AtCoder Beginner Contest 150

 

AB没啥好说的

 

C - Count Order / 

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300 points

Problem Statement

We have two permutations (1, 2, ..., N)).

There are |a−b|.

Notes

For two sequences Xk<Yk.

Constraints

• 2≤N≤8
• N.

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Input

Input is given from Standard Input in the following format:

N
PN
QN

Output

Print |a−b|.

 

用康托展开，这里数据很小也可以直接STL跑next_permutation。。

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 
 5 int jc[10]={1};
 6 int a[10],b[10],n;
 7 int cal(int a[],int n){
 8     int res=0;
 9     for(int i=1;i<=n;++i){
10         int c=0;
11         for(int j=i+1;j<=n;++j)c+=(a[j]<a[i]);
12         res+=c*jc[n-i];
13     }return res;
14 }
15 int main(){
16     for(int i=1;i<=9;++i)jc[i]=i*jc[i-1];
17     cin>>n;
18     for(int i=1;i<=n;++i)cin>>a[i];
19     for(int i=1;i<=n;++i)cin>>b[i];
20     cout<<abs(cal(b,n)-cal(a,n))<<endl;
21     return 0;
22 }

 

 

 

D - Semi Common Multiple / 

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400 points

Problem Statement

Given are a sequence M.

Let a semi-common multiple of (1≤k≤N):

• There exists a non-negative integer X=ak×(p+0.5).

Find the number of semi-common multiples of M (inclusive).

Constraints

• 1≤N≤105
• 1≤M≤109
• 2≤ai≤109
• ai is an even number.
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

M
aN

Output

Print the number of semi-common multiples of M (inclusive).

 

X=a_i*(p+0.5)可以看作是X=a_i/2*(2p+1)，令b_i=a_i/2的话，那么显然X满足X%b_i==0且X/b_i是个奇数。

令g=lcm(b₁,b₂...b_n)的话，满足条件的数只可能是k*g，假设g不满足题意那么k*g也不会满足题意（g不满足，即存在一个b使得g/b是偶数，那显然k*g/b也是偶数，也不符合）。

假设g满足题意那么只有在k为奇数得情况下k*g满足题意，这时候统计下小于等于M且满足上述条件的数即可。

还要注意求lcm得过程可能会爆LL，为此只要当前的lcm>M那么答案显然就是零了，特判下。

这种数学规律题神烦，代码很丑= =

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 
 5 LL gcd(LL a,LL b){return b==0?a:gcd(b,a%b);}
 6 LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
 7 LL N,M,a[100010];
 8 int main(){
 9     cin>>N>>M;
10     LL g=1,ok=1;
11     for(int i=1;i<=N;++i){
12         cin>>a[i];
13         a[i]/=2;
14         if(i==1)g=a[i];
15         g=lcm(g,a[i]);
16         if(g>M)ok=0;
17     }
18     for(int i=1;i<=N;++i){
19         if(g/a[i]%2==0){ok=0;break;}
20     }
21     if(!ok){
22         cout<<0<<endl;
23         return 0;
24     }
25     LL ans=0;
26     LL b=M/g;
27     ans=b/2;
28     if(b&1)ans++;
29     cout<<ans<<endl;
30     return 0;
31 }