2016 ACM/ICPC Asia Regional Dalian Online

太懶，強迫自己一波。MDZZ
A
題意：有N 個人，可以選擇任意數目的人來參與遊戲。
遊戲規則要求選出來的人圍成一個圓且任意兩個人到圓心的連線夾角必須是 2∗πN 的倍數且不能是 2∗πN ，而且通過旋轉得到的方案認爲是相同的。

我們把N 個人當做N 個點，這樣第i 個點坐人的話記爲黑點，反之記爲白點。
問題就變成−> 給定環上N 個點，黑白染色，要求黑點不能相鄰，問你旋轉同構的方案數。

記f[i] ：i 個點黑白染色，不考慮旋轉同構的方案數。
這樣poyla 定理：N 個點旋轉同構的方案數ans=∑Ni=1f[gcd(i,N)] ，用歐拉函數處理一下就OK 了。

一開始推出公式f[i]=f[i−1]+f[i−3] ，然後莫名GG ？後來發現公式f[i]=f[i−1]+f[i−2] 可以對上數據，why ？
不管了，過了再說。。。但是1 的時候輸出2 ？不懂出題人

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 2e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
LL euler(int n) {
    LL ans = n;
    for(int i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            ans = ans / i * (i - 1);
            while(n % i == 0) n /= i;
        }
    }
    if(n > 1) ans = ans / n * (n - 1);
    return ans;
}
struct Matrix {
    LL a[2][2];
};
Matrix multi(Matrix x, Matrix y) {
    Matrix z; CLR(z.a, 0);
    for(int i = 0; i < 2; i++) {
        for(int k = 0; k < 2; k++) {
            if(x.a[i][k] == 0) continue;
            for(int j = 0; j < 2; j++) {
                z.a[i][j] = (z.a[i][j] + x.a[i][k] * y.a[k][j] % MOD) % MOD;
            }
        }
    }
    return z;
}
Matrix Pow(Matrix x, int n) {
    Matrix ans;
    for(int i = 0; i < 2; i++) {
        for(int j = 0; j < 2; j++) {
            if(i == j) {
                ans.a[i][j] = 1;
            }
            else {
                ans.a[i][j] = 0;
            }
        }
    }
    while(n) {
        if(n & 1) {
            ans = multi(ans, x);
        }
        x = multi(x, x);
        n >>= 1;
    }
    return ans;
}
LL Work(int n) {
    if(n == 1) return 1;
    if(n == 2) return 3;
    Matrix x;
    x.a[0][0] = 1; x.a[0][1] = 1; x.a[1][0] = 1; x.a[1][1] = 0;
    Matrix y = Pow(x, n - 2);
    return (1LL * 3 * y.a[0][0] % MOD + 1LL * 1 * y.a[1][0] % MOD) % MOD;
}
LL pow_mod(LL a, int n) {
    LL ans = 1;
    while(n) {
        if(n & 1) {
            ans = ans * a % MOD;
        }
        n >>= 1;
        a = a * a % MOD;
    }
    return ans;
}
int main()
{
//    for(int i = 1; i <= 8; i++) {
//        cout << Work(i) << endl;
//    }
    int n;
    while(scanf("%d", &n) != EOF) {
        if(n == 1) {
            printf("2\n");
            continue;
        }
        LL ans = 0;
        for(int i = 1; i * i <= n; i++) {
            if(n % i == 0) {
                add(ans, Work(i) * euler(n / i) % MOD);
                if(i * i != n) {
                    add(ans, Work(n / i) * euler(i) % MOD);
                }
            }
        }
        printf("%lld\n", ans * pow_mod(n, MOD - 2) % MOD);
    }
    return 0;
}

