2019 Multi-University Training Contest 5

2019 Multi-University Training Contest 5

Problem 6624~6633

A.fraction

先解决这样一个问题：找到最小的\(x,y\)，求解\(\frac{a}{b}\leq \frac{x}{y}\leq \frac{c}{d}\)。
这个问题可以递归求解：
若\(\lceil\frac{a}{b}\rceil\leq \lfloor\frac{c}{d}\rfloor\)，直接令\(x=\lceil\frac{a}{b}\rceil,y=1\)即可；
否则，令式子减去左边的整数部分，然后取倒数递归即可，即：
令\(t=\lceil\frac{a}{b}\rceil\),有

\[\frac{a-(t-1)*b}{b}\leq \frac{x-(t-1)*y}{y}\leq \frac{c-(t-1)*d}{d} \]

取反后即为：

\[\frac{c-(t-1)*d}{d}\leq \frac{y}{x-(t-1)*y}\leq \frac{b}{a-(t-1)*b} \]

之后递归求解即可。
代码如下：

void gao(ll a, ll b, ll c, ll d, ll &x, ll &y) { // a/b < x/y < c/d
    ll t = (a + b - 1) / b;
    if(c / d >= t) {
        y = 1; x = t;
        return;
    }
    a -= (t - 1) * b;
    c -= (t - 1) * d;
    gao(d, c, b, a, y, x);
    x += (t - 1) * y;
}

递归中传入的是\(y,x\)，也就是说有：

\[x'=x-(t-1)*y,y'=y \]

即

\[x=x'+(t-1)*y,y=y' \]

所以这个问题就解决了。
回到此题，题目要求\(a=bx\mod p\)，变换一下即有：\(a=bx-py\)。
根据题目条件:\(0<a<b\)，式子即可化为:

\[\frac{p}{x}<\frac{b}{y}<\frac{p}{x-1} \]

之后问题就转化为刚才的问题了。
代码如下：

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;
int T;
void gao(ll a, ll b, ll c, ll d, ll &x, ll &y) { // a/b < x/y < c/d
    ll t = (a + b - 1) / b;
    if(c / d >= t) {
        y = 1; x = t;
        return;
    }
    a -= (t - 1) * b;
    c -= (t - 1) * d;
    gao(d, c, b, a, y, x);
    x += (t - 1) * y;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> T;
    ll p, x;
    while(T--) {
        cin >> p >> x;
        ll b, y;
        gao(p, x, p, x - 1, b, y);
        ll a = b * x - p * y;
        cout << a << '/' << b << '\n';
    }
    return 0;
}

B.three arrays

考虑分别对\(a,b\)数组建两颗字典树，然后贪心取最长相同前缀即可。
题解的做法比较神奇，就贪心找一条\(a->b->a\cdots\)这样的链，每个数选择一个异或最小的连边。可以证明最终会形成长度不超过\(2\)的环，并且这样的一对是最恩爱优的。
所以找到环后将它们取出来继续重复上面过程即可。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int T, n;
int a[N], b[N];
struct Trie{
    int ch[N * 30][2], cnt[N * 30];
    int tot;
    void init() {
        tot = 0;
        memset(ch[tot], 0, sizeof(ch[tot]));
    }
    int New_node() {
        memset(ch[++tot], 0, sizeof(ch[tot]));
        return tot;
    }
    void Insert(int x) {
        int u = 0;
        for(int i = 29; i >= 0; i--) {
            int p = ((x >> i & 1) ? 1 : 0);
            if(!ch[u][p]) ch[u][p] = New_node();
            u = ch[u][p];
            cnt[u]++;
        }
    }
}t1, t2;
vector <int> work() {
    int u1, u2;
    vector <int> ans;
    while(n--) {
        u1 = u2 = 0;
        int x = 0;
        for(int p = 29; p >= 0; p--) {
            if(t1.cnt[t1.ch[u1][0]] && t2.cnt[t2.ch[u2][0]]) {
                t1.cnt[t1.ch[u1][0]]--;
                t2.cnt[t2.ch[u2][0]]--;
                u1 = t1.ch[u1][0]; u2 = t2.ch[u2][0];
            } else if(t1.cnt[t1.ch[u1][1]] && t2.cnt[t2.ch[u2][1]]) {
                t1.cnt[t1.ch[u1][1]]--;
                t2.cnt[t2.ch[u2][1]]--;
                u1 = t1.ch[u1][1]; u2 = t2.ch[u2][1];
            } else if(t1.cnt[t1.ch[u1][0]] && t2.cnt[t2.ch[u2][1]]) {
                t1.cnt[t1.ch[u1][0]]--;
                t2.cnt[t2.ch[u2][1]]--;
                u1 = t1.ch[u1][0]; u2 = t2.ch[u2][1];
                x ^= (1 << p);
            } else {
                t1.cnt[t1.ch[u1][1]]--;
                t2.cnt[t2.ch[u2][0]]--;
                u1 = t1.ch[u1][1]; u2 = t2.ch[u2][0];
                x ^= (1 << p);
            }
        }
        ans.push_back(x);
    }
    return ans;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> T;
    while(T--) {
        cin >> n;
        for(int i = 1; i <= n; i++) cin >> a[i];
        for(int i = 1; i <= n; i++) cin >> b[i];
        t1.init(); t2.init();
        for(int i = 1; i <= n; i++) t1.Insert(a[i]);
        for(int i = 1; i <= n; i++) t2.Insert(b[i]);
        vector <int> ans = work();
        sort(ans.begin(), ans.end());
        for(int i = 0; i < ans.size(); i++) cout << ans[i] << " \n"[i == ans.size() - 1];
    }
    return 0;
}

