HDU-6703 array

Description

You are given an array a1,a2,...,an(∀i∈[1,n],1≤ai≤n). Initially, each element of the array is unique.

Moreover, there are m instructions.

Each instruction is in one of the following two formats:

1. (1,pos),indicating to change the value of apos to apos+10,000,000;
2. (2,r,k),indicating to ask the minimum value which is not equal to any ai ( 1≤i≤r ) and **not less ** than k.

Please print all results of the instructions in format 2.

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, there are two integers n(1≤n≤100,000),m(1≤m≤100,000) in the first line, denoting the size of array a and the number of instructions.

In the second line, there are n distinct integers a1,a2,...,an (∀i∈[1,n],1≤ai≤n),denoting the array.

For the following m lines, each line is of format (1,t1) or (2,t2,t3).

The parameters of each instruction are generated by such way :

For instructions in format 1 , we defined pos=t1⊕LastAns . (It is promised that 1≤pos≤n)

For instructions in format 2 , we defined r=t2⊕LastAns,k=t3⊕LastAns. (It is promised that 1≤r≤n,1≤k≤n )

(Note that ⊕ means the bitwise XOR operator. )

Before the first instruction of each test case, LastAns is equal to 0 .After each instruction in format 2, LastAns will be changed to the result of that instruction.

(∑n≤510,000,∑m≤510,000 )

Output

For each instruction in format 2, output the answer in one line.

Sample Input

3
5 9
4 3 1 2 5
2 1 1
2 2 2
2 6 7
2 1 3
2 6 3
2 0 4
1 5
2 3 7
2 4 3
10 6
1 2 4 6 3 5 9 10 7 8
2 7 2
1 2
2 0 5
2 11 10
1 3
2 3 2
10 10
9 7 5 3 4 10 6 2 1 8
1 10
2 8 9
1 12
2 15 15
1 12
2 1 3
1 9
1 12
2 2 2
1 9

Sample Output

1
5
2
2
5
6
1
6
7
3
11
10
11
4
8
11

题解

首先理解对题意...题意给了两种操作，一种是将位于pos的数变成\(a_{pos}+10000000\),另一种是询问不等于\([1,r]\)中的任何一个数的，并且大于等于k的最小的数。也就是说这个询问的答案不一定是数组中的数。

因为题中给了限制，一开始数组中的数都满足\(1<=a_i<=n\)，且值各不相同，所以一旦进行1操作，这个数就可以被任何答案包含了，那我们就直接维护权值为下标，出现的位置为值得线段树，每次修改就是单点修改，将\(a_{pos}\)的值修改为n+1，询问就是查询区间[k,n+1]内，第一个值大于r的下标，直接利用线段树的二分性查找即可

AC代码

#include <bits/stdc++.h>
#define lson (o << 1)
#define rson (o << 1 | 1)
using namespace std;
const int N = 1e5 + 50;
int a[N], pos[N];
int maxv[N << 2];
int n, m;
void pushup(int o) {
    maxv[o] = max(maxv[lson], maxv[rson]);
}
void build(int o, int l, int r) {
    if (l == r) {
        maxv[o] = pos[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid); build(rson, mid + 1, r);
    pushup(o);
}
int query(int o, int l, int r, int ql, int qr, int v) {
    if (ql > qr) return n + 1;
    if (l == r) {
        if (maxv[o] >= v) {
            return l;
        }
        else return n + 1;
    }
    if (ql <= l && r <= qr) {
        if (maxv[o] < v) return n + 1;
    }
    int mid = (l + r) >> 1;
    if (qr <= mid) return query(lson, l, mid, ql, qr, v);
    else if (ql > mid) return query(rson, mid + 1, r, ql, qr, v);
    else {
        int x = query(lson, l, mid, ql, mid, v);
        if (x != n + 1) return x;
        return query(rson, mid + 1, r, mid + 1, qr, v);
    }
}
void update(int o, int l, int r, int pos) {
    if (l == r) {
        maxv[o] = n + 1;
        return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) update(lson, l, mid, pos);
    else update(rson, mid + 1, r, pos);
    pushup(o);
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            pos[a[i]] = i;
        }
        pos[n + 1] = n + 1;
        build(1, 1, n + 1);
        int ans = 0;
        for (int i = 1; i <= m; i++) {
            int t;
            scanf("%d", &t);
            if (t == 2) {
                int r, k; scanf("%d%d", &r, &k);
                r ^= ans, k ^= ans;
                printf("%d\n", ans = query(1, 1, n + 1, k, n + 1, r + 1));
            }
            if (t == 1) {
                int x;
                scanf("%d", &x);
                x ^= ans;
                update(1, 1, n + 1, a[x]);
            }
        }
    }
    return 0;
}