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,pos),indicating to change the value of apos to apos+10,000,000;
- (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
", 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;
}