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
3 4
Sample Output 1
6
There are six blocks, as shown below:
![9179be829dc9810539213537d4c7398c.png](https://atcoder.jp/img/abc069/9179be829dc9810539213537d4c7398c.png)
Sample Input 2
2 2
Sample Output 2
1
There are one block, as shown below:
![997bfafa99be630b54d037225a5c68ea.png](https://atcoder.jp/img/abc069/997bfafa99be630b54d037225a5c68ea.png)
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:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 inline void write(int x) 22 { 23 if(x<0) 24 { 25 putchar('-'); 26 x=-x; 27 } 28 if(x>9) 29 { 30 write(x/10); 31 } 32 putchar(x%10+'0'); 33 } 34 int main() 35 { 36 int a,b; 37 cin>>a>>b; 38 cout<<(a-1)*(b-1)<<endl; 39 return 0; 40 }
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:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 inline void write(int x) 22 { 23 if(x<0) 24 { 25 putchar('-'); 26 x=-x; 27 } 28 if(x>9) 29 { 30 write(x/10); 31 } 32 putchar(x%10+'0'); 33 } 34 char s[105]; 35 int main() 36 { 37 cin>>s; 38 int len=strlen(s); 39 cout<<s[0]<<len-2<<s[len-1]<<endl; 40 return 0; 41 }
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:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 inline void write(int x) 22 { 23 if(x<0) 24 { 25 putchar('-'); 26 x=-x; 27 } 28 if(x>9) 29 { 30 write(x/10); 31 } 32 putchar(x%10+'0'); 33 } 34 int main() 35 { 36 int n; 37 cin>>n; 38 int a=0,b=0,c=0; 39 for(int i=0;i<n;i++) 40 { 41 ll x; 42 x=read(); 43 if(x%4==0) 44 a++; 45 else if(x%2==0) 46 b++; 47 else c++; 48 } 49 if(b>0) 50 c++; 51 if(a+1>=c) 52 cout<<"Yes"<<endl; 53 else 54 cout<<"No"<<endl; 55 return 0; 56 }
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:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 inline void write(int x) 22 { 23 if(x<0) 24 { 25 putchar('-'); 26 x=-x; 27 } 28 if(x>9) 29 { 30 write(x/10); 31 } 32 putchar(x%10+'0'); 33 } 34 int w,h,n; 35 int s[105][105]; 36 int cnt[10005]; 37 int main() 38 { 39 h=read(); 40 w=read(); 41 n=read(); 42 for(int i=1;i<=n;i++) 43 cnt[i]=read(); 44 int k=1; 45 for(int i=1;i<=h;i++) 46 { 47 if(i%2==1) 48 { 49 for(int j=1;j<=w;j++) 50 { 51 if(cnt[k]==0) 52 k++; 53 cnt[k]--; 54 s[i][j]=k; 55 } 56 } 57 else 58 { 59 for(int j=w;j>0;j--) 60 { 61 if(cnt[k]==0) 62 k++; 63 cnt[k]--; 64 s[i][j]=k; 65 } 66 } 67 } 68 for(int i=1;i<=h;i++) 69 { 70 for(int j=1;j<=w;j++) 71 printf("%d ",s[i][j]); 72 printf("\n"); 73 } 74 return 0; 75 }
作 者:Angel_Kitty
出 处:https://www.cnblogs.com/ECJTUACM-873284962/
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