noip5

11.10

为什么noip模拟赛从5开始？

前面的不想写（懒）。

---

分了个div1/2

不是你题目难度也不对应啊?

div2版

t1

抽象状压。

赛后帮Gon_Tata hack 他的假状压，获得金牌辅助。

首先\(\ldots\)

然后\(\ldots\)

最后\(\ldots\)

懒。

直接看吧

code

。。。

#include <bits/stdc++.h>
using namespace std;
int n, m;
int dp[110][260];
char c[110][10];

signed main()
{
    freopen("merging.in", "r", stdin);
    freopen("merging.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            cin >> c[i][j];
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    int ans = 0x3f3f3f3f;
    for (int i = 1, tmp, cnt, len; i <= n; ++i)
    {
        tmp = len = 0;
        for (int j = 1; j <= m; ++j)
        {
            if (c[i][j] == '1' && c[i][j - 1] != '1')
                tmp |= 1 << (j - 1);
            else if (c[i][j] == '1')
                ++len;
        }
        int now;
        for (int k = 0; k < 1 << len; ++k)
        {
            now = tmp;
            cnt = 0;
            for (int j = 1; j <= m; ++j)
            {
                if (c[i][j] == '1' && !(1 & (tmp >> (j - 1))))
                    now |= (1 & (k >> (cnt++))) << (j - 1);
            }
            cnt = __builtin_popcount(now);
            for (int o = 0; o < 1 << m; ++o)
            {
                if (dp[i - 1][o] > 2000)
                    continue;
                int res = cnt;
                int flag = 0;
                for (int p = 1; p <= m; ++p)
                {
                    if ((c[i][p] == '0' || 1 & now >> (p - 1)) && (c[i - 1][p] == '0' || 1 & o >> (p - 1)))
                    {
                        res -= flag;
                        if (1 & now >> (p - 1) & 1 & o >> (p - 1))
                            flag = 1;
                    }
                    if (c[i][p] == '0' || c[i - 1][p] == '0' || (1 & now >> (p - 1)) ^ (1 & o >> (p - 1)))
                        flag = 0;
                }
                res -= flag;
                dp[i][now] = min(dp[i][now], dp[i - 1][o] + res);
                ans = min(ans, dp[n][now]);
            }
        }
    }
    cout << ans << "\n";
    return 0;
}

t2

水。

一眼二分答案然后文件名写错？？？

应该上午太困了导致的（中间差点睡着）

代码没必要放

t3

巴巴博弈。

性质比较好想。

所以只看性质吧（

code

巴巴博弈

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 5e4 + 10;
int n, k, sum, cnt;
int a[N];

inline void solve1()
{
    if (sum % 2 == 0)
    {
        for (int i = 1; i <= n; ++i)
            if (a[i] >= 2)
                cout << i << ' ' << 2 << "\n";
    }
    else
    {
        for (int i = 1; i <= n; ++i)
        {
            cout << i << ' ' << 1 << "\n";
            if (a[i] == 3)
                if ((cnt - 2) & 1)
                    cout << i << ' ' << 3 << "\n";
        }
    }
    exit(0);
}

inline void solve2()
{
    if (sum % 2 == 0)
    {
        for (int i = 1; i <= n; ++i)
            if (a[i] >= 2)
                cout << i << ' ' << 2 << "\n";
    }
    else
    {
        for (int i = 1; i <= n; ++i)
        {
            cout << i << ' ' << 1 << "\n";
            if (a[i] >= 3)
            {
                if (a[i] % 2 == 0 && (cnt - 2) % 2 == 0)
                    cout << i << ' ' << 3 << "\n";
                else if ((a[i] & 1) && (cnt - 1) % 2 == 0)
                    cout << i << ' ' << 3 << "\n";
            }
        }
    }
    exit(0);
}

signed main()
{
    freopen("nim.in", "r", stdin);
    freopen("nim.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> k;
    bool f = 0;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        a[i] > 3 ? f = 1 : 0;
        sum += a[i];
        cnt += a[i] >> 1;
    }
    if (sum % 2 == 0)
    {
        if (cnt % 2 == 0)
        {
            cout << 0;
            exit(0);
        }
    }
    cout << "1\n";
    if (!f)
        solve1();
    if (k <= 3)
        solve2();
    return 0;
}

t4

性质： border的boeder也是border

于是border构成树型关系，建出树即失配树（KMP求nxt以建树）。

树型dp即可。

code

border

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6 + 10;
const int mod = 998244353;
int n, k;
string s;
int jc[N], inv[N], nxt[N], dep[N], siz[N], dp[N];
vector<int> e[N];

inline int km(int a, int b)
{
    int ans = 1;
    while (b)
    {
        if (b & 1)
            (ans *= a) %= mod;
        (a *= a) %= mod;
        b >>= 1;
    }
    return ans;
}

inline int C(int a, int b)
{
    if (a < b)
        return 0;
    return jc[a] * inv[b] % mod * inv[a - b] % mod;
}

inline void get_nxt()
{
    nxt[1] = 0;
    for (int i = 2, j = 0; i <= n; ++i)
    {
        while (j && s[i] != s[j + 1])
            j = nxt[j];
        if (s[i] == s[j + 1])
            ++j;
        nxt[i] = j;
        e[j].emplace_back(i);
    }
}

inline void dfs(int x, int f)
{
    siz[x] = 1, dep[x] = dep[f] + 1;
    for (auto y : e[x])
    {
        if (y == f)
            continue;
        dfs(y, x);
        dp[x] -= dep[x] * dep[x] % mod * C(siz[y], k) % mod;
        // (ans += mod) %= mod;
        // (ans -=) %= mod;
        siz[x] += siz[y];
    }
    (dp[x] += dep[x] * dep[x] % mod * C(siz[x], k)) %= mod;
    // cout << "x=" << x << " ans=" << ans << "\n";
}

signed main()
{
    freopen("string.in", "r", stdin);
    freopen("string.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> k;
    cin >> s;
    n = s.size();
    s = " " + s;
    jc[0] = 1, inv[0] = 1;
    for (int i = 1; i <= n + 1; ++i)
        jc[i] = jc[i - 1] * i % mod;
    inv[n + 1] = km(jc[n + 1], mod - 2);
    for (int i = n; i; --i)
        inv[i] = inv[i + 1] * (i + 1) % mod;

    get_nxt();
    e[0].emplace_back(1);
    dfs(0, n + 2);
    int ans = 0;
    for (int i = 0; i <= n; ++i)
        (ans += dp[i]) %= mod;
    cout << ans;
    return 0;
}
/*
多个串的border相关问题:
    建fail树，树上做（若最长公共border长为lca）
*/

解释的很少，但其实每道题细讲都不少（除了t2）。

大抵是累了。

或许马上就要退役了？

不知道。