AtCoder Beginner Contest 069【A，水，B，水，C，数学，D，暴力】

A - K-City

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

In K-city, there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?

Constraints

• 2≤n,m≤100

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Input

Input is given from Standard Input in the following format:

n m

Output

Print the number of blocks in K-city.

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Sample Input 1

Copy

3 4

Sample Output 1

Copy

6

There are six blocks, as shown below:

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Sample Input 2

Copy

2 2

Sample Output 2

Copy

1

There are one block, as shown below:

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题目链接：http://abc069.contest.atcoder.jp/tasks/abc069_a

分析：结论就是ans=(a-1)*(b-1)

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
     {
         write(x/);
     }
     putchar(x%+'');
 }
 int main()
 {
     int a,b;
     cin>>a>>b;
     cout<<(a-)*(b-)<<endl;
     return ;
 }

B - i18n

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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

The word internationalization is sometimes abbreviated to i18n. This comes from the fact that there are 18 letters between the first i and the last n.

You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

Constraints

• 3≤|s|≤100 (|s| denotes the length of s.)
• s consists of lowercase English letters.

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Input

Input is given from Standard Input in the following format:

s

Output

Print the abbreviation of s.

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Sample Input 1

Copy

internationalization

Sample Output 1

Copy

i18n

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Sample Input 2

Copy

smiles

Sample Output 2

Copy

s4s

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Sample Input 3

Copy

xyz

Sample Output 3

Copy

x1z

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题目链接：http://abc069.contest.atcoder.jp/tasks/abc069_b

分析：输出第一个，最后一个就好咯

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
     {
         write(x/);
     }
     putchar(x%+'');
 }
 char s[];
 int main()
 {
     cin>>s;
     int len=strlen(s);
     cout<<s[]<<len-<<s[len-]<<endl;
     return ;
 }

C - 4-adjacent

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a sequence of length N, a=(a₁,a₂,…,a_N). Each a_i is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

• For each 1≤i≤N−1, the product of a_i and a_i+1 is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

• 2≤N≤10⁵
• a_i is an integer.
• 1≤a_i≤10⁹

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Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.

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Sample Input 1

Copy

3
1 10 100

Sample Output 1

Copy

Yes

One solution is (1,100,10).

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Sample Input 2

Copy

4
1 2 3 4

Sample Output 2

Copy

No

It is impossible to permute a so that the condition is satisfied.

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Sample Input 3

Copy

3
1 4 1

Sample Output 3

Copy

Yes

The condition is already satisfied initially.

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Sample Input 4

Copy

2
1 1

Sample Output 4

Copy

No

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Sample Input 5

Copy

6
2 7 1 8 2 8

Sample Output 5

Copy

Yes

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题目链接：http://abc069.contest.atcoder.jp/tasks/arc080_a

分析：统计4的倍数，2的倍数还有不是这两个的倍数的数，然后2个2的倍数等于4的倍数，然后就这样了！

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
     {
         write(x/);
     }
     putchar(x%+'');
 }
 int main()
 {
     int n;
     cin>>n;
     int a=,b=,c=;
     for(int i=;i<n;i++)
     {
         ll x;
         x=read();
         if(x%==)
             a++;
         else if(x%==)
             b++;
         else c++;
     }
     if(b>)
         c++;
     if(a+>=c)
         cout<<"Yes"<<endl;
     else
     cout<<"No"<<endl;
     return ;
 }

D - Grid Coloring

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, …, N. Here, the following conditions should be satisfied:

• For each i (1≤i≤N), there are exactly a_i squares painted in Color i. Here, a₁+a₂+…+a_N=HW.
• For each i (1≤i≤N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.

Constraints

• 1≤H,W≤100
• 1≤N≤HW
• a_i≥1
• a₁+a₂+…+a_N=HW

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Input

Input is given from Standard Input in the following format:

H W
N
a1 a2 … aN

Output

Print one way to paint the squares that satisfies the conditions. Output in the following format:

c11 … c1W
:
cH1 … cHW

Here, c_ij is the color of the square at the i-th row from the top and j-th column from the left.

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Sample Input 1

Copy

2 2
3
2 1 1

Sample Output 1

Copy

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.

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Sample Input 2

Copy

3 5
5
1 2 3 4 5

Sample Output 2

Copy

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

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Sample Input 3

Copy

1 1
1
1

Sample Output 3

Copy

1

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题目链接：http://abc069.contest.atcoder.jp/tasks/arc080_b

分析：从一个点可以到达其它所有的点，直接来一个水平填充好像就过了

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
     {
         write(x/);
     }
     putchar(x%+'');
 }
 int w,h,n;
 int s[][];
 int cnt[];
 int main()
 {
     h=read();
     w=read();
     n=read();
     for(int i=;i<=n;i++)
         cnt[i]=read();
     int k=;
     for(int i=;i<=h;i++)
     {
         if(i%==)
         {
             for(int j=;j<=w;j++)
             {
                 if(cnt[k]==)
                 k++;
                 cnt[k]--;
                 s[i][j]=k;
             }
         }
     else
     {
         for(int j=w;j>;j--)
         {
             if(cnt[k]==)
               k++;
             cnt[k]--;
             s[i][j]=k;
         }
     }
     }
     for(int i=;i<=h;i++)
     {
         for(int j=;j<=w;j++)
             printf("%d ",s[i][j]);
         printf("\n");
     }
     return ;
 }