B
題意：求解區間不同GCD 數目。
思路：枚舉右端點，二分找到連續區間GCD 的最右區間端點。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int g[MAXN][21], a[MAXN];
map<int, int> lp;
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}
void RMQ_init(int N) {
    for(int i = 1; i <= N; i++) {
        g[i][0] = a[i];
    }
    for(int j = 1; (1 << j) <= N; j++) {
        for(int i = 1; i + (1 << j) - 1 <= N; i++) {
            g[i][j] = gcd(g[i][j-1], g[i + (1 << (j-1))][j-1]);
        }
    }
}
int Query(int L, int R) {
    int k = 0;
    while((1 << (k + 1)) <= R - L + 1) k++;
    return gcd(g[L][k], g[R - (1 << k) + 1][k]);
}
struct Node {
    int l, r, id;
};
Node num[MAXN];
bool cmp1(Node a, Node b) {
    return a.r < b.r;
}
LL ans[MAXN];
LL C[MAXN];
int lowbit(int x) {
    return x & (-x);
}
int n;
void add(int x, int d) {
    while(x <= n) {
        C[x] += d;
        x += lowbit(x);
    }
}
LL Sum(int x) {
    LL s = 0;
    while(x > 0) {
        s += C[x];
        x -= lowbit(x);
    }
    return s;
}
struct GCD {
    int l, r, val;
};
GCD gnum[MAXN * 30];
bool cmp2(GCD a, GCD b) {
    return a.r < b.r;
}
int main()
{
    int m;
    while(scanf("%d%d", &n, &m) != EOF) {
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        RMQ_init(n); int top = 0;
        for(int i = 1; i <= n; i++) {
            int j = i;
            while(j >= 1) {
                int v = Query(j, i);
                int l = 1, r = j, next;
                gnum[top].l = j; gnum[top].r = i; gnum[top].val = v;
                //cout << gnum[top].l << ' ' << gnum[top].r << ' ' << gnum[top].val << endl;
                top++;
                while(r >= l) {
                    int mid = (l + r) >> 1;
                    if(Query(mid, i) == v) {
                        next = mid;
                        r = mid - 1;
                    }
                    else {
                        l = mid + 1;
                    }
                }
                j = next - 1;
            }
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &num[i].l, &num[i].r);
            num[i].id = i;
        }
        sort(num + 1, num + m + 1, cmp1);
        sort(gnum, gnum + top, cmp2);
        int j = 0; CLR(C, 0); lp.clear();
        for(int i = 1; i <= m; i++) {
            while(j < top && gnum[j].r <= num[i].r) {
                int v = gnum[j].val;
                if(lp[v]) {
                    add(lp[v], -1);
                }
                lp[v] = gnum[j].l;
                add(lp[v], 1);
                j++;
            }
            ans[num[i].id] = Sum(num[i].r) - Sum(num[i].l - 1);
        }
        for(int i = 1; i <= m; i++) {
            printf("%lld\n", ans[i]);
        }
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int a[20000 + 10];
int main()
{
    int t;
    while(scanf("%d", &t) != EOF) {
        while(t--) {
            int n; scanf("%d", &n);
            int sum = 0;
            for(int i = 1; i <= n; i++) {
                scanf("%d", &a[i]);
                sum += a[i];
            }
            if(sum != n * (n - 1)) {
                printf("F\n");
                continue;
            }
            sort(a + 1, a + n + 1);
            sum = 0; int s = 0;
            bool flag = true;
            for(int i = n; i >= 1; i--) {
                s += a[i];
                sum += 2 * (i - 1);
                if(s > sum) {
                    flag = false; break;
                }
            }
            printf(flag ? "T\n" : "F\n");
        }
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef unsigned long long LL;
typedef pair<int, int> pii;
const int MAXN = 2e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int main()
{
    LL n, m;
    while(scanf("%llu%llu", &m, &n) != EOF) {
        LL sum = floor(m * m * 1.0 / 4);
        printf(n >= sum ? "T\n" : "F\n");
    }
    return 0;
}

H
思路：二分套個RMQ 找到最小的j 滿足a[j]<a[i] &&j>i 。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int a[MAXN];
int Amin[MAXN][20];
void RMQ_init(int N) {
    for(int i = 1; i <= N; i++)
        Amin[i][0] = a[i];
    for(int j = 1; (1<<j) <= N; j++) {
        for(int i = 1; i + (1<<j)-1 <= N; i++)
            Amin[i][j] = min(Amin[i][j-1], Amin[i+(1<<(j-1))][j-1]);
    }
}
int Query(int L, int R) {
    int k = 0;
    while((1<<(k+1)) <= R-L+1) k++;
    return min(Amin[L][k], Amin[R-(1<<k)+1][k]);
}
int Next[MAXN];
int main()
{
    int t; scanf("%d", &t);
    while(t--) {
        int n; scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        RMQ_init(n);
        for(int i = 1; i <= n; i++) {
            int l = i + 1, r = n, pos = -1;
            while(r >= l) {
                int mid = l + r >> 1;
                if(Query(l, mid) < a[i]) {
                    pos = mid;
                    r = mid - 1;
                }
                else {
                    l = mid + 1;
                }
            }
            Next[i] = pos;
        }
        int m; scanf("%d", &m);
        while(m--) {
            int l, r; scanf("%d%d", &l, &r);
            int ans = a[l];
            int pos = l;
            for(;;) {
                pos = Next[pos];
                if(pos > r || pos == -1) break;
                ans %= a[pos];
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}

