array
Time Limit: 1500ms Memory Limit: 256M Description
You are given an array . Initially, each element of the array is unique.
Moreover, there are instructions. Each instruction is in one of the following two formats: 1. ,indicating to change the value of to ; 2. ,indicating to ask the minimum value which is not equal to any ( ) and not less than . Please print all results of the instructions in format .
Input
The first line of the input contains an integer , denoting the number of test cases. In each test case, there are two integers , in the first line, denoting the size of array and the number of instructions.
In the second line, there are distinct integers ,denoting the array. For the following lines, each line is of format or . The parameters of each instruction are generated by such way : For instructions in format , we defined . (It is promised that )
For instructions in format , we defined . (It is promised that ) (Note that means the bitwise XOR operator. ) Before the first instruction of each test case, is equal to .After each instruction in format , will be changed to the result of that instruction.
( ) Output
For each instruction in format , 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 [hint] note: After the generation procedure ,the instructions of the first test case are : 2 1 1, in format 2 and r=1 , k=1 2 3 3, in format 2 and r=3 , k=3 2 3 2, in format 2 and r=3 , k=2 2 3 1, in format 2 and r=3 , k=1 2 4 1, in format 2 and r=4 , k=1 2 5 1, in format 2 and r=5 , k=1 1 3 , in format 1 and pos=3 2 5 1, in format 2 and r=5 , k=1 2 5 2, in format 2 and r=5 , k=2
the instructions of the second test case are : 2 7 2, in format 2 and r=7 , k=2 1 5 , in format 1 and pos=5 2 7 2, in format 2 and r=7 , k=2 2 8 9, in format 2 and r=8 , k=9 1 8 , in format 1 and pos=8 2 8 9, in format 2 and r=8 , k=9 the instructions of the third test case are : 1 10 , in format 1 and pos=10 2 8 9 , in format 2 and r=8 , k=9 1 7 , in format 1 and pos=7 2 4 4 , in format 2 and r=4 , k=4 1 8 , in format 1 and pos=8 2 5 7 , in format 2 and r=5 , k=7 1 1 , in format 1 and pos=1 1 4 , in format 1 and pos=4 2 10 10, in format 2 and r=10 , k=10 1 2 , in format 1 and pos=2 [/hint]
#include <bits / STDC ++ H.>
#pragma Optimize the GCC (. 3)
the using namespace STD;
typedef int LL;
const int = 1E5 + MAXN. 7;
const LL = INF 1E9;
int n-, m, str2 [MAXN]; // str2 original value is saved
struct Inti {
int Data, POI;
} STR [MAXN]; // save the original value after sorting, to facilitate the establishment of segment tree
struct Node {
int L, R & lt;
int MAXR;
} ARR [MAXN. 4 * ]; // segment tree structure array
BOOL CMP (Inti A, B Inti) {
return a.data <b.data;
}
void Build (int P, L int, int R & lt) {// contribution
arr [p]. L = L, ARR [P] .r = R & lt;
IF (L == R & lt) {
ARR [P] .Maxr STR = [L] .poi;
return;
}
int MID = (L + R & lt) / 2;
Build (p * 2, l, mid );
Build (p * 2 + 1, mid + 1, r);
arr [p] .Maxr = max ( arr [p * 2] .Maxr, arr [p * 2 + 1] .Maxr); // update from the bottom
}
// void Change (P int, int X) { the value of the subscript x is updated. 1 + n-
IF (ARR [P] .L == ARR [P] .r) {
ARR [P] = n-.Maxr +. 1;
return;
}
int = MID (ARR [ P] .L + ARR [P] .r) / 2;
IF (X <= MID) Change (* P 2, X);
the else Change (. 1 + 2 * P, X);
ARR [P] = .Maxr max (arr [p * 2] .Maxr, arr [p * 2 + 1] .Maxr); // from the maximum interval to update the index
}
(R & lt, R & lt int int P, L int, int) int ASK lr {// interrogation interval minimum weight and the subscript value larger than R, where r seems useless in fact to as +. 1 are n-
IF (ARR [P] .Maxr <= R). 1 + n-return ;
IF (ARR [P] .L == ARR [P] .r && ARR [P] .L> = L) {
IF (ARR [P] .Maxr> R & lt) return ARR [P] .L;
return. 1 + n- ;
}
int = MID (ARR [P] .L + ARR [P] .r) / 2;
int Val = n + 1; n + 1 is a maximum //
if(l<=mid)val=min(val,ask(p*2,l,r,R));//取最小值
if(r>mid)val=min(val,ask(p*2+1,l,r,R));//取最小值
return val;
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
str[n+1].data=n+1,str[n+1].poi=n+1;
for(int i=1;i<=n;++i)
scanf("%d",&str[i].data),str[i].poi=i,str2[i]=str[i].data;
sort(str+1,str+n+1,cmp);
build(1,1,n+1);
int id,a,b,lastans=0;
while(m--){
scanf("%d",&id);
if(id==1){
scanf("%d",&a);
change(1,str2[a^lastans]);
}
else{
scanf("%d %d",&a,&b);
int r=a^lastans,k=b^lastans;
int ans=ask(1,k,n+1,r);
printf("%d\n",ans);
lastans=ans;//更新lastans
}
}
}
return 0;
}