2018 Multi-University Training Contest 10

 

Recently, TeaTree acquire new knoledge gcd (Greatest Common Divisor), now she want to test you.
As we know, TeaTree is a tree and her root is node 1, she have n nodes and n-1 edge, for each node i, it has it’s value v[i].
For every two nodes i and j (i is not equal to j), they will tell their Lowest Common Ancestors (LCA) a number : gcd(v[i],v[j]).
For each node, you have to calculate the max number that it heard. some definition:
In graph theory and computer science, the lowest common ancestor (LCA) of two nodes u and v in a tree is the lowest (deepest) node that has both u and v as descendants, where we define each node to be a descendant of itself.

InputOn the first line, there is a positive integer n, which describe the number of nodes.
Next line there are n-1 positive integers f[2] ,f[3], …, f[n], f[i] describe the father of node i on tree.
Next line there are n positive integers v[2] ,v[3], …, v[n], v[i] describe the value of node i.
n<=100000, f[i]<i, v[i]<=100000OutputYour output should include n lines, for i-th line, output the max number that node i heard.
For the nodes who heard nothing, output -1.Sample Input

4
1 1 3
4 1 6 9

Sample Output

2
-1
3
-1

 

题意 ： 给你一颗树，每个结点上面都有一个权值，询问你一任意一个结点作为树根其子树上任意两点的 gcd 的最大值是多少

思路分析 ：

　　1 . 用 vector 去模拟归并排序，当合并时遇到相同的就记录一下最大值

#define ll long long
const int maxn = 1e5+5;

int n;
vector<int>d[maxn], ve[maxn], f[maxn];
void init(){
    for(int i = 1; i <= 100000; i++){
        for(int j = i; j <= 100000; j += i){
            d[j].push_back(i);
        }
    } 
}
int val[maxn], ans[maxn];
vector<int>temp;

void merge(int x, int y){
    temp.clear();    
    if (f[x].size() == 0) f[x] = d[val[x]];
    if (f[y].size() == 0) f[y] = d[val[y]];
    
    int i = 0, j = 0;
    while((i < f[x].size()) && (j < f[y].size())){
        if (f[x][i] < f[y][j]) temp.push_back(f[x][i]), i++;
        else if (f[x][i] > f[y][j]) temp.push_back(f[y][j]), j++;
        else {
            temp.push_back(f[x][i]);
            ans[x] = max(ans[x], f[x][i]); i++, j++;
        }
    }
    while (i < f[x].size()) {
        temp.push_back(f[x][i]);
        i++;
    }
    while(j < f[y].size()) {
        temp.push_back(f[y][j]);
        j++;
    }
    f[x].clear(); f[y].clear();
    f[x] = temp; 
}

void dfs(int x){
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i];
        dfs(to);
        merge(x, to);
    }
}

int main() {
    init();
    cin >> n;
    
    int x;
    memset(ans, -1, sizeof(ans));
    for(int i = 2; i <= n; i++){
        scanf("%d", &x);
        ve[x].push_back(i);
    }
    for(int i = 1; i <= n; i++){
        scanf("%d", &val[i]);
    }
    dfs(1);    
    for(int i = 1; i <= n; i++) printf("%d
", ans[i]);
    return 0;
}

2 .

 线段树合并

 

Count the number of cyclic permutations of length n with no continuous subsequence [i, i + 1 mod n].
Output the answer modulo 998244353.

InputThe first line of the input contains an integer T , denoting the number of test cases.
In each test case, there is a single integer n in one line, denoting the length of cyclic permutations.
1 ≤ T ≤ 20, 1 ≤ n ≤ 100000OutputFor each test case, output one line contains a single integer, denoting the answer modulo 998244353.Sample Input

3
4
5
6

Sample Output

1
8
36

题意 ： 有一个循环全排列，求相邻的位置不存在 [i, i+1] 以及 [n, 1] 的排列的方案数有多少个？
思路分析 ：
　　　　好菜啊..学的假的组合数学吧....
　

首先先说明什么是循环排列：

即把1-n这n个数随意地放到一个圆圈上，循环排列的不同仅仅取决于这n个数的相对位置的不同。

例如1234，2341，3412，4123这些数为相同的循环排列数。

循环排列没有首末之分，这四个元素随便从哪一个元素开始，绕一个方向转过去，都不改变它们的相对顺序；直线排列则首末分明，原来排末位，调换排首位，已改变它们的相对顺序。循环排列与直线排列的主要区别就在这一点上。

从例子看出，直线排列的个数是循环排列个数的n倍

由直线排列个数为n!可推知循环排列个数为(n-1)!。

讲完了循环排列，再来看此题是求不含子串[i,i+1]或[n,1]（以下简称顺序子串，共有n个）的循环排列个数

因为一个排列中可能含有多个顺序子串，所以我们列举至少含有0个，1个，...n个的情况  （注意是至少，因为无法保证恰好含有i个）

包含至少一个顺序子串的循环排列数为C(n,1)*(n-2)!

包含至少两个顺序子串的循环排列数为C(n,2)*(n-3)!

...

包含至少k个顺序子串的循环排列数为C(n,k)*(n-k-1)! 

（为什么是(n-k-1)! 当你选出了k个子串之后，至少有k+1个数相对位置已被确定，我们让剩下的(n-k-1)个数全排列即可。）

同时注意到包含n个顺序子串的循环排列数一定是1个。

事件之间相互包含，所以用到容斥原理：

 ∑(k从0到n-1)(-1)^k*C(n,k)*(n-k-1)!+(-1)^n*1