Contest1900 - 2019 Nian 6 Yue multi-layer test a joint training school 1

Portal  Password: waxadyt

T1

Hit the table to find the law

Starting from n> = 5, the difference between two adjacent numbers there is a tolerance of 28 arithmetic sequence

Directly to the bar code

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        unsigned long long n;
        cin>>n;
        if(n==0){printf("1\n");continue;}
        if(n==1){printf("9\n");continue;}
        if(n==2){printf("41\n");continue;}
        if(n==3){printf("109\n");continue;}
        if(n==4){printf("205\n");continue;}
        if(n==5){printf("325\n");continue;}
        cout<<325+(148+(n-6)*14)*(n-5);
    }
}

T1

T2

dfs bare title

Plus a vis

Code

#include<bits/stdc++.h>
using namespace std;
int vis[(1<<11)+1];
int a[1001001],n,p;
void dfs(int pos)
{
    if(p)return;
    if(pos==(1<<n)+1-n)
    {
        for(int i=1;i<=(1<<n);i++)
        {
            printf("%d",a[i]);
        }
        p= 1 ;
        return ; 
    } 
    Int num = 0 ;
    for ( int i = pos-m + 1 ; i <= pos; i ++) whether = (whether << 1 ) + a [i];
    if (! force [whether]) 
    { 
        from [pos] = 0 ; 
        force [num] = 1 ; 
        DFS (pos + 1 ); 
        force [num] = 0 ; 
    } 
    If (b) return ; 
    force [num + 1 ] =1;
    a[pos]=1;dfs(pos+1);
}
int main()
{
    scanf("%d",&n);
    for(int i=(1<<n)-n+1;i<=1<<n;i++)a[i]=1;
    vis[0]=1;
    int num=(1<<n)-1;
    for(int i=0;i<n;i++)
    {
        vis[num]=1;
        num^=(1<<i);
    }
    printf("%d ",1<<n);
    dfs(n+1);
}

T2

T3

Leave a pit