Codeforces Round #307 (Div. 2)E. GukiZ and GukiZiana（分块）

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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 number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is noy 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:

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.
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 test(s)
input
4 31 2 3 41 1 2 11 1 1 12 3
output
2
input
2 31 21 2 2 12 32 4
output
0-1

比较裸的分块题，开始我是分块然后用一个map套set搞的，结果超时了，这种静态的就应该用数组加二分嘛！，复杂度是q*b*log(b),b = sqrt(n),自己写的很挫，借用cf上别人的代码一下，感觉那个query写的很机智呀。

#include <bits/stdc++.h>using namespace std;typedef long long ll;vector<int> s[1000];ll add[1000], a[512346], pos[512346];ll n, q, bk, N , ti;bool cmp( int x, int y ){    if( a[x] == a[y]  ) return x < y;    return a[x] < a[y];}inline void modify( ll l, ll r, ll x ){    int k = pos[l], t = pos[r];    if( k == t ) {        for( ll i = l; i <= r; ++i )            a[i] += x;        sort( s[k].begin(), s[k].end(), cmp );        return ;    }    for( ll i = k + ( pos[l - 1] == k ); i <= t - ( pos[r + 1] == t ) ; ++i )        add[i] += x;    if( pos[l - 1] == k ) {        for( ll i = l; pos[i] == k; ++i ) {            a[i] += x;        }        sort( s[k].begin(), s[k].end(), cmp );    }    if( pos[r + 1] == t ) {        for( ll i = r; pos[i] == t; --i ) {            