Codeforces Round #460 (Div. 2) 919A. Supermarket 919B. Perfect Number 919C. Seat Arrangements

Posted 2020-10-23 Persistent.

这场cf有点意思，hack场，C题等于1的特判hack很多人(我hack成功3个人，上分了，哈哈哈，咳咳。。。) 

D题好像是树形dp，E题好像是中国剩余定理，F题好像还是dp，具体的不清楚，最近dp的题目好多，一会滚去学dp。

写A,B,C的题解。

 

 

A. Supermarket

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We often go to supermarkets to buy some fruits or vegetables, and on the tag there prints the price for a kilo. But in some supermarkets, when asked how much the items are, the clerk will say that a yuan for b kilos (You don‘t need to care about what "yuan" is), the same as a?/?b yuan for a kilo.

Now imagine you‘d like to buy m kilos of apples. You‘ve asked n supermarkets and got the prices. Find the minimum cost for those apples.

You can assume that there are enough apples in all supermarkets.

Input

The first line contains two positive integers n and m (1?≤?n?≤?5?000, 1?≤?m?≤?100), denoting that there are n supermarkets and you want to buy m kilos of apples.

The following n lines describe the information of the supermarkets. Each line contains two positive integers a,?b (1?≤?a,?b?≤?100), denoting that in this supermarket, you are supposed to pay a yuan for b kilos of apples.

Output

The only line, denoting the minimum cost for m kilos of apples. Please make sure that the absolute or relative error between your answer and the correct answer won‘t exceed 10?-?6.

Formally, let your answer be x, and the jury‘s answer be y. Your answer is considered correct if .

Examples

input

3 5
1 2
3 4
1 3

output

1.66666667

input

2 1
99 100
98 99

output

0.98989899

Note

In the first sample, you are supposed to buy 5 kilos of apples in supermarket 3. The cost is 5?/?3 yuan.

In the second sample, you are supposed to buy 1 kilo of apples in supermarket 2. The cost is 98?/?99 yuan.

 

 大水题。

代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdlib>
 6 #include<string.h>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<cmath>
13 using namespace std;
14 const int inf=0x3f3f3f3f;
15 int main(){
16     double a,b,ans;
17     double minn=inf;
18     int n,m;
19     cin>>n>>m;
20     for(int i=0;i<n;i++){
21         cin>>a>>b;
22         if(minn>a/b) minn=1.0*a/b;
23     }
24     ans=minn*(double)m;
25     printf("%.8lf",ans);
26 }

 

 

 

 

B. Perfect Number

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-th smallest perfect positive integer.

Input

A single line with a positive integer k (1?≤?k?≤?10?000).

Output

A single number, denoting the k-th smallest perfect integer.

Examples

input

1

output

19

input

2

output

28

Note

The first perfect integer is 19 and the second one is 28.

 

 这个题想吐槽一下，我一开始写的多组输入，交上RE了，改了多组输入过了，本来以为自己多组输入初始化什么的写挫了，赛后改了改交上去还是RE，后来直接WA，放弃。多组输入不知道为什么错。。。

代码:

 1 //B-这个题多组输入就不行。。。
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<string.h>
 8 #include<set>
 9 #include<vector>
10 #include<queue>
11 #include<stack>
12 #include<map>
13 #include<cmath>
14 using namespace std;
15 int main(){
16     int n,cnt=0;
17     ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
18     cin>>n;
19     for(int i=0;i<30000000;i++){
20         int temp=i;
21         int ans=0;
22         while(temp){
23             int x=temp%10;
24             ans+=x;
25             temp/=10;
26         }
27         if(ans==10)cnt++;
28         if(cnt==n){
29             cout<<i<<endl;
30             break;
31         }
32     }
33 }

 

 

 

C. Seat Arrangements

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Suppose that you are in a campus and have to go for classes day by day. As you may see, when you hurry to a classroom, you surprisingly find that many seats there are already occupied. Today you and your friends went for class, and found out that some of the seats were occupied.

The classroom contains n rows of seats and there are m seats in each row. Then the classroom can be represented as an n?×?m matrix. The character ‘.‘ represents an empty seat, while ‘*‘ means that the seat is occupied. You need to find k consecutive empty seats in the same row or column and arrange those seats for you and your friends. Your task is to find the number of ways to arrange the seats. Two ways are considered different if sets of places that students occupy differs.

Input

The first line contains three positive integers n,?m,?k (1?≤?n,?m,?k?≤?2?000), where n,?m represent the sizes of the classroom and k is the number of consecutive seats you need to find.

Each of the next n lines contains m characters ‘.‘ or ‘*‘. They form a matrix representing the classroom, ‘.‘ denotes an empty seat, and ‘*‘ denotes an occupied seat.

Output

A single number, denoting the number of ways to find k empty seats in the same row or column.

Examples

input

2 3 2
**.
...

output

3

input

1 2 2
..

output

1

input

3 3 4
.*.
*.*
.*.

output

0

Note

In the first sample, there are three ways to arrange those seats. You can take the following seats for your arrangement.

• (1,?3), (2,?3)
• (2,?2), (2,?3)
• (2,?1), (2,?2)

 

这个题，特判1那里hack一堆人，改了就可以，不用搜索，直接暴力就可以。

代码:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 const int maxn=1e4+5;
10 const double eps=1e6;
11 const int inf=1<<30;
12 char s[maxn][maxn];
13 int main(){
14     int n,m,k;
15     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
16     while(cin>>n>>m>>k){
17         memset(s,0,sizeof(s));
18         for(int i=1;i<=n;i++)
19             cin>>s[i];
20         int ans=0,l=0;
21         for(int i=1;i<=n;i++){
22             l=0;
23             for(int j=0;j<=m-1;j++){
24                 if(s[i][j]==‘.‘)l++;
25                 else{
26                     ans+=max(0,l-k+1);
27                     l=0;
28                 }
29                 if(j==(m-1)&&s[i][j]!=‘*‘)
30                     ans+=max(0,l-k+1);
31             }
32         }
33         for(int j=0;j<=m-1;j++){
34             l=0;
35             for(int i=1;i<=n;i++){
36                 if(s[i][j]==‘.‘) l++;
37                 else{
38                     ans+=max(0,l-k+1);
39                     l=0;
40                 }
41                 if(i==n&&s[i][j]!=‘*‘)
42                 ans+=max(0,l-k+1);
43             }
44         }
45         if(k==1) ans/=2;
46         cout<<ans<<endl;
47     }
48 }

 

 

 

滚去学dp，学会了补后面的题|??ω?` )