数组模拟链表

双向链表的删除操作

#include<bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define pw(x) (1ll << (x))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rep(i,l,r) for(int i=(l);i<(r);i++)
#define per(i,r,l) for(int i=(r);i>=(l);i--)
#define FOR(i,l,r) for(int i=(l);i<=(r);i++)
#define eps 1e-9
#define PIE acos(-1)
#define cl(a,b) memset(a,b,sizeof(a))
#define fastio ios::sync_with_stdio(false);cin.tie(0);
#define lson l , mid , ls
#define rson mid + 1 , r , rs
#define ls (rt<<1)
#define rs (ls|1)
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define lowbit(x) (x&(-x))
#define sqr(a) a*a
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int, int>
#define dd(x) cout << #x << " = " << (x) << ", "
#define de(x) cout << #x << " = " << (x) << "
"
#define endl "
"
using namespace std;
//**********************************

//**********************************
const int maxn=5e3+7;
int le[maxn],ri[maxn],a[maxn];
inline void link(int l,int r)
{
    ri[l]=r;le[r]=l;
}
//**********************************
int main()
{
    int n,m,t,cnt,n1;
    scanf("%d %d",&n,&m);n1=n;
    cl(le,-1);cl(ri,-1);
    FOR(i,1,n)scanf("%d",&a[i]),ri[i-1]=i,le[i]=i-1;
    FOR(i,1,m){
        scanf("%d",&t);
        cnt=0;
        for(int i=ri[0];~i&&cnt<=n;i=ri[i]){
            cnt++;
//            de(cnt);
            if(cnt%t==0)link(le[i],ri[i]),n1--;
        }
    }
    printf("%d
",n1);
    cnt=0;
    for(int i=ri[0];~i;i=ri[i])printf("%d%c",a[i],++cnt==n1?'
':' ');
    return 0;
}