Submission #967956

#	Time	Username	Problem	Language	Result	Execution time	Memory
967956		Darren0724	Binaria (CCO23_day1problem1)	C++17	0 / 25	109 ms	39588 KiB

Main

#include <bits/stdc++.h>
using namespace std;
#define all(x) x.begin(),x.end()
#define ll long long 
#define LCBorz ios_base::sync_with_stdio(false);cin.tie(0);
const int N=1000005;
const int mod=1e6+3;
vector<ll> fac(N),inv(N);
ll pw(ll a,int b){
    ll ans=1;
    while(b>0){
        if(b&1)ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
int32_t main(){
    LCBorz;
    fac[0]=inv[0]=1;
    for(int i=1;i<N;i++){
        fac[i]=fac[i-1]*i%mod;
        inv[i]=pw(fac[i],mod-2);
    }
    int n,m;cin>>n>>m;
    int p=n-m+1;
    vector<int> v(p);
    for(int i=0;i<p;i++){
        cin>>v[i];
        if(v[i]>m){
            cout<<0<<endl;
            return 0;
        }
    }
    vector<int> ans(n,-1);
    auto upd=[&](int i,int j){
        if(ans[i]==-1){
            ans[i]=j;
        }
        else if(ans[i]!=j){
            cout<<0<<endl;
            exit(0);
        }
    };
    vector<int> adj[N];
    auto dfs=[&](auto dfs1,int k,int pa,int val)->void {
        upd(k,val);
        for(int j:adj[k]){
            if(j==pa)continue;
            dfs1(dfs1,j,k,val);
        }
    }; 
    for(int i=0;i<p-1;i++){
        if(abs(v[i]-v[i+1])>1){
            cout<<0<<endl;
            return 0;
        }
        if(v[i]==v[i+1]+1){
            upd(i,1);
            upd(i+m,0);
        }
        else if(v[i]==v[i+1]-1){
            upd(i,0);
            upd(i+m,1);
        }
        else{
            adj[i].push_back(i+m);
            adj[i+m].push_back(i);
        }
    }
    for(int i=0;i<n;i++){
        if(adj[i].size()==1){
            dfs(dfs,i,i,ans[i]);
        }
    }
    int a=0,b=v[0];
    for(int i=0;i<m;i++){
        if(ans[i]==-1){
            a++;
        }
        if(ans[i]==1){
            b--;
        }
    }
    if(b>a||b<0){
        cout<<0<<endl;
        return 0;
    }
    cout<<fac[a]*inv[b]%mod*inv[a-b]%mod<<endl;

    return 0;
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 3
• Subtask 3: 0 / 4
• Subtask 4: 0 / 4
• Subtask 5: 0 / 4
• Subtask 6: 0 / 7