预处理
    solve(n);//hash表预处理
    manacher(n);//manacher
    printf("%lld\n",ans);
    return 0;
}

模拟甩锅给队友

posted @ 2018-09-07 11:14 暴力都不会的蒟蒻阅读(375) 评论(0) 编辑收藏举报

会员力量，点亮园子希望

刷新页面返回顶部

BobHuang

ICPC南京补题

Input

Output

Hint

样例输入

样例输出

题目来源

Input

Output

样例输入1

样例输出1

样例输入2

样例输出2

题目来源

公告

BobHuang

ICPC南京补题

Input

Output

Hint

样例输入复制

样例输出复制

题目来源

Input

Output

样例输入1复制

样例输出1复制

样例输入2复制

样例输出2复制

题目来源

公告

样例输入

样例输出

样例输入1

样例输出1

样例输入2

样例输出2