AtCoder Beginner Contest 131 题解

原创

mb5f5b1df7f1e34 2023-02-08 09:16:22 ©著作权

文章标签 Apple Problem i++ 文章分类 JavaScript 前端开发

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A - Security

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100 points

Problem Statement

The door of Snuke's laboratory is locked with a security code.

The security code is a 44-digit number. We say the security code is hard to enter when it contains two consecutive digits that are the same.

You are given the current security code SS. If SS is hard to enter, print Bad; otherwise, print Good.

Constraints

SS is a 44-character string consisting of digits.

Input

Input is given from Standard Input in the following format:

SS

Output

If SS is hard to enter, print Bad; otherwise, print Good.

Sample Input 1 Copy

Copy

Sample Output 1 Copy

Copy

Bad

The second and third digits are the same, so 37763776 is hard to enter.

Sample Input 2 Copy

Copy

Sample Output 2 Copy

Copy

Good

There are no two consecutive digits that are the same, so 80808080 is not hard to enter.

Sample Input 3 Copy

Copy

Sample Output 3 Copy

Copy

Bad

Sample Input 4 Copy

Copy

Sample Output 4 Copy

Copy

Bad

太水

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
int a[N];
int n;
int main()
{
    // scanf("%d",&n);
    string s;
    cin>>s;
    for(int i=0;i<s.length()-1;i++)
    {
        if(s[i]==s[i+1])
        {
            cout<<"Bad"<<endl;
            return 0;
        }
    }
    cout<<"Good"<<endl;
    return 0;
 
}

B - Bite Eating

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 200200 points

Problem Statement

You have NN apples, called Apple 11, Apple 22, Apple 33, ..., Apple NN. The flavor of Apple ii is L+i−1L+i−1, which can be negative.

You can make an apple pie using one or more of the apples. The flavor of the apple pie will be the sum of the flavors of the apples used.

You planned to make an apple pie using all of the apples, but being hungry tempts you to eat one of them, which can no longer be used to make the apple pie.

You want to make an apple pie that is as similar as possible to the one that you planned to make. Thus, you will choose the apple to eat so that the flavor of the apple pie made of the remaining N−1N−1 apples will have the smallest possible absolute difference from the flavor of the apple pie made of all the NN apples.

Find the flavor of the apple pie made of the remaining N−1N−1 apples when you choose the apple to eat as above.

We can prove that this value is uniquely determined.

Constraints

2≤N≤2002≤N≤200
−100≤L≤100−100≤L≤100
All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN LL

Output

Find the flavor of the apple pie made of the remaining N−1N−1 apples when you optimally choose the apple to eat.

Sample Input 1 Copy

Copy

5 2

Sample Output 1 Copy

Copy

The flavors of Apple 11, 22, 33, 44, and 55 are 22, 33, 44, 55, and 66, respectively. The optimal choice is to eat Apple 11, so the answer is 3+4+5+6=183+4+5+6=18.

Sample Input 2 Copy

Copy

3 -1

Sample Output 2 Copy

Copy

The flavors of Apple 11, 22, and 33 are −1−1, 00, and 11, respectively. The optimal choice is to eat Apple 22, so the answer is (−1)+1=0(−1)+1=0.

Sample Input 3 Copy

Copy

30 -50

Sample Output 3 Copy

Copy

-1044

也水

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
int a[N];
int n,L;
int main()
{
    scanf("%d%d",&n,&L);
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        a[i]=L+i-1;
        sum+=a[i];
    }
    int sum1=sum;
    int ans;
    int cha=1e9;
    for(int i=1;i<=n;i++)
    {
        sum1=sum;
        sum1-=a[i];
        if(abs(sum1-sum)<cha)
        {
            ans=sum1;
            cha=abs(sum1-sum);
        }
    }
    
    cout<<ans<<endl;

    return 0;
 
}

C - Anti-Division

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300 points

Problem Statement

You are given four integers AA, BB, CC, and DD. Find the number of integers between AA and BB (inclusive) that can be evenly divided by neither CC nor DD.

Constraints

1≤A≤B≤10181≤A≤B≤1018
1≤C,D≤1091≤C,D≤109
All values in input are integers.

Input

Input is given from Standard Input in the following format:

AA BB CC DD

Output

Print the number of integers between AA and BB (inclusive) that can be evenly divided by neither CC nor DD.

Sample Input 1 Copy

Copy

4 9 2 3

Sample Output 1 Copy

Copy

55 and 77 satisfy the condition.

