题目描述
Recently, Miss Huang want to receive a Tree as her birthday gift! (What a interesting person!) As a interesting person, Miss Huang likes interesting tree.
The interesting tree is a tree with the Max interesting degree. The interesting degree equals LCM(H(1),H(2), .. , H(N)), H(x) means the height of node x in the tree and LCM means least common multiple.
Now Mr Chen has some non-rooted tree, he wants you to choose a root of the non-rooted tree and make the strange degree of tree be maximum;
The answer may be very large please print the answer mod 10^9+7.
输入
Your program will be tested on one or more test cases. In each test case:
First line a number N(1<=N<=10^5) represent the number of node.
The next N-1 line has two number u, v indicates that there is a edge between u, v.
输出
For every test case, print the following line:
answer
where answer is the maximum LCM.(mod 10^9+7)
样例输入
41 21 31 421 2
样例输出
62
题解:问题分两步:
1.求树的直径:即树种距离最长的两个点之间的距离
2.求从1到n这n个数的最小公倍数.
先说第一个问题,有两种方法:深搜和广搜.
先说深搜:对于树的某个节点,它的最深的两个儿子m1,m2.
那么很显然m1+m2+1有可能是树的直径.用这个数更新树的直径
再说广搜:1.随机选取一个节点A进行广搜,从而找到距离它最远的结点B.当然B可能不唯一.
2.B必然是此树某个直径的一个端点(树的直径可能不唯一).
从B开始广搜找到离B最远的C结点.BC极为直径.
再说第二个问题:有两种方法,其中一种是错的.
先说正确的:比如从1到9.既然有8,那么4,2就没法说话了,山上有老虎,猴子怎能称大王.
既然有9,那么3就没法说话了.
规律出来了:全体质数打表求出来,2,3,5,7.
再说错误的: 对于任意一个序列a[],求其全部元素的最小公倍数.怎么求?
求其全部元素的最大公约数怎么求?这个你会.两两求,从头扫描到尾.
但是这道题里不行,必须进行质因数分解,因为有取模运算!!!!!!!!!! 取模之后就没法再求最大公倍数了!!!!!!!!!!!!!!!!
比赛的时候,我就错在了这里,我真傻真的
#include<iostream>
#include<list>
#include<math.h>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn = 1e5 + 7;
const int big = 1e9 + 7;
int N;
struct Edge{
int to, next;
}e[maxn*2];
int g[maxn],ei;
bool vis[maxn];
int maxLen;
int prime[maxn / 5], psize;
void init(){
bool is[maxn];
psize = 0;
memset(is, 1, sizeof(is));
for (int i = 2; i < maxn; i++){
if (is[i]){
prime[psize++] = i;
}
for (int j = 0; i*prime[j] < maxn; j++){
is[i*prime[j]] = 0;
if (i%prime[j] == 0)break;
}
}
}
void push_back(int from, int to){
e[ei].next = g[from];
e[ei].to = to;
g[from] = ei++;
}
int deep(int x){
int m1, m2;
vis[x] = 1, m1 = m2 = 0;
for (int i = g[x]; i; i = e[i].next){
int t = e[i].to;
if (vis[t] == false){
int d = deep(t);
if (m2<d){
if (m1<d){ m2 = m1, m1 = d; }
else m2 = d;
}
}
}
int len = m1 + m2 + 1;
if (len > maxLen)maxLen = len;
return m1 + 1;
}
int main(){
//freopen("in.txt", "r", stdin);
init();
while (~scanf("%d", &N)){
ei = 1, memset(g, 0, sizeof(int)*(N + 1));
int x, y; for (int i = 1; i < N; i++)scanf("%d%d", &x, &y), push_back(x, y), push_back(y, x);
memset(vis, 0, sizeof(bool)*(N + 1));
maxLen = 0;
deep(1);
long long int ans = 1;
for (int i = 0; i < psize; i++){
int mi = floor(log(maxLen) / log(prime[i]));
if (mi == 0)break;
ans *= pow(prime[i], mi);
ans %= big;
}
cout << ans << endl;
}
return 0;
}
/**************************************************************
Problem: 1578
User: 20124003
Language: C++
Result: 正确
Time:181 ms
Memory:9568 kb
****************************************************************/