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

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=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
write(x/10);
}
putchar(x%10+'0');
}
int main()
{
int a,b;
cin>>a>>b;
cout<<(a-1)*(b-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=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
write(x/10);
}
putchar(x%10+'0');
}
char s[105];
int main()
{
cin>>s;
int len=strlen(s);
cout<<s[0]<<len-2<<s[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

62 7 1 8 2 8

Sample Output 5

Copy

Yes

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=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
write(x/10);
}
putchar(x%10+'0');
}
int main()
{
int n;
cin>>n;
int a=0,b=0,c=0;
for(int i=0;i<n;i++)
{
ll x;
x=read();
if(x%4==0)
a++;
else if(x%2==0)
b++;
else c++;
}
if(b>0)
c++;
if(a+1>=c)
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 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

ai≥1a1+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 23
2 1 1

Sample Output 1

Copy

1 12 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

1 111

Sample Output 3

Copy

1

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=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9)
{
write(x/10);
}
putchar(x%10+'0');
}
int w,h,n;
int s[105][105];
int cnt[10005];
int main()
{
h=read();
w=read();
n=read();
for(int i=1;i<=n;i++)
cnt[i]=read();
int k=1;
for(int i=1;i<=h;i++)
{
if(i%2==1)
{
for(int j=1;j<=w;j++)
{
if(cnt[k]==0)
k++;
cnt[k]--;
s[i][j]=k;
}
}
else
{
for(int j=w;j>0;j--)
{
if(cnt[k]==0)
k++;
cnt[k]--;
s[i][j]=k;
}
}
}
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
printf("%d ",s[i][j]);
printf("\n");
}
return 0;
}