Sample Input 2 Copy

Copy

10 40 6 8

Sample Output 2 Copy

Copy

Sample Input 3 Copy

Copy

314159265358979323 846264338327950288 419716939 937510582

Sample Output 3 Copy

Copy

532105071133627368

题意：
求[A,B]里既不能整除C又不能整除D的数。
分析：
(A，B](左边取不到）里可以整除C的数有B/C-A/C个，那答案就出来了
B/C-A/C+B/D-A/D-（B/e-A/e）（e为cd最小公倍数）
在判断一下左边界A，总的一减即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
//int a[N];
int n,L ;
LL gcd(LL a,LL b)
{
   return a == 0 ? b : gcd(b % a, a);
}
LL  a,b,c,d;
int main()
{
     cin>>a>>b>>c>>d;
     LL ans=b/c-a/c+b/d-a/d-(b/(c*d/gcd(c,d))-a/(c*d/gcd(c,d)));
     if(a%c==0) ans++;
     if(a%d==0) ans++;
     if(a%(c*d)==0) ans--;
   //cout<<ans<<endl;
     ans=b-a+1-ans;
     cout<<ans<<endl;

   return 0;
 
}

D - Megalomania

Time Limit: 2 sec / Memory Limit: 1024 MB

Score: 400400 points

Problem Statement

Kizahashi, who was appointed as the administrator of ABC at National Problem Workshop in the Kingdom of AtCoder, got too excited and took on too many jobs.

Let the current time be time 00. Kizahashi has NN jobs numbered 11 to NN.

It takes AiAi units of time for Kizahashi to complete Job ii. The deadline for Job ii is time BiBi, and he must complete the job before or at this time.

Kizahashi cannot work on two or more jobs simultaneously, but when he completes a job, he can start working on another immediately.

Can Kizahashi complete all the jobs in time? If he can, print Yes; if he cannot, print No.

Constraints

All values in input are integers.
1≤N≤2×1051≤N≤2×105
1≤Ai,Bi≤109(1≤i≤N)1≤Ai,Bi≤109(1≤i≤N)

Input

Input is given from Standard Input in the following format:

NN
A1A1 B1B1
..
..
..
ANAN BNBN

Output

If Kizahashi can complete all the jobs in time, print Yes; if he cannot, print No.

Sample Input 1 Copy

Copy

Sample Output 1 Copy

Copy

Yes

He can complete all the jobs in time by, for example, doing them in the following order:

Do Job 22 from time 00 to 11.
Do Job 11 from time 11 to 33.
Do Job 44 from time 33 to 77.
Do Job 33 from time 77 to 88.
Do Job 55 from time 88 to 1111.

Note that it is fine to complete Job 33 exactly at the deadline, time 88.

Sample Input 2 Copy

Copy

Sample Output 2 Copy

Copy

No

He cannot complete all the jobs in time, no matter what order he does them in.

Sample Input 3 Copy

Copy

Sample Output 3 Copy

Copy

Yes

水

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=200005;
//int a[N];
int n,L ;
struct node
{
    int id;
    int A,B;
}a[N];
bool cmp(node x,node y)
{
    if(x.B==y.B)
        return x.A<y.A;
    return x.B<y.B;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i].A>>a[i].B;
        a[i].id=i;
    }
    sort(a+1,a+n+1,cmp);
    int t=0;
    for(int i=1;i<=n;i++)
    {
    //  cout<<a[i].A<<" "<<a[i].B<<" "<<t<<endl;
        if(t+a[i].A<=a[i].B)
        {
            t+=a[i].A;
        }
        else
        {
            cout<<"No"<<endl;
            return 0;
        }
    }
    cout<<"Yes"<<endl;
    return 0;
 
}

E - Friendships

Time Limit: 2 sec / Memory Limit: 1024 MB

Score: 500500 points

Problem Statement

Does there exist an undirected graph with NN vertices satisfying the following conditions?

The graph is simple and connected.
The vertices are numbered 1,2,...,N1,2,...,N.
Let MM be the number of edges in the graph. The edges are numbered 1,2,...,M1,2,...,M, the length of each edge is 11, and Edge ii connects Vertex uiui and Vertex vivi.
There are exactly KK pairs of vertices (i, j) (i<j)(i, j) (i<j) such that the shortest distance between them is 22.

If there exists such a graph, construct an example.

Constraints

All values in input are integers.
2≤N≤1002≤N≤100
0≤K≤N(N−1)20≤K≤N(N−1)2

Input

Input is given from Standard Input in the following format:

NN KK

Output

If there does not exist an undirected graph with NN vertices satisfying the conditions, print -1.

If there exists such a graph, print an example in the following format (refer to Problem Statement for what the symbols stand for):

MM
u1u1 v1v1
::
uMuM vMvM

If there exist multiple graphs satisfying the conditions, any of them will be accepted.

Sample Input 1 Copy

Copy

5 3

Sample Output 1 Copy

Copy

This graph has three pairs of vertices such that the shortest distance between them is 22: (1, 4)(1, 4), (2, 4)(2, 4), and (3, 5)(3, 5). Thus, the condition is satisfied.

Sample Input 2 Copy

Copy

5 8

Sample Output 2 Copy

Copy

-1

There is no graph satisfying the conditions.

