题意：两棵树，求不同树上两个节点的最近公共祖先
思路：比赛时看过的队伍不是很多，没有仔细想。今天补题才发现有个 倍增算法，自己竟然不知道。
解法来自 qscqesze ，这个其实之前如果了解过倍增的话还是不是很难，不过这题的数据也不是很给力，极限数据理论上是过不了的。
其他解法有树链剖分？并不是很清楚。就这样水过了吧。。。
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define eps 1e-8
#define INF 0x3f3f3f3f
#define MAXN 100005
using namespace std; int f1[MAXN][], f2[MAXN][];
int deep1[MAXN], deep2[MAXN];
int step1, step2, ans;
void work(int x, int y){
step1 = step2 = ;
while (x != y){
if (x < y){
//x < y means y is not x's ancestor, so let y up
for (int i = ; i >= ; i--){
if (f2[y][i] > x){
y = f2[y][i];
step2 += << i;
break;
}
}
y = f2[y][];
step2++;
}
else{
for (int i = ; i >= ; i--){
if (f1[x][i] > y){
x = f1[x][i];
step1 += << i;
break;
}
}
x = f1[x][];
step1++;
}
}
ans = x;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
int n, m;
while (~scanf("%d%d", &n, &m)){
int x, y;
deep1[] = deep2[] = ;
for (int i = ; i <= ; i++){
f1[][i] = f2[][i] = ;
}
for (int i = ; i <= n; i++){
scanf("%d", &x);
f1[i][] = x;
deep1[i] = deep1[x] + ;
for (int j = ; j <= ; j++){
f1[i][j] = f1[f1[i][j - ]][j - ];
}
}
for (int i = ; i <= n; i++){
scanf("%d", &x);
f2[i][] = x;
deep2[i] = deep2[x] + ;
for (int j = ; j <= ; j++){
f2[i][j] = f2[f2[i][j - ]][j - ];
}
}
ans = ;
for (int i = ; i <= m; i++){
scanf("%d%d", &x, &y);
x = (x + ans) % n + ;
y = (y + ans) % n + ;
work(x, y);
printf("%d %d %d\n", ans, step1, step2);
}
}
}