답안 #52484

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
52484 2018-06-26T05:44:04 Z 노영훈(#1364) 비밀 (JOI14_secret) C++11
0 / 100
697 ms 16392 KB
#include "secret.h"
#include <bits/stdc++.h>
using namespace std;

const int MX=1001;

int n, A[MX];

vector<int> P;

// p <= x < q, p < y <= q
int D[MX][MX]; // [x, y)
int E[MX][MX]; // [x, q)
int F[MX][MX]; // [p, y)

void filld(){
    for(int i=0; i<(int)P.size()-1; i++){
        int p=P[i], q=P[i+1];
        for(int x=p; x<q; x++){
            D[x][x+1]=A[x];
            for(int y=x+2; y<=q; y++)
                D[x][y]=Secret(D[x][y-1], A[y-1]);
        }
    }
    /*
    for(int i=0; i<n; i++){
        D[i][i+1]=A[i];
        for(int j=i+2; j<=n; j++){
            D[i][j]=Secret(D[i][j-1], A[j-1]);
        }
    }
    for(int i=0; i<n; i++){
        for(int j=i+1; j<=n; j++){
            cout<<D[i][j]<<' ';
        }
        cout<<'\n';
    }
    */
}

void fille(){
    for(int i=1; i<(int)P.size(); i++){
        int p=P[i-1], q=P[i];
        E[q-1][q]=A[q-1];
        for(int x=q-2; x>=p; x--){
            E[x][q]=Secret(A[x], E[x+1][q]);
        }
    }
    for(int i=0; i<(int)P.size(); i+=2){
        int q=P[i];
        E[q-1][q]=A[q-1];
        for(int x=q-2; x>=0; x--){
            E[x][q]=Secret(A[x], E[x+1][q]);
        }
        if(q>n/2) break;
    }
}

void fillf(){
    for(int i=0; i<(int)P.size()-1; i++){
        int p=P[i], q=P[i+1];
        F[p][p+1]=A[p];
        for(int y=p+2; y<=q; y++)
            F[p][y]=Secret(F[p][y-1], A[y]);
    }
    for(int i=(int)P.size()-1; i>=0; i-=2){
        int p=P[i];
        F[p][p+1]=A[p];
        for(int y=p+2; y<=n; y++)
            F[p][y]=Secret(F[p][y-1], A[y-1]);
        if(p<=n/2) break;
    }
}

void Init(int N, int _A[]) {
    n=N;
    for(int i=0; i<n; i++) A[i]=_A[i];
    for(int p=0; p<n; p+=32) P.push_back(p);
    P.push_back(n);

    for(int i=0; i<n; i++)
        for(int j=0; j<=n; j++)
            D[i][j]=E[i][j]=F[i][j]=-1;

    filld();
    fille();
    fillf();
}

int Query(int L, int R) {
    int l=L, r=R+1;
    for(int p:P){
        // [l, p), [p, r)
        if(l<p && p<r && E[l][p]>=0 && F[p][r]>=0)
            return Secret(E[l][p], F[p][r]);
        if(l<p && p<r && D[l][p]>=0 && D[p][r]>=0)
            return Secret(D[l][p], D[p][r]);
    }
    return D[l][r];
}
# 결과 실행 시간 메모리 Grader output
1 Incorrect 173 ms 8440 KB Wrong Answer: Query(222, 254) - expected : 34031541, actual : 860882111.
2 Incorrect 172 ms 8444 KB Wrong Answer: Query(102, 157) - expected : 32612619, actual : 794329227.
3 Incorrect 168 ms 8764 KB Wrong Answer: Query(111, 503) - expected : 353540466, actual : -1.
4 Incorrect 597 ms 16204 KB Wrong Answer: Query(73, 368) - expected : 75711012, actual : -1.
5 Incorrect 645 ms 16204 KB Wrong Answer: Query(587, 915) - expected : 752404486, actual : -1.
6 Incorrect 654 ms 16376 KB Wrong Answer: Query(250, 258) - expected : 907404, actual : 297015796.
7 Incorrect 661 ms 16376 KB Output isn't correct - number of calls to Secret by Init = 22291, maximum number of calls to Secret by Query = 1
8 Incorrect 609 ms 16376 KB Output isn't correct - number of calls to Secret by Init = 22291, maximum number of calls to Secret by Query = 1
9 Incorrect 638 ms 16392 KB Output isn't correct - number of calls to Secret by Init = 22291, maximum number of calls to Secret by Query = 1
10 Incorrect 697 ms 16392 KB Output isn't correct - number of calls to Secret by Init = 22291, maximum number of calls to Secret by Query = 1