题意：
给你n与k，要求构造n个点，恰有k条距离为2 的路径的图
分析：
我们选取一个点当作中间节点，假设n=5

注意最多有3+2+1（即最多就有（n-2）*（n-1）/2）
连上一条你变会消除一条，模拟一下思路即可
接下来给出徐老爷的代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=200+10;
int vis[maxn][maxn];
struct Node
{
    int x,y;
    Node(){}
    Node(int a,int b):x(a),y(b){}
};
vector<Node >G;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    if(n==0)
    {
         printf("-1\n");
        return 0;
    }
    int m=n-1;
    int sum=(m)*(m-1)/2;
    if(k>sum)
    {
        printf("-1\n");
        return 0;
    }
    for(int i=1;i<=m;i++)
    {
        G.push_back(Node(i,n));
    }

    int i=1,j=2;
    while(sum!=k)
    {
        sum--;
        G.push_back(Node(i,j));
        j++;
        if(j==m+1) i++,j=i+1;
    }

    printf("%d\n",G.size());
    for(int i=0;i<G.size();i++)
    {
        printf("%d %d\n",G[i].x,G[i].y);
    }
}

F - Must Be Rectangular!

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 600600 points

Problem Statement

There are NN dots in a two-dimensional plane. The coordinates of the ii-th dot are (xi,yi)(xi,yi).

We will repeat the following operation as long as possible:

Choose four integers aa, bb, cc, dd (a≠c,b≠d)(a≠c,b≠d) such that there are dots at exactly three of the positions (a,b)(a,b), (a,d)(a,d), (c,b)(c,b) and (c,d)(c,d), and add a dot at the remaining position.

We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

1≤N≤1051≤N≤105
1≤xi,yi≤1051≤xi,yi≤105
If i≠ji≠j, xi≠xjxi≠xj or yi≠yjyi≠yj.
All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN
x1x1 y1y1
::
xNxN yNyN

Output

Print the maximum number of times we can do the operation.

Sample Input 1 Copy

Copy

Sample Output 1 Copy

Copy

By choosing a=1a=1, b=1b=1, c=5c=5, d=5d=5, we can add a dot at (1,5)(1,5). We cannot do the operation any more, so the maximum number of operations is 11.

Sample Input 2 Copy

Copy

2
10 10
20 20

Sample Output 2 Copy

Copy

There are only two dots, so we cannot do the operation at all.

Sample Input 3 Copy

Copy

Sample Output 3 Copy

Copy

We can do the operation for all choices of the form a=1a=1, b=1b=1, c=ic=i, d=jd=j (2≤i,j≤5)(2≤i,j≤5), and no more. Thus, the maximum number of operations is 1616.

题意：
给你n个点，如果有三个点如下图红点所示，则蓝色的就是可以扩充的点，问有多少个这样的点。
分析：

我们发现这个连通区域内的点数就是平行与x线条个数乘以平行与y线条个数，如果我们求出来这个，则减去已有的即可。
首先，连通区域的定义是x坐标与y坐标任意一点在这个集合内，这时候可以用并查集，但我们又不能完全的混合在一起，因为最后需要x的线条乘以y的线条个数，所以规定x在0~N，y在N+1~2*N（N为坐标最大值）
举一个例子，
N=4
（2，1）（1，1）（1，2）
则x坐标 1 2 3 4 y：5 6 7 8
fa 1 2 3 4 5 6 7 8
1.
添加（2，1）fa[find(x)]=find(y+N) fa[2]=5
x坐标 1 2 3 4 y：5 6 7 8
fa 1 5 3 4 5 6 7 8
2.
添加（1，1）fa[find(x)]=find(y+N) fa[1]=5
x坐标 1 2 3 4 y：5 6 7 8
fa 5 5 3 4 5 6 7 8
3.
添加（1，2）fa[find(x)]=find(y+N) fa[5]=6
x坐标 1 2 3 4 y：5 6 7 8
fa 5 5 3 4 6 6 7 8

#include<bits/stdc++.h>
using namespace std;
#define N 100100
typedef long long ll;
int n,f[N<<1];
int getf(int x)
{
    return x==f[x]?x:f[x]=getf(f[x]);
}
int sz1[N<<1],sz2[N<<1];

int main()
{
    ll ans=0;
    cin>>n;
    for(int i=1; i<N+N; ++i)
        f[i]=i;
    int u,v;
    for(int i=1; i<=n; ++i)
    {
        cin>>u>>v;
        f[getf(u)]=getf(v+N);
    }
    for(int i=1; i<=N; ++i)
        sz1[getf(i)]+=1;
    for(int i=N+1; i<N+N; ++i)
        sz2[getf(i)]+=1;
    for(int i=1; i<N+N; ++i)
        ans+=1ll*sz1[i]*sz2[i];
    cout<<ans-n<<endl;
    return 0;
}