Submission #116747

#	Time	Username	Problem	Language	Result	Execution time	Memory
116747		JohnTitor	Calvinball championship (CEOI15_teams)	C++11	100 / 100	926 ms	632 KiB

teams

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, j, k) for(int i=(j); i<=(k); i++)
#define FFOR(i, j, k) for(int i=(j); i<(k); i++)
#define DFOR(i, j, k) for(int i=(j); i>=(k); i--)
#define bug(x) cerr<<#x<<" = "<<(x)<<'\n'
#define pb push_back
#define mp make_pair
#define bit(s, i) (((s)>>(i))&1LL)
#define mask(i) ((1LL<<(i)))
#define builtin_popcount __builtin_popcountll
#define __builtin_popcount __builtin_popcountll
using ll=long long; using ld=long double;
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count()); const ld pi=acos(0)*2;
template <typename T> inline void read(T &x){char c; bool nega=0; while((!isdigit(c=getchar()))&&(c!='-')); if(c=='-'){nega=1; c=getchar();} x=c-48; while(isdigit(c=getchar())) x=x*10+c-48; if(nega) x=-x;}
template <typename T> inline void writep(T x){if(x>9) writep(x/10); putchar(x%10+48);}
template <typename T> inline void write(T x){if(x<0){ putchar('-'); x=-x;} writep(x);}
template <typename T> inline void writeln(T x){write(x); putchar('\n');}
#define taskname "Calvinball"
int n;
const ll base=1000007;
ll f[10002];
ll g[10002];
int main(){
    #ifdef Aria
        if(fopen(taskname".in", "r"))
            freopen(taskname".in", "r", stdin);
    #endif // Aria
    read(n);
    FOR(i, 1, n) f[i]=1;
    FOR(i, 1, n){
        FOR(j, 0, n) f[j]=(f[j]*j+f[j+1])%base;
        g[i]=f[0];
    }
    int team=0, a;
    ll ans=1;
    FOR(i, 1, n){
        ll oldj=f[0];
        FOR(j, 1, n){
            ll temp=f[j];
            f[j]=(oldj-f[j-1]*(j-1))%base;
            oldj=temp;
        }
        f[0]=g[n-i];
        read(a);
        ans+=(a-1)*f[team];
        team=max(team, a);
    }
    ans%=base;
    if(ans<0) ans+=base;
    writeln(ans);
}
//f[i][j] -> have i more people and now have j team
///f[i][j]=f[i-1][j]*j+f[i-1][j+1];
///f[i+1][j]=f[i][j]*j+f[i][j+1];
///f[i+1][j-1]=f[i][j-1]*(j-1)+f[i][j];
///f[i][j]=f[i+1][j-1]-f[i][j-1]*(j-1);

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