Submission #527101

#	Time	Username	Problem	Language	Result	Execution time	Memory
527101		zaneyu	Calvinball championship (CEOI15_teams)	C++14	10 / 100	24 ms	23628 KiB

teams

/*input
3
1 2 2
*/ 
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

//#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e17;
const int INF=0x3f3f3f3f;
const ll MOD=1e6+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]+1<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=1e3+5;
ll dp[maxn][maxn][2];
int arr[maxn];
int main(){
    int n;
    cin>>n;
    REP1(i,n) cin>>arr[i];
    dp[0][0][1]=1;
   	REP1(i,n){
   		REP1(j,arr[i]){
   			if(j==arr[i]){
   				dp[i][j][1]=dp[i-1][j][1]+dp[i-1][j-1][1];
   				dp[i][j][0]=dp[i-1][j][0]*j+dp[i-1][j][1]*(j-1)+dp[i-1][j-1][0];
   			}
   			else{
   				dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1])*j+(dp[i-1][j-1][0]+dp[i-1][j-1][1]);
   			}
   			REP(k,2) dp[i][j][k]%=MOD;
   		}
    }
    ll ans=1;
    REP1(j,arr[n]) ans+=dp[n][j][0],ans%=MOD;
    cout<<ans;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 10 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10