AtCoder Petrozavodsk Contest 001

时间 2019-11-30

标签 atcoder petrozavodsk contest 繁體版

原文原文链接

第一场apc，5H的持久战，我固然水几个题就睡了ios

A - Two Integers

Time limit : 2sec / Memory limit : 256MBc++

Score : 100 pointsapp

Problem Statement

You are given positive integers X and Y. If there exists a positive integer not greater than 1018 that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print −1.ide

Constraints

1≤X,Y≤109
X and Y are integers.

Input

Input is given from Standard Input in the following format:优化

X Y

Output

Print a positive integer not greater than 1018 that is a multiple of X but not a multiple of Y, or print −1 if it does not exist.ui

Sample Input 1

Copy

8 6

Sample Output 1

Copy

For example, 16 is a multiple of 8 but not a multiple of 6.this

Sample Input 2

Copy

3 3

Sample Output 2

Copy

-1

A multiple of 3 is a multiple of 3.spa

这个是特判题的rest

输出一个数是x的倍数但并非y的倍数code

emmmm，模拟一下？不，a是b的倍数的话不存在，不然输出a就行了啊

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    if (x%y==0)printf("-1");
    else printf("%d",x);
}

给了16，因此我再猜最大公约数*a，太naive了，想了数据把本身hack了

话说直接暴力模拟就是直接在输出x

B - Two Arrays

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

You are given two integer sequences of length N: a1,a2,..,aN and b1,b2,..,bN. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to ai.
Add 1 to bj.

Constraints

1≤N≤10 000
0≤ai,bi≤109 (1≤i≤N)
All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a1 a2 .. aN
b1 b2 .. bN

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

Copy

3
1 2 3
5 2 2

Sample Output 1

Copy

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a={3,2,3}, b={5,3,2}.
Second operation: i=1 and j=2. Now we have a={5,2,3}, b={5,4,2}.
Third operation: i=2 and j=3. Now we have a={5,4,3}, b={5,4,3}.

Sample Input 2

Copy

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

Copy

No

Sample Input 3

Copy

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

Copy

No

说是能够加任意次，可是你要让两个序列相等，最屡次数就是sumb-suma，每次操做会让suma+2，sumb+1，至关于差少了1

而后去模拟让两个相等，须要分下奇偶。

#include<bits/stdc++.h>
using namespace std;
const int N=1e4+5;
int a[N],b[N];
int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>a[i];
    long long s=0;
    for(int i=0; i<n; i++)
        cin>>b[i],s+=a[i]-b[i];
    if(s>0)
        printf("No");
    else
    {
        long long ta=-s,tb=-s;
        for(int i=0; i<n; i++)
        {
            if(a[i]>b[i])
                tb-=a[i]-b[i];
            else if(a[i]<b[i])
            {
                if((b[i]-a[i])&1)
                    ta-=(b[i]-a[i])/2+1,tb--;
                else ta-=(b[i]-a[i])/2;
            }
        }
        if(2*ta==tb&&ta>=0&&tb>=0)
            printf("Yes");
        else printf("No");
    }
    return 0;
}

固然ta,tb也能够省掉，我就是模拟这个次数就能够了

#include<bits/stdc++.h>
using namespace std;
const int N=1e4+5;
int a[N],b[N];
int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>a[i];
    long long s=0;
    for(int i=0; i<n; i++)
        cin>>b[i];
    for(int i=0; i<n; i++)
    {
        if(a[i]>b[i])
            s-=a[i]-b[i];
        else s+=(b[i]-a[i])/2;
    }
    if(s>=0)
        printf("Yes");
    else printf("No");
    return 0;
}

C - Vacant Seat

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

This is an interactive task.

Let N be an odd number at least 3.

There are N seats arranged in a circle. The seats are numbered 0 through N−1. For each i (0≤i≤N−2), Seat i and Seat i+1are adjacent. Also, Seat N−1 and Seat 0 are adjacent.

Each seat is either vacant, or oppupied by a man or a woman. However, no two adjacent seats are occupied by two people of the same sex. It can be shown that there is at least one empty seat where N is an odd number at least 3.

You are given N, but the states of the seats are not given. Your objective is to correctly guess the ID number of any one of the empty seats. To do so, you can repeatedly send the following query:

Choose an integer i (0≤i≤N−1). If Seat i is empty, the problem is solved. Otherwise, you are notified of the sex of the person in Seat i.

Guess the ID number of an empty seat by sending at most 20 queries.

Constraints

N is an odd number.
3≤N≤99 999

Input and Output

First, N is given from Standard Input in the following format:

Then, you should send queries. A query should be printed to Standart Output in the following format. Print a newline at the end.

The response to the query is given from Standard Input in the following format:

Here, s is Vacant, Male or Female. Each of these means that Seat i is empty, occupied by a man and occupied by a woman, respectively.

Notice

Flush Standard Output each time you print something. Failure to do so may result in TLE.
Immediately terminate the program when s is Vacant. Otherwise, the verdict is indeterminate.
The verdict is indeterminate if more than 20 queries or ill-formatted queries are sent.