I
題意：求解補圖最短路。
前面到達的點後面就沒有用了，那就維護兩個集合，一個當前點可以到達且沒有被用過，一個當前點不可以到達且沒有被用過。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 2e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
vector<int> G[MAXN];
int d[MAXN];
set<int> use, have;
set<int> :: iterator it;
int n;
int ans[MAXN];
void Solve(int s) {
    queue<int> Q; Q.push(s); CLR(d, INF);
    d[s] = 0; have.clear(); use.clear();
    for(int i = 1; i <= n; i++) {
        if(i != s) {
            have.insert(i);
        }
    }
    while(!Q.empty()) {
        int u = Q.front(); Q.pop();
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i];
            if(!have.count(v)) continue;
            have.erase(v); use.insert(v);
        }
        for(it = have.begin(); it != have.end(); ++it) {
            Q.push(*it);
            d[*it] = d[u] + 1;
        }
        have.swap(use); use.clear();
    }
}
int main()
{
    int t; scanf("%d", &t);
    while(t--) {
        int m; scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) G[i].clear();
        while(m--) {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        int s; scanf("%d", &s); Solve(s);
        int top = 0;
        for(int i = 1; i <= n; i++) {
            if(i == s) continue;
            ans[top++] = (d[i] != INF ? d[i] : -1);
        }
        for(int i = 0; i < top; i++) {
            if(i > 0) printf(" ");
            printf("%d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}

J
思路：我們處理出DFS 序，然後對於第i 個點找到孩子區間有多少個合法a[j] 即a[j]<=h=k/a[i] 。N 次查詢，我們離線做，在對應的h 值限制下插入合法的解，然後用樹狀數組去統計，是可以做到O(nlog(n)) 的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cstdlib>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<LL, LL> pii;
const int MAXN = 1e5 + 10;
const int INF = 1e9 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
struct Edge {
    int from, to, next;
};
Edge edge[MAXN * 2];
int head[MAXN], edgenum;
void init() {CLR(head, -1); edgenum = 0;}
void addEdge(int u, int v) {
    Edge E = {u, v, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
LL vs[MAXN];
int top;
int son[MAXN];
LL a[MAXN];
int id[MAXN];
void DFS(int u, int fa) {
    son[u] = 1; id[u] = ++top; vs[top] = a[u];
    for(int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if(v == fa) continue;
        DFS(v, u);
        son[u] += son[v];
    }
}
int C[MAXN], n;
int lowbit(int x) {
    return x & (-x);
}
void Update(int x, int d) {
    while(x <= n) {
        C[x] += d;
        x += lowbit(x);
    }
}
int Sum(int x) {
    int s = 0;
    while(x > 0) {
        s += C[x];
        x -= lowbit(x);
    }
    return s;
}
struct Node{
    int l, r;
    LL h;
};
Node num[MAXN];
bool cmp2(Node a, Node b){
    return a.h < b.h;
}

int in[MAXN];
vector<pii> G;
vector<pii> :: iterator it;
bool cmp1(pii a, pii b){
    return a.first < b.first;
}

int main()
{
    int t, kcase = 1; scanf("%d", &t);
    while(t--) {
        LL k; scanf("%d%lld", &n, &k);
        init();
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            in[i] = 0;
        }
        for(int i = 0; i < n - 1; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            addEdge(u, v);
            in[v]++;
        }
        int root;
        for(int i = 1; i <= n; i++) {
            if(in[i] == 0) {
                root = i; break;
            }
        }
        top = 0; DFS(root, -1);
        G.clear();
        for(int i = 1; i <= top; i++) {
            G.push_back(pii(vs[i], i));
        }
        sort(G.begin(), G.end(), cmp1);
        for(int i = 1; i <= n; i++) {
            num[i].l = id[i];
            num[i].r = id[i] + son[i] - 1;
            num[i].h = k / a[i];
        }
        sort(num + 1, num + n + 1, cmp2);
        LL ans = 0; CLR(C, 0); it = G.begin();
        for(int i = 1; i <= n; i++) {
            while(it != G.end() && it->first <= num[i].h) {
                Update(it->second, 1);
                it++;
            }
            ans += Sum(num[i].r) - Sum(num[i].l - 1);
        }
        for(int i = 1; i <= n; i++) {
            if(a[i] * a[i] <= k) {
                ans--;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}

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