2024 (ICPC) Jiangxi Provincial Contest -- Official Contest

2024 (ICPC) Jiangxi Provincial Contest -- Official Contest

A. Maliang Learning Painting

題意：a + b + c

void solve(){
    ll a, b, c;
    cin >> a >> b >> c;
    ll ans = a + b + c;
    cout << ans << '\n';
}

C. Liar

題意：n 個 數字總和是 s ，每個人說出自己得到的數字，有人可能撒謊，問最多有多少人說真話

思路：首先總和就是 s 的話，最多所有人說的都是真話，再或者假設 1~i 的人都不撒謊，只有第 n 個人撒謊

void solve(){
    ll n, s;
    cin >> n >> s;
    ll sum = 0;
    for(int i = 1; i <= n; i ++){
        ll x;
        cin >> x;
        sum += x;
    }
    if(sum == s) cout << n << '\n';
    else cout << n - 1 << '\n';
}

D. Magic LCM

題意：給定 n 個數，每次可以選定兩個數，把其中一個改為 gcd(x, y)，另外一個改為 lcm(x, y)，求最大的數列和

思路：可以發現，透過操作，可以把二者共有的質因子的最大的冪交給一方，如果是非共有的質因子，會交給較大的一方，相當於共有質因子的較大冪和非共有質因子交給一方，共有質因子的較小冪交給另外一方，我們可以先處理 1e3 以內的質數，求出每個數的質因子，然後對共有質因子的冪進行排序，然後找出項數最多的質因子，對該質因子的所有冪排序，其它的質因子會全部轉移當這個項數最多的質因子上，同時我們要貪心的其它質因子的每一項的冪按照大和大匹配來運算，最後不要忘了除了最多項數的質因子之外其它的全部變成了1，要把它們加上

const int N = 1000010, mod = 998244353;
const int Mod = 1e9 + 7, INF = 0x3f3f3f3f;
 
bool vis[1010];
vector<ll> pri;
void ola_s(ll n){
    for(int i = 2; i <= n; i ++){
        if(!vis[i]) pri.push_back(i);
        for(auto v : pri){
            if(i * v > n) break;
            vis[i * v] = true;
            if(i % v == 0) break;
        }
    }
}
 
vector<ll> v[N];
set<ll> st;
 
void get_fac(ll x){
    for(auto p : pri){
        if(x < p) break;
        if(x % p == 0){
            ll t = 1;
            while(x % p == 0) {
                t *= p;
                x /= p;
            }
            v[p].push_back(t);
            st.insert(p);
        }
    }
    if(x > 1){
        v[x].push_back(x);
        st.insert(x);
    }
}
 
void solve(){
    ll n;
    cin >> n;
    vector<ll> a(n + 1);
    for(int i = 1; i <= n; i ++){
        cin >> a[i];
        get_fac(a[i]);
    }
    ll len = 0, ma = 0;;
    for(auto i : st){
        if(v[i].size() > len) len = v[i].size(), ma = i;
        sort(v[i].begin(), v[i].end());
    }
    for(auto i : st){
        if(i == ma) continue;
        for(int j = len - 1, k = v[i].size() - 1; k >= 0; j --, k --){
            v[ma][j] = v[ma][j] * v[i][k] % mod;
        }
    }
    ll ans = n - v[ma].size();
    for(auto i : v[ma]) ans = (ans + i + mod) % mod;
    cout << ans << '\n';
    for(auto i : st) v[i].clear();
    st.clear();
}
 
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    ola_s(1000);    
    int t = 1;
    cin >> t;
    while(t --){
        solve();
    }
    return 0;
}

G. Multiples of 5

題意：給定一個11進位制數，求它是不是 5 的倍數

思路：首先這個11進位制數非常大，肯定不能轉化成十進位制的數來求，我們手推一下 11 的沒一項冪會發現它們的最後一個數字都是 1 ，也只有這個 1 會影響答案，我們求出 11 進位制下每一位有多少個，然後加一下看是不是 5 的整數倍就行了。（又把‘A’ == 10當成'A' == 11 wa了一發）

void solve(){
    ll n;
    cin >> n;
    ll ans = 0;
    for(int i = 1; i <= n; i ++){
        ll a1, a2;
        char b;
        cin >> a1 >> b;
        if(b != 'A') a2 = (b - '0');
        else a2 = 10;
        ans = (ans + a1 * a2) % 5;
    }
    if(ans == 5 || ans == 0) cout << "Yes\n";
    else cout << "No\n";
}