Sample Input / Output 1

In this sample, N=3, and Seat 0, 1, 2 are occupied by a man, occupied by a woman and vacant, respectively.

Input	Output
3
	0
Male
	1
Female
	2
Vacant

C是个交互题，记得在每次输出后fflush(stdout);

#include<bits/stdc++.h>
using namespace std;
map<string,int>M;
int x[999999];
int main()
{
    M["Male"]=1;
    M["Female"]=0;
    int n;
    scanf("%d",&n);
    string s;
    cout<<0<<endl;
    fflush(stdout);
    cin>>s;
    if(s=="Vacant")return 0;
    int lval=M[s];
    cout<<n-1<<endl;
    fflush(stdout);
    cin>>s;
    if(s=="Vacant")return 0;
    int rval=M[s];
    int l=0,r=n-1;
    while(true)
    {
        int mi=(l+r)/2;
        cout<<mi<<endl;
        fflush(stdout);
        cin>>s;
        if(s=="Vacant")return 0;
        int val=M[s];
        if(!((mi-l+1)%2&1)&&val==lval)
        {
            r=mi;
            rval=val;
        }
        else if((mi-l+1)&1&&val!=lval)
        {
            r=mi;
            rval=val;
        }
        else if(!((r-mi+1)%2)&&val==rval)
        {
            l=mi;
            lval=val;
        }
        else if((r-mi+1)&1&&val!=rval)
        {
            l=mi;
            lval=val;
        }
    }
    return 0;
}

大神的牛逼代码

#include<iostream>
using namespace std;
string s,t;
main()
{
    int n,f,l,m;cin>>n;f=0;l=n;
    cout<<0<<endl;
    cin>>s;
    if(s=="Vacant")return 0;
    while(t!="Vacant"){
        m=(f+l)/2;
        cout<<m<<endl;
        cin>>t;
        if(m%2^s==t)f=m;
        else l=m;
    }
}

D - Forest

Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

You are given a forest with N vertices and M edges. The vertices are numbered 0 through N−1. The edges are given in the format (xi,yi), which means that Vertex xi and yi are connected by an edge.

Each vertex i has a value ai. You want to add edges in the given forest so that the forest becomes connected. To add an edge, you choose two different vertices i and j, then span an edge between i and j. This operation costs ai+aj dollars, and afterward neither Vertex i nor j can be selected again.

Find the minimum total cost required to make the forest connected, or print Impossible if it is impossible.

Constraints

1≤N≤100,000
0≤M≤N−1
1≤ai≤109
0≤xi,yi≤N−1
The given graph is a forest.
All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
a0 a1 .. aN−1
x1 y1
x2 y2
:
xM yM

Output

Print the minimum total cost required to make the forest connected, or print Impossible if it is impossible.

Sample Input 1

Copy

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

Copy

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1+6=7 dollars.

Sample Input 2

Copy

5 0
3 1 4 1 5

Sample Output 2

Copy

Impossible

We can't make the graph connected.

Sample Input 3

Copy

1 0
5

Sample Output 3

Copy

The graph is already connected, so we do not need to add any edges.

并查集+优先队列的一个很优秀的题目

有一个森林，如今要把这个森林合并成一棵树。每一个节点都回有一个权值。每一个添加一条边所付出的代价就是所链接的两个权值的和。每一个节点只能链接一次。

每组最小的是必定要的，可是这个只能加一次

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
int a[N],fa[N],vis[N],mi[N];
pair<int,int>t;
priority_queue<pair<int,int>, vector<pair<int,int> >, greater<pair<int,int> > >Q;
int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        cin>>a[i],fa[i]=i,mi[i]=INF;
    for(int i=0,u,v; i<m; i++)
    {
        cin>>u>>v,u=find(u),v=find(v);
        if(u!=v)fa[u]=v;
    }
    for(int i=0; i<n; i++)find(i);
    for(int i=0; i<n; i++)
        mi[fa[i]]=min(a[i],mi[fa[i]]),Q.push(make_pair(a[i],fa[i]));
    sort(fa,fa+n);
    int tn=unique(fa,fa+n)-fa;
    if(tn==1)
        cout<<0;
    else
    {
        long long s=0;
        for(int i=0; i<tn; i++)
            s+=mi[fa[i]];
        while(!Q.empty())
        {
            if(tn==2)break;
            t=Q.top();
            Q.pop();
            if(!vis[t.second])
            {
                vis[t.second]=1;
                continue;
            }
            s+=t.first,--tn;

        }
        if(tn==2)cout<<s;
        else cout<<"Impossible";
    }
    return 0;
}

