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
3 4
Sample Output 1
6
There are six blocks, as shown below:
Sample Input 2
2 2
Sample Output 2
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=,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
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
internationalization
Sample Output 1
i18n
Sample Input 2
smiles
Sample Output 2
s4s
Sample Input 3
xyz
Sample Output 3
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=,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
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
3
1 10 100
Sample Output 1
Yes
One solution is (1,100,10).
Sample Input 2
4
1 2 3 4
Sample Output 2
No
It is impossible to permute a so that the condition is satisfied.
Sample Input 3
3
1 4 1
Sample Output 3
Yes
The condition is already satisfied initially.
Sample Input 4
2
1 1
Sample Output 4
No
Sample Input 5
6
2 7 1 8 2 8
Sample Output 5
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=,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
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≥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
2 2
3
2 1 1
Sample Output 1
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
3 5
5
1 2 3 4 5
Sample Output 2
1 4 4 4 3
2 5 4 5 3
2 5 5 5 3
Sample Input 3
1 1
1
1
Sample Output 3
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=,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 ;
}