H. Convolution

題意：給定一個n * m的矩陣 a ，求一個k * l的矩陣 b 和 a 的乘積的最小值，b 只包含(-1, 0, 1)

思路：手畫一下發現，矩陣 b 的每一項，會和矩陣 a 左上角 （i, j) ~ (n - k + i, m - l + j) 的子矩陣相乘，處理一下二維字首和，正數取1，否則取-1

void solve(){
    ll n, m, k, l;
    cin >> n >> m >> k >> l;
    vector<vector<ll>> I(3010, vector<ll>(3010));
    vector<vector<ll>> IQ(3010, vector<ll>(3010));
    
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            cin >> I[i][j];
            ll x = I[i][j];            
            IQ[i][j] = x + IQ[i - 1][j] + IQ[i][j - 1] - IQ[i - 1][j - 1];
        }
    }
    auto query =[&](ll x1, ll y1, ll x2, ll y2) -> ll{
        ll ans = IQ[x2][y2] - IQ[x1 - 1][y2] - IQ[x2][y1 - 1] + IQ[x1 - 1][y1 - 1];
        return ans; 
    };
    
    ll res = 0;
    for(int i = 1; i <= k; i ++){
        for(int j = 1; j <= l; j ++){
            res += abs(query(i, j, n - k + i, m - l + j));
        }
    }
    cout << res << '\n';
}

I. Neuvillette Circling

題意：給定 n 個點，所有點之間都有邊相連線，問包含 1 ~ n * (n - 1) / 2 條邊分別至少需要的半徑

思路：分類討論，兩種情況，兩個點確定一個圓，三個點確定一個圓（外接圓），然後分別求一下包含多少個點

using db = double;
const db eps = 1e-6;
int sign(db a) { return a < -eps ? -1 : a > eps; } //< :-1 =:0 >:1
struct Point {
    db x, y;
    Point() {}
    Point(db x, db y) : x(x), y(y) {}
    Point operator+(Point p) { return {x + p.x, y + p.y}; }
    Point operator-(Point p) { return {x - p.x, y - p.y}; }
    Point operator*(db d) { return {x * d, y * d}; }
    Point operator/(db d) { return {x / d, y / d}; }
    bool operator==(Point &p) const { return !sign(x - p.x) && !sign(y - p.y); }
    int toleft(Point b) { // 點線上段的方向
        ll res = x * b.y - y * b.x;
        if (res == 0)
            return 0; // 直線上
        else if (res > 0)
            return 1; // 左
        return -1;    // 右
    }
    friend istream &operator>>(istream &is, Point &p) {
        return is >> p.x >> p.y;
    }
    friend ostream &operator<<(ostream &os, Point p) {
        return os << "(" << p.x << ", " << p.y << ")";
    }
};
struct Circle {
    Point o;
    db r;
    Circle() {}
    Circle(Point _o, db _r) : o(_o), r(_r) {}
};
db dot(Point a, Point b) {
    return a.x * b.x + a.y * b.y;
}
db cross(Point a, Point b) {
    return a.x * b.y - a.y * b.x;
}
db dis1(Point a, Point b) { // 兩點距離的平方
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    // return sqrtl((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));//long double ver
}
db dis2(Point a, Point b) { // 兩點距離的平方
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool ver(Point a, Point b) { // 是否垂直
    return sign(dot(a, b)) == 0;
}
db pw(db x) {
    return x * x;
}
db fun(db a1, db a2, db a3, db b1, db b2, db b3, db c1, db c2, db c3) {
    return a1 * b2 * c3 +
           b1 * c2 * a3 +
           c1 * a2 * b3 -
           a3 * b2 * c1 -
           b3 * c2 * a1 -
           c3 * a2 * b1;
}
Circle three_p_get_cir(Point a, Point b, Point c) { // 求三角形的外接圓
    db A = dis1(a, b), B = dis1(b, c), C = dis1(a, c);
    db L = (A + B + C) / 2;                       // 三角形半周長
    db S = sqrt(L * (L - A) * (L - B) * (L - C)); // 由海倫公式求得三角形面積
    db r = A * B * C / (4 * S);                   // 外接圓半徑
 
