CodeForces 551E GukiZ and GukiZiana

GukiZ and GukiZiana

Time Limit: 10000ms

Memory Limit: 262144KB

This problem will be judged on  CodeForces. Original ID: 551E
64-bit integer IO format: %I64d      Java class name: (Any)

 

Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and numbery, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:

1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
2. Second type has form 2 y and asks you to find value of GukiZiana(a, y).

For each query of type 2, print the answer and make GukiZ happy!

 

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

 

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

 

Sample Input

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3

Output

2

Input

2 3
1 2
1 2 2 1
2 3
2 4

Output

0
-1

Source

Codeforces Round #307 (Div. 2)

 

解题：分块搞

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 1010;
 5 LL a[maxn*maxn],lazy[maxn],x;
 6 vector<int>block[maxn];
 7 int b_size,N,pos[maxn*maxn],n,q,cmd,L,R;
 8 bool cmp(const int x,const int y) {
 9     if(a[x] == a[y]) return x < y;
10     return a[x] < a[y];
11 }
12 void update(int L,int R,LL x) {
13     int k = pos[L],t = pos[R];
14     if(k == t) {
15         for(int i = L; i <= R; ++i) a[i] += x;
16         sort(block[k].begin(),block[k].end(),cmp);
17         return;
18     }
19     for(int i = k + (pos[L-1] == k); i <= t - (pos[R + 1] == t); ++i) lazy[i] += x;
20     if(pos[L-1] == k) {
21         for(int i = L; pos[i] == k; ++i) a[i] += x;
22         sort(block[k].begin(),block[k].end(),cmp);
23     }
24     if(pos[R+1] == t) {
25         for(int i = R; pos[i] == t; --i) a[i] += x;
26         sort(block[t].begin(),block[t].end(),cmp);
27     }
28 }
29 LL query(LL x) {
30     int L = -1,R = -1,i;
31     for(i = 1; i <= N; ++i){
32         a[0] = x - lazy[i];
33         vector<int>::iterator it = lower_bound(block[i].begin(),block[i].end(),0,cmp);
34         if(it == block[i].end()) continue;
35         if(a[*it] + lazy[i] == x){
36             L = *it;
37             break;
38         }
39     }
40     if(L == -1) return -1;
41     for(int j = N; j >= i; --j){
42         a[n+1] = x - lazy[j];
43         vector<int>::iterator it = lower_bound(block[j].begin(),block[j].end(),n+1,cmp);
44         if(it == block[j].begin()) continue;
45         --it;
46         if(a[*it] + lazy[j] == x){
47             R = *it;
48             break;
49         }
50     }
51     return R - L;
52 }
53 int main() {
54     ios::sync_with_stdio(false);
55     cin.tie(0);
56     cin>>n>>q;
57     b_size = ceil(sqrt(n*1.0));
58     for(int i = 1; i <= n; ++i) {
59         cin>>a[i];
60         pos[i] = (i - 1)/b_size + 1;
61         block[pos[i]].push_back(i);
62     }
63     N = (n - 1)/b_size + 1;
64     for(int i = 1; i <= N; ++i) sort(block[i].begin(),block[i].end(),cmp);
65     while(q--) {
66         cin>>cmd;
67         if(cmd == 1) {
68             cin>>L>>R>>x;
69             update(L,R,x);
70         } else {
71             cin>>x;
72             cout<<query(x)<<endl;
73         }
74     }
75     return 0;
76 }

View Code

 

转载于:https://www.cnblogs.com/crackpotisback/p/4796020.html