D.equation

高中数学知识即可解决，按照每部分的零点排序，然后去掉绝对值依次统计答案即可。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
const int MAXN = 1e5 + 5, INF = 0x3f3f3f3f;
int t, n, C;
bool cmp(ll a, ll b, ll c, ll d) {
    if (a < 0)a = -a, b = -b;
    if (c < 0)c = -c, d = -d;
    return b * c <= a * d;
}
bool cmp2(ll a, ll b, ll c, ll d) {
    if (a < 0)a = -a, b = -b;
    if (c < 0)c = -c, d = -d;
    return b * c < a * d;
}
struct node {
    int a, b;
    bool operator <(const node &rhs)const {
        return cmp2(a, -b, rhs.a, -rhs.b);
    }
}s[MAXN];
struct res {
    int a, b;
    bool operator <(const res &rhs)const {
        return cmp2(b, a, rhs.b, rhs.a);
    }
};
vector<res> ans;
vector <pair<ll, ll>> cc;
ll gcd(ll a, ll b) {
    return !b ? a : gcd(b, a%b);
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> t;
    while (t--) {
        cin >> n >> C;
        ans.clear(); cc.clear();
        for (int i = 1; i <= n; i++) {
            cin >> s[i].a >> s[i].b;
        }
        sort(s + 1, s + 1 + n);
        s[0].a = 1; s[0].b = 2000; s[n + 1].a = 1; s[n + 1].b = -2000;
        ll A = 0, B = 0;
        for (int i = 1; i <= n; i++) {
            A -= s[i].a;
            B -= s[i].b;
        }
        bool is_infinite = false;
        for (int i = 1; i <= n + 1; i++) {
            if (A == 0 && C - B == 0) {
                is_infinite = true;
                cout << "-1\n";
                break;
            }
            if (A == 0 && C - B != 0) {
                A += 2 * s[i].a;
                B += 2 * s[i].b;
                continue;
            }
            if (cmp2(s[i - 1].a, -1 * s[i - 1].b, A, C - B)) {
                if (cmp(A, C - B, s[i].a, -1 * s[i].b)) {
                    ans.push_back({ (int)(C - B),(int)A });
                }
            }
            A += 2 * s[i].a;
            B += 2 * s[i].b;
        }
        if (is_infinite) {
            continue;
        }
        ll g;
        for (int i = 0; i < ans.size(); i++) {
            g = gcd(ans[i].a, ans[i].b);
            ans[i].a /= g; ans[i].b /= g;
            if (ans[i].a < 0 && ans[i].b < 0)ans[i].a = -ans[i].a, ans[i].b = -ans[i].b;
            if (ans[i].a > 0 && ans[i].b < 0)ans[i].a = -ans[i].a, ans[i].b = -ans[i].b;
        }
        sort(ans.begin(), ans.end());
        for (int i = 0; i < ans.size(); i++) {
            cc.push_back({ ans[i].a,ans[i].b });
        }
        cc.erase(unique(cc.begin(), cc.end()), cc.end());

        cout << cc.size() << " \n"[cc.size() == 0];
        for (int i = 0; i < cc.size(); i++) {
            if (cc[i].first == 0)cout << "0/1" << '\n';
            else {
                cout << cc[i].first << "/" << cc[i].second << " \n"[i == cc.size() - 1];
            }
        }
    }
    return 0;
}

E.permutation 1

注意\(n,k\)的范围，因为有\(k\leq min(10^4,N!),n\leq 20\)，那么当\(n\leq 9\)时直接枚举所有全排列暴力即可；当\(n>9\)时，打表可以发现，这种情况下的差值字典序大小与原来的字典序大小相对应，并且前两位一定是\(n,1\)，所以直接暴力枚举后面每一位即可。
代码写得有点丑陋...