    // cout << a << " " << b << " " << c << endl;
    // cout << A << " " << B << " " << C << endl;
    // cout << r << endl;
    // 求圓心座標
    Point res;
    res.x = 0.5 * fun(1, 1, 1, pw(a.x) + pw(a.y), pw(b.x) + pw(b.y), pw(c.x) + pw(c.y), a.y, b.y, c.y) / fun(1, 1, 1, a.x, b.x, c.x, a.y, b.y, c.y);
    res.y = 0.5 * fun(1, 1, 1, a.x, b.x, c.x, pw(a.x) + pw(a.y), pw(b.x) + pw(b.y), pw(c.x) + pw(c.y)) / fun(1, 1, 1, a.x, b.x, c.x, a.y, b.y, c.y);
    return Circle(res, r);
}
Circle two_p_get_cir(Point a, Point b) { // 已知直徑求圓
    db r = dis1(a, b) / 2;
    Point res;
    res.x = (a.x + b.x) / 2;
    res.y = (a.y + b.y) / 2;
    return Circle(res, r);
}

void solve(){
    ll n;
    cin >> n;
    vector<Point> p(n);
    vector<Circle> a;
    for(int i = 0; i < n; i ++) cin >> p[i];
    for(int i = 0; i < n; i ++){
        for(int j = i + 1; j < n; j ++){
            a.push_back(two_p_get_cir(p[i], p[j]));
            for(int k = j + 1; k < n; k ++){
                a.push_back(three_p_get_cir(p[i], p[j], p[k]));
            }
        }
    }
    auto cal =[&](Circle c) -> int{
        auto [x1, y1] = c.o;
        db r2 = c.r * c.r;
        ll ans = 0;
        for(auto &[x2, y2] : p){
            db res = pw(x2 - x1) + pw(y2 - y1);
            if(sign(res - r2) == 0 || sign(res - r2) == -1) ans ++;
        }
        return ans * (ans - 1) / 2;
    };
    vector<db> ans(n * (n - 1) / 2 + 1, INF);
    for(auto &t : a){
        ll num = cal(t);
        ans[num] = min(ans[num], t.r);
    }
    for(int i = n * (n - 1) / 2 - 1; i >= 1; i --) ans[i] = min(ans[i], ans[i + 1]);
    for(int i = 1; i <= n * (n - 1) / 2; i ++){
        cout << fixed << setprecision(8) << ans[i] << '\n';
    }
}

J. Magic Mahjong

題意：給定一副麻將牌，判斷手上是否有十三么和七對（不打麻將英語還不行的有難了）

思路：模擬即可，包含七個對子或者十三種給定牌

string win[] = {"1z", "2z", "3z", "4z", "5z", "6z", "7z", "1p", "9p", "1s", "9s", "1m", "9m"};
 
void solve(){
    string s;
    cin >> s;
    vector<string> pai;
    vector<ll> ws(10);
    
    for(int i = 0; i < s.size(); i += 2){
        string s1;
        s1 += s[i];
        s1 += s[i + 1];
        pai.push_back(s1);
    }
    
    map<string, ll> mp;
    for(auto v : pai) mp[v] ++;
    
    if(mp.size() == 7){
        bool ok = 1;
        for(auto [x, y] : mp){
            if(y != 2) ok = 0;
        }
        if(ok == 1){
            cout << "7 Pairs\n";
            return;
        }
    }    
    
    
    if(mp.size() == 13){
        vector<ll> se(13);
        for(auto [x, y] : mp){
            bool ok = 0;
            for(int i = 0; i < 13; i ++){
                if(x == win[i]){
                    ok = 1;
                    se[i] = 1;
                    break;
                }
            }
            if(!ok){
                cout << "Otherwise\n";    
                return;            
            }
        }
        ll sum = 0;
        for(auto x : se) sum += x;
        if(sum == 13) cout << "Thirteen Orphans\n";        
        else cout << "Otherwise\n";
    }
    else cout << "Otherwise\n";
}

