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 : $300300$ points

Problem Statement

We have two permutations $\mathrm{PP}$ and $\mathrm{QQ}$ of size $\mathrm{NN}$ (that is, $\mathrm{PP}$ and $\mathrm{QQ}$ are both rearrangements of $(1, 2, ..., N)(1, 2, ..., N)$).

There are $\mathrm{N!N!}$ possible permutations of size $\mathrm{NN}$. Among them, let $\mathrm{PP}$ and $\mathrm{QQ}$ be the $\mathrm{aa}$-th and $\mathrm{bb}$-th lexicographically smallest permutations, respectively. Find $|a-b||a-b|$.

Notes

For two sequences $\mathrm{XX}$ and $\mathrm{YY}$, $\mathrm{XX}$ is said to be lexicographically smaller than $\mathrm{YY}$ if and only if there exists an integer $\mathrm{kk}$ such that ${\mathrm{Xi=Yi (1\le i<k)Xi=Yi (1\le i<k)}}^{}$ and ${\mathrm{Xk<YkXk<Yk}}^{}$.

Constraints

• $\mathrm{2\le N\le 82\le N\le 8}$
• $\mathrm{PP}$ and $\mathrm{QQ}$ are permutations of size $\mathrm{NN}$.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{P1P1}}^{}$ ${\mathrm{P2P2}}^{}$ $......$ ${\mathrm{PNPN}}^{}$
${\mathrm{Q1Q1}}^{}$ ${\mathrm{Q2Q2}}^{}$ $......$ ${\mathrm{QNQN}}^{}$

Output

Print $|a-b||a-b|$.

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

#include<bits/stdc++.h>
using namespace std;
#define LL long long

int jc[10]={1};
int a[10],b[10],n;
int cal(int a[],int n){
    int res=0;
    for(int i=1;i<=n;++i){
        int c=0;
        for(int j=i+1;j<=n;++j)c+=(a[j]<a[i]);
        res+=c*jc[n-i];
    }return res;
}
int main(){
    for(int i=1;i<=9;++i)jc[i]=i*jc[i-1];
    cin>>n;
    for(int i=1;i<=n;++i)cin>>a[i];
    for(int i=1;i<=n;++i)cin>>b[i];
    cout<<abs(cal(b,n)-cal(a,n))<<endl;
    return 0;
}

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 $\mathrm{A=a1,a2,......aNA=a1,a2,......aN}$ of $\mathrm{NN}$ positive even numbers, and an integer $\mathrm{MM}$.

Let a semi-common multiple of $\mathrm{AA}$ be a positive integer $\mathrm{XX}$ that satisfies the following condition for every $\mathrm{kk}$ $(1\le k\le N)(1\le k\le N)$:

• There exists a non-negative integer $\mathrm{pp}$ such that $\mathrm{X=ak\times (p+0.5)X=ak\times (p+0.5)}$.

Find the number of semi-common multiples of $\mathrm{AA}$ among the integers between $11$ and $\mathrm{MM}$ (inclusive).

Constraints

• $\mathrm{1\le N\le 1051\le N\le 105}$
• $\mathrm{1\le M\le 1091\le M\le 109}$
• $\mathrm{2\le ai\le 1092\le ai\le 109}$
• ${\mathrm{aiai}}^{}$ 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:

$\mathrm{NN}$ $\mathrm{MM}$
${\mathrm{a1a1}}^{}$ ${\mathrm{a2a2}}^{}$ $......$ ${\mathrm{aNaN}}^{}$

Output

Print the number of semi-common multiples of $\mathrm{AA}$ among the integers between $11$ and $\mathrm{MM}$ (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= =

#include<bits/stdc++.h>
using namespace std;
#define LL long long

LL gcd(LL a,LL b){return b==0?a:gcd(b,a%b);}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL N,M,a[100010];
int main(){
    cin>>N>>M;
    LL g=1,ok=1;
    for(int i=1;i<=N;++i){
        cin>>a[i];
        a[i]/=2;
        if(i==1)g=a[i];
        g=lcm(g,a[i]);
        if(g>M)ok=0;
    }
    for(int i=1;i<=N;++i){
        if(g/a[i]%2==0){ok=0;break;}
    }
    if(!ok){
        cout<<0<<endl;
        return 0;
    }
    LL ans=0;
    LL b=M/g;
    ans=b/2;
    if(b&1)ans++;
    cout<<ans<<endl;
    return 0;
}