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 25;
int T;
int n, k;
int a[N], b[N];
ll fac[N];
bool vis[N];
void work(int n) {
    vector <int> ans;
    ans.push_back(n); ans.push_back(1);
    memset(vis, 0, sizeof(vis));
    vis[1] = vis[n] = 1;
    for(int i = 3; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            if(vis[j]) continue;
            if(fac[n - i] < k) {
                k -= fac[n - i];
            } else {
                ans.push_back(j);
                vis[j] = 1;
                break;
            }
        }
    }
    for(int i = 0; i < n - 1; i++) cout << ans[i] << ' ';
    cout << ans[n - 1] << '\n';
}
void solve(int n) {
    for(int i = 1; i <= n; i++) a[i] = i;
    vector <int> ans[fac[n] + 1];
    int tot = 0;
    do {
        tot++;
        for(int i = 1; i < n; i++) ans[tot].push_back(a[i + 1] - a[i]);
    } while(next_permutation(a + 1, a + n + 1));
    sort(ans + 1, ans + tot + 1);
    sort(a + 1, a + n + 1);
    do {
        int ok = 1;
        for(int i = 1; i < n; i++) {
            if(a[i + 1] - a[i] != ans[k][i - 1]) {
                ok = 0; break;
            }
        }
        if(ok) {
            for(int i = 1; i < n; i++) cout << a[i] << ' ';
            cout << a[n] << '\n';
            break;
        }
    } while(next_permutation(a + 1, a + n + 1));
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    fac[0] = 1;
    for(int i = 1; i < N; i++) fac[i] = fac[i - 1] * i;
    cin >> T;
    while(T--) {
        cin >> n >> k;
        if(n >= 10) work(n);
        else solve(n);
    }
    return 0;
}

F.string matching

扩展\(KMP\)模板题。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 5;
struct ExKmp{
    int Next[N];
    int extend[N];
    void get_next(char *s) {
        int len = strlen(s + 1), p = 1, pos;
        Next[1] = len;
        while(p + 1 <= len && s[p] == s[p + 1]) p++;
        Next[pos = 2] = p - 1;
        for(int i = 3; i <= len; i++) {
            int l = Next[i - pos + 1];
            if(i + l < p + 1) Next[i] = l;
            else {
                int j = max(p - i + 1, 0);
                while(i + j <= len && s[j + 1] == s[i + j]) ++j;
                p = i + (Next[pos = i] = j) - 1;
            }
        }
    }
    void work(char *s, char *t) {
        get_next(t);
        int lens = strlen(s + 1), lent = strlen(t + 1), p = 1, pos;
        while(p <= lent && s[p] == t[p]) ++p;
        p = extend[pos = 1] = p - 1;
        for(int i = 2; i <= lens; i++) {
            int len = Next[i - pos + 1];
            if(len + i < p + 1) extend[i] = len;
            else {
                int j = max(p - i + 1, 0);
                while(i + j <= lens && j <= lent && t[j + 1] == s[i + j]) j++;
                p = i + (extend[pos = i]) - 1;
            }
        }
    }
}exkmp;
int T, n;
char s[N];
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> T;
    while(T--) {
        cin >> s + 1;
        exkmp.get_next(s);
        n = strlen(s + 1);
        ll ans = 0;
        for(int i = 2; i <= n; i++) {
            if(i + exkmp.Next[i] - 1 == n) ans += n - i + 1;
            else ans += exkmp.Next[i] + 1;
        }
        cout << ans << '\n';
    }
    return 0;
}

G.permutation 2

注意到只有中间上升的某段会对答案产生贡献，之后对上升的情况打表找规律即可。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5, MOD = 998244353;
int n;
int x, y;
int T;
ll f[N];
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> T;
    f[1] = f[2] = f[3] = 1;
    for(int i = 4; i < N; i++) f[i] = (f[i - 1] + f[i - 3]) % MOD;
    while(T--) {
        cin >> n >> x >> y;
        int d;
        if(x == 1 && y == n) {
            d = n;
        } else if(x == 1) {
            d = y - 1;
        } else if(y == n) {
            d = n - x;
        } else {
            d = y - 1 - (x + 1) + 1;
        }
        cout << f[d] << '\n';
    }
    return 0;
}