涂抹果酱

涂抹果酱

 

 题目分析：这道题跟上一道题有点像，不过这题有三种，所以想到三进制，而不是二进制了，然后把1,2,3化成0,1,2；然后，相应的比较部分换一下就好了

AC_Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e6;
const int maxn = 1e4 + 10;

int n, m, k;
int maxx, tot, sta[250], zk, pos, ans;
int f[10100][250];

bool check(int x) {
    int tmp = 0x3f;
    for (int i = 1; i <= m; i++) {
        if (tmp == x % 3)
            return false;
        tmp = x % 3;
        x /= 3;
    }
    return true;
}

bool judge(int u, int v) {
    for (int i = 1; i <= m; i++) {
        if (u % 3 == v % 3)
            return false;
        u /= 3;
        v /= 3;
    }
    return true;
}

int qpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1)
            res *= a;
        a *= a;
        b >>= 1;
    }
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++) {
        int x;
        cin >> x;
        zk = zk * 3 + (x - 1);
    }

    maxx = qpow(3, m);
    for (int i = 0; i < maxx; i++) {
        if (check(i))
            sta[++tot] = i;
    }
    for (int i = 1; i <= tot; i++) {
        if (zk == sta[i]) {
            pos = i;
            break;
        }
    }
    if (!pos) {  //第k行本身就不符合条件
        cout << "0" << '
';
        return 0;
    }
    for (int i = 1; i <= n; i++) {
        if (i == k) {  //此行状态已经固定
            if (i == 1)
                f[i][pos] = 1;
            else {
                for (int j = 1; j <= tot; j++) {
                    if (judge(sta[pos], sta[j])) {
                        f[i][pos] = (f[i][pos] + f[i - 1][j]) % mod;
                    }
                }
            }
        } else {
            for (int j = 1; j <= tot; j++) {  //此行状态没有固定,所以多一重循环
                if (i == 1)
                    f[i][j] = 1;
                else {
                    for (int k = 1; k <= tot; k++) {
                        if (judge(sta[k], sta[j])) {
                            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
                        }
                    }
                }
            }
        }
    }
    for (int i = 1; i <= tot; i++) ans = (ans + f[n][i]) % mod;
    cout << ans << '
';
    return 0;
}