codeforces com contest 855 problem D(数位DP)

Rowena Ravenclaw's Diadem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Harry, upon inquiring Helena Ravenclaw's ghost, came to know that she told Tom Riddle or You-know-who about Rowena Ravenclaw's diadem and that he stole it from her.

Harry thought that Riddle would have assumed that he was the only one to discover the Room of Requirement and thus, would have hidden it there. So Harry is trying to get inside the Room of Requirement to destroy the diadem as he knows that it is a horcrux.

But he has to answer a puzzle in order to enter the room. He is given n objects, numbered from 1 to n. Some of the objects have a parent object, that has a lesser number. Formally, objecti may have a parent object parent_i such that parent_i < i.

There is also a type associated with each parent relation, it can be either of type1 or type 2. Type 1 relation means that the child object is like a special case of the parent object. Type2 relation means that the second object is always a part of the first object and all its special cases.

Note that if an object b is a special case of objecta, and c is a special case of objectb, then c is considered to be a special case of objecta as well. The same holds for parts: if objectb is a part of a, and objectc is a part of b, then we say that objectc is a part of a. Also note, that if objectb is a part of a, and objectc is a special case of a, then b is a part of c as well.

An object is considered to be neither a part of itself nor a special case of itself.

Now, Harry has to answer two type of queries:

• 1 u v: he needs to tell if object v is a special case of object u.
• 2 u v: he needs to tell if object v is a part of object u.

Input

First line of input contains the number n (1 ≤ n ≤ 10⁵), the number of objects.

Next n lines contain two integer parent_i and type_i ( - 1 ≤ parent_i < iparent_i ≠ 0, - 1 ≤ type_i ≤ 1), implying that thei-th object has the parent parent_i. (If type_i = 0, this implies that the objecti is a special case of object parent_i. If type_i = 1, this implies that the objecti is a part of object parent_i). In case the i-th object has no parent, both parent_i andtype_i are-1.

Next line contains an integer q (1 ≤ q ≤ 10⁵), the number of queries.

Next q lines each represent a query having three space separated integerstype_i, u_i, v_i (1 ≤ type_i ≤ 2, 1 ≤ u, v ≤ n).

Output

Output will contain q lines, each containing the answer for the corresponding query as "YES" (affirmative) or "NO" (without quotes).

You can output each letter in any case (upper or lower).

Examples

Input

3-1 -11 02 021 1 32 1 3

Output

YESNO

Input

3-1 -11 01 122 2 32 3 2

Output

YESNO

Note

In test case 1, as object 2 is a special case of object 1 and object 3 is a special case of object 2, this makes object3 a special case of object 1.

In test case 2, as object 2 is a special case of object 1 and object 1 has object 3, this will mean that object 2 will also have object 3. This is because when a general case (object1) has object 3, its special case (object2) will definitely have object 3.

题意：给出三个数b, l, r，求出从l到r中有多少数是魔法数
魔法数的定义是：b进制下所有数字出现次数都是偶数次

解：一开始多开了一维状态记录 每一位是否出现偶数次 结果T到死  因为最多只有10进制 直接用异或状态是否为0 就能快速判断 

还是太菜 没有关注到 进制只有10位这个有利条件

#include <bits/stdc++.h>using namespace std;typedef long long LL;int pos[100];LL dp[100][11][2048];int tmp[11];LL  l, r;int b;LL dfs(int px,int pre,int limit,int k,int num){    if(px==0)    {        if(k==0) return 0;        if(num!=0) return 0;        return 1;    }    if(!limit&&k!=0&&dp[px][b][num]!=-1) return dp[px][b][num];    int m=limit?pos[px]:b-1;    LL ans=0;    for(int i=0; i<=m; i++)    {        if(k==0)        {            if(i==0) ans+=dfs(px-1,pre,limit&&i==m,0,num);            else   ans+=dfs(px-1,i,limit&&i==m,1,num^(1<<i));        }        else   ans+=dfs(px-1,i,limit&&i==m,1,num^(1<<i));    }    if(!limit&&k!=0) dp[px][b][num]=ans;    return ans;}LL solve(LL x,int b){    if(x==0) return 0;    int k=0;    while(x)    {        pos[++k]=x%b;        x/=b;    }    return dfs(k,0,1,0,0);}int main(){    memset(dp,-1,sizeof(dp));    int q;    scanf("%d", &q);    while(q--)    {        scanf("%d %I64d %I64d",&b,&l,&r);        printf("%I64d\n",solve(r,b)-solve(l-1,b));    }    return 0;}