特判下impossible速度并无更快

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
int a[N],fa[N],vis[N],mi[N];
typedef pair<int,int> pii;
priority_queue<pii, vector<pii >, greater<pii > >Q;
int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        cin>>a[i],fa[i]=i,mi[i]=INF;
    for(int i=0,u,v; i<m; i++)
    {
        cin>>u>>v,u=find(u),v=find(v);
        if(u!=v)fa[u]=v;
    }
    for(int i=0; i<n; i++)find(i);
    for(int i=0; i<n; i++)
        mi[fa[i]]=min(a[i],mi[fa[i]]),Q.push(make_pair(a[i],fa[i]));
    sort(fa,fa+n);
    int tn=unique(fa,fa+n)-fa;
    if(tn==1)
        cout<<0;
    else if((tn-1)*2>n)
        cout<<"Impossible";
    else
    {
        long long s=0;
        for(int i=0; i<tn; i++)
            s+=mi[fa[i]];
        while(!Q.empty())
        {
            if(tn==2)break;
            pii t=Q.top();
            Q.pop();
            if(!vis[t.second])
            {
                vis[t.second]=1;
                continue;
            }
            s+=t.first,--tn;
        }
        cout<<s;
    }
    return 0;
}

优化不了了，最后的代码也很棒

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
typedef pair<int,int> pii;
int fa[N],vis[N],mi[N],V[N];
vector<pii >Q(N);
int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{
    ios::sync_with_stdio(false);
    int n,m,tn=0;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        cin>>Q[i].first,fa[i]=i,mi[i]=INF;
    for(int i=0,u,v; i<m; i++)
    {
        cin>>u>>v,u=find(u),v=find(v);
        if(u!=v)fa[u]=v;
    }
    for(int i=0; i<n; i++)
    {
        find(i);
        if(!vis[fa[i]])V[tn++]=fa[i];
        vis[fa[i]]=1,Q[i].second=fa[i],mi[fa[i]]=min(Q[i].first,mi[fa[i]]);
    }
    if(tn==1)cout<<0;
    else if((tn-1)*2>n)cout<<"Impossible";
    else
    {
        long long s=0;
        for(int i=0; i<tn; i++)s+=mi[V[i]];
        if(tn!=2)sort(Q.begin(),Q.begin()+n);
        for(int i=0; i<n&&tn!=2; i++)
        {
            if(vis[Q[i].second])
            {
                vis[Q[i].second]=0;
                continue;
            }
            s+=Q[i].first,tn--;
        }
        cout<<s;
    }
    return 0;
}

E - Antennas on Tree

Time limit : 2sec / Memory limit : 256MB

Score : 900 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 0 through N−1, and the i-th edge (0≤i<N−1) comnnects Vertex aiand bi. For each pair of vertices u and v (0≤u,v<N), we define the distance d(u,v) as the number of edges in the path u-v.

It is expected that one of the vertices will be invaded by aliens from outer space. Snuke wants to immediately identify that vertex when the invasion happens. To do so, he has decided to install an antenna on some vertices.

First, he decides the number of antennas, K (1≤K≤N). Then, he chooses K different vertices, x0, x1, ..., xK−1, on which he installs Antenna 0, 1, ..., K−1, respectively. If Vertex v is invaded by aliens, Antenna k (0≤k<K) will output the distance d(xk,v). Based on these K outputs, Snuke will identify the vertex that is invaded. Thus, in order to identify the invaded vertex no matter which one is invaded, the following condition must hold:

For each vertex u (0≤u<N), consider the vector (d(x0,u),…,d(xK−1,u)). These N vectors are distinct.

Find the minumum value of K, the number of antennas, when the condition is satisfied.

Constraints

2≤N≤105
0≤ai,bi<N
The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a0 b0
a1 b1
:
aN−2 bN−2

Output

Print the minumum value of K, the number of antennas, when the condition is satisfied.

Sample Input 1

Copy

Sample Output 1

Copy

For example, install an antenna on Vertex 1 and 3. Then, the following five vectors are distinct:

(d(1,0),d(3,0))=(1,1)
(d(1,1),d(3,1))=(0,2)
(d(1,2),d(3,2))=(2,2)
(d(1,3),d(3,3))=(2,0)
(d(1,4),d(3,4))=(3,1)

Sample Input 2

Copy

2
0 1

Sample Output 2

Copy

For example, install an antenna on Vertex 0.

Sample Input 3

Copy

Sample Output 3

Copy

For example, install an antenna on Vertex 0, 4, 9.

一颗树让你找最少k个点放监控，使得全部节点到这个节点的k维空间向量（无方向）两两不一样

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
vector<int>G[N];
int n,dp[N],f=-1;
void dfs(int i,int p)
{
    int F=0;
    for(auto j:G[i])
    {
        if(j==p)continue;
        dfs(j,i);
        dp[i]+=dp[j];
        if(!dp[j])
        {
            if(!F)F=1;
            else dp[i]++;
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1,u,v; i<n; i++)
        scanf("%d%d",&u,&v),G[u].push_back(v),G[v].push_back(u);
    for(int i=0; i<n; i++)
        if(G[i].size()>2)
        {
            f=i;
            break;
        }
    if(f==-1)printf("1\n");
    else dfs(f,-1),printf("%d\n",dp[f]);
    return 0;
}