K. Magic Tree

題意：2 * n的路有幾條不同的填充方式

思路：手畫一下發現 2^（n - 1) 即可

ll fastpow(ll a, ll n){
    ll res = 1, base = a;
    while(n){
        if(n & 1) res = (res * base) % mod;
        base = (base * base) % mod;
        n >>= 1;
    }
    return res;
}
void solve(){
    ll n;
    cin >> n;
    ll ans = fastpow(2, n - 1);
    cout << ans << '\n';
}

L. Campus

題意：一個 n 個點 m 條邊的無向圖，每個節點上有 ai 個人。n 個節點中有 k 個 節點為出口，每個出口有一個開放時間 [li , ri ]。求第 1 ∼ T 時刻所有人到 最近的開放出口的路徑長度之和。

思路：預處理出每一種出口對每個人的最短路，然後分情況列舉取min即可

struct Node{
    ll y, v;
    Node(ll _y, ll _v){
        y = _y, v = _v;
    }
};
 
struct gate{
    ll id, l, r;
    bool operator < (gate & a1)const{
        return l < a1.l;
    }
};
 
bool cmp(gate & a1, gate & a2){
    return a1.l < a2.l;
}
 
vector<Node> edge[N];
ll a[N], res = 0;
 
void solve(){
    ll n, m, k, T;
    cin >> n >> m >> k >> T;
    string s1;
    for(int i = 1; i <= k; i ++) s1 += '0';
    
    vector<gate> v(k + 1);
    for(int i = 1; i <= n; i ++) cin >> a[i];
    for(int i = 1; i <= k; i ++){
        ll id, l, r;
        cin >> id >> l >> r;
        v[i] = {id, l, r}; 
    }
    for(int i = 1; i <= m; i ++){
        ll u, v, w;
        cin >> u >> v >> w;
        edge[u].push_back({v, w});
        edge[v].push_back({u, w});
    }
    
    auto dijkstra =[&](ll s, vector<ll> & p) -> void{
        vector<ll> dist(n + 1, INF);
        vector<bool> b(n + 1, false);
        priority_queue<PLL, vector<PLL>, greater<PLL>> q;
        dist[s] = 0;
        for(int i = 1; i <= n; i ++) q.push({dist[i], i});
        while(!q.empty()){
            ll x = q.top().second;
            q.pop();
            if(b[x]) continue;
            b[x] = true;
            for(auto i : edge[x]){
                if(dist[x] + i.v < dist[i.y]){
                    dist[i.y] = dist[x] + i.v;
                    q.push({dist[i.y], i.y});
                }
            }
        }
        p = dist;
    };
    vector<vector<ll>> d(k + 1, vector<ll>(n + 1));
    for(int i = 1; i <= k; i ++){
        dijkstra(v[i].id, d[i]);
    }
    
//    for(int i = 1; i <= k; i ++){
//        cout << i << '\n';
//        for(int j = 1; j <= n; j ++){
//            cout << d[i][j] << ' ';
//        }
//        cout << '\n';
//    }
 
 
    map<string, vector<ll>> mp;
    vector<ll> ans(T + 1);
    
    for(int i = 1; i <= T; i ++){
        string s = s1;
        for(int j = 1; j <= k; j ++){
            if(v[j].l <= i && i <= v[j].r) s[j - 1] = '1';
        }
        mp[s].push_back(i);
    }
    
    for(auto [x, y] : mp){
        vector<ll> q(n + 1, INF);
        bool ok = 0;
        for(int i = 1; i <= k; i ++){
            if(x[i - 1] == '1'){
                ok = 1;
                for(int j = 1; j <= n; j ++){
                    q[j] = min(q[j], d[i][j]);
                }
            }
        }
        ll res = 0;
        if(!ok){
            res = -1;
        }
        else{
            for(int i = 1; i <= n; i ++) res += q[i] * a[i];            
        }
        for(auto v : y) ans[v] = res;
    }
    for(int i = 1; i <= T; i ++) cout << ans[i] << '\n';
}