AtCoder Beginner Contest 069 ABCD题

题目链接：http://abc069.contest.atcoder.jp/assignments

A - K-City

---

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

---

Input

Input is given from Standard Input in the following format:

n m

Output

Print the number of blocks in K-city.

---

Sample Input 1

Copy

3 4

Sample Output 1

Copy

6

There are six blocks, as shown below:

---

Sample Input 2

Copy

2 2

Sample Output 2

Copy

1

There are one block, as shown below:

题解：水题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main()
{
    int n,m;
    while(cin>>n>>m){
        cout<<(n-1)*(m-1)<<endl;
    }
    return 0;
}

B - i18n

---

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.

---

Input

Input is given from Standard Input in the following format:

s

Output

Print the abbreviation of s.

---

Sample Input 1

Copy

internationalization

Sample Output 1

Copy

i18n

---

Sample Input 2

Copy

smiles

Sample Output 2

Copy

s4s

---

Sample Input 3

Copy

xyz

Sample Output 3

Copy

x1z

 题解：水题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main()
{
    string a;
    while(cin>>a){
        int len=a.length();
        cout<<a[0]<<len-2<<a[len-1]<<endl;
    }
    return 0;
}

C - 4-adjacent

---

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a sequence of length N, a=(a1,a2,…,aN). Each ai 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 ai and ai+1 is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

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

---

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.

---

Sample Input 1

Copy

3
1 10 100

Sample Output 1

Copy

Yes

One solution is (1,100,10).

---

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.

---

Sample Input 3

Copy

3
1 4 1

Sample Output 3

Copy

Yes

The condition is already satisfied initially.

---

Sample Input 4

Copy

2
1 1

Sample Output 4

Copy

No

---

Sample Input 5

Copy

6
2 7 1 8 2 8

Sample Output 5

Copy

Yes

 题解：找出4和2的倍数即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int N=100005;
int a[N];
int main()
{
    int n;
    cin>>n;
    int t1=0,t2=0;
    for(int i=0;i<n;i++){
        cin>>a[i];
        if(a[i]%4==0)
            t1++;
        else if(a[i]%2==0)
            t2++;
    }
    int flag=0;
    int m;
    if(t2%2==0) m=t2;
    else m=t2-1;
    if((n-m)/2<=t1)flag=1;
    if(flag) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
}

D - Grid Coloring

---

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 ai squares painted in Color i. Here, a1+a2+…+aN=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 iby 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
• ai≥1
• a1+a2+…+aN=HW

---

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

---

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.

---

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

---

Sample Input 3

Copy

1 1
1
1

Sample Output 3

Copy

1

 题解：看半天不懂撒意思 题解说是蛇形填数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
int r,c,n,x,y,ans[200][200];
int main(void)
{
    scanf("%d%d%d",&r,&c,&n);
    x=1,y=1;
    for(int i=1,cnt;i<=n;i++){
        scanf("%d",&cnt);
        while(cnt--){
            ans[x][y]=i;
            if(y==c&&x%2==1)
                y=c,x++;
            else if(y==1&&x%2==0)
                y=1,x++;
            else if(x&1)
                y++;
            else
                y--;
        }
    }
    for(int i=1;i<=r;i++)
    for(int j=1;j<=c;j++)
        printf("%d%c",ans[i][j],j==c?'
':' ');
    return 0;
}