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.
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 9Sample 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\n", ans[i]); return 0; }
2 .
Count the number of cyclic permutations of length n with no continuous subsequence [i, i + 1 mod n].
Output the answer modulo 998244353.
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 6Sample 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
东北日出西边雨 道是无情却有情