oj.uz

답안 #23299

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
23299 2017-05-06T04:00:41 Z duongthoi1999 Skyscraper (JOI16_skyscraper) C++14
15 / 100
 0 ms 4764 KB 
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long ll;
typedef long double ld;
typedef pair<ll, int> ii;
const int mod = (int) 1e9 + 7;
const ll inf = 1LL << 60;
const int maxn = (int) 105;
const ld eps = 1e-9;

void add(int &a, ll b) {
    b %= mod;
    a += b;
    while (a >= mod) a -= mod;
}

int n, L, a[maxn];
int f[2][maxn][maxn * 10][3];

int main() {
    //freopen("test.txt", "r", stdin);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> L;
    rep(i, 1, n + 1) cin >> a[i];
    sort(a + 1, a + 1 + n);
    a[n + 1] = 12345;
    if(a[2] - a[1] <= L) f[1][1][a[2] - a[1]][1] = 2;
    if(2 * (a[2] - a[1]) <= L) f[1][1][2 * (a[2] - a[1])][0] = 1;

    rep(i, 1, n) {
        int d = a[i + 2] - a[i + 1];
        rep(j, 1, i + 1) {
            rep(k, 0, L + 1) {
                rep(e, 0, min(j + 1, 3)) {
                    if(!f[i & 1][j][k][e]) continue;
                    if(e < 2) {
                        if(k + (2 * (j + 1) - (e + 1)) * d <= L)
                            add(f[(i + 1) & 1][j + 1][k + (2 * (j + 1) - (e + 1)) * d][e + 1], (2 - e) * f[i & 1][j][k][e]);
                        if(k + (2 * j - (e + 1)) * d <= L)
                            add(f[(i + 1) & 1][j][k + (2 * j - (e + 1)) * d][e + 1], (2 - e) * f[i & 1][j][k][e]);
                    }
                    if(k + (2 * (j + 1) - e) * d <= L)
                        add(f[(i + 1) & 1][j + 1][k + (2 * (j + 1) - e) * d][e], 1LL * (j - e + 1) * f[i & 1][j][k][e]);
                    if(k + (2 * j - e) * d <= L)
                        add(f[(i + 1) & 1][j][k + (2 * j - e) * d][e], 1LL * (2 * j - e) * f[i & 1][j][k][e]);
                    if(j > 1 && k + (2 * (j - 1) - e) * d <= L)
                        add(f[(i + 1) & 1][j - 1][k + (2 * (j - 1) - e) * d][e], 1LL * (j - 1) * f[i & 1][j][k][e]);
                }
            }
        }
        memset(f[i & 1], 0, sizeof f[i & 1]);
    }
    int res = 0;
    rep(i, 0, L + 1) add(res, f[n & 1][1][i][2]);
    cout << res << endl;
}

Subtask #1
0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 0 ms 4764 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
15.0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 4764 KB Output is correct
2 Correct 0 ms 4764 KB Output is correct
3 Correct 0 ms 4764 KB Output is correct
4 Correct 0 ms 4764 KB Output is correct
5 Correct 0 ms 4764 KB Output is correct
6 Correct 0 ms 4764 KB Output is correct
7 Correct 0 ms 4764 KB Output is correct
8 Correct 0 ms 4764 KB Output is correct
9 Correct 0 ms 4764 KB Output is correct
10 Correct 0 ms 4764 KB Output is correct

Subtask #3
0 / 80.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 0 ms 4764 KB Output isn't correct
2 Halted 0 ms 0 KB -