The 2019 ACM-ICPC China Shannxi Provincial Programming Contest-L.Swap

https://nanti.jisuanke.com/t/39279

题意：给一个数组操作1：前后部分调换(n为偶数中间的不变)；操作2：偶数位置与前面位置的交换(n为奇数最后一个不变)

思路：打表找规律，n偶数时如果是4倍数就是4，不是就是数组长度；n为奇数时发现每个一个奇数是12，其余是2*n;

#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)const int maxn=1e6+9;
const int mod=1e9+7;inline ll read()
{ll f=1,x=0;char ss=getchar();while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}    return f*x;
}/*map<vector<int>,bool> mp;
ll cnt=0;
int n;
void dfs(vector<int> s,int len)
{if(mp[s]) return ;//for(int i=0;i<len;i++) cout<<s[i]<<" ";//cout<<endl;mp[s]=1;cnt++;vector<int> temp=s;int mid=(len+1)/2;for(int i=0;i<len/2;i++){swap(temp[i],temp[mid+i]);}//cout<<temp<<endl;dfs(temp,len);temp.clear();temp=s;for(int i=1;i<len;i+=2){swap(temp[i],temp[i-1]);}//cout<<temp<<endl;dfs(temp,len);
}*/
int main()
{//FAST_IO;//freopen("input.txt","r",stdin);ll n;cin>>n;//vector<int> s;//mp.clear();for(int i=0;i<n;i++){string x;cin>>x;//s.pb(x);//s.pb(i);}if(n==1){cout<<1<<endl;}else if(n==2){cout<<2<<endl;}else if(n==3){cout<<6<<endl;}else if(n%2==1){if((n-5)%4==0)cout<<n*2<<endl;else cout<<12<<endl;}else{if(n%4==0) cout<<4<<endl;else cout<<n<<endl;}/*for(int i=2;i<=200;i++){int n=i;vector<int> s;for(int j=0;j<n;j++) s.pb(j+1);if(n<=2)cout<<n<<endl;else {cnt=0;dfs(s,s.size());if(i%2==1)cout<<i<<":"<<cnt<<endl<<endl;}}*/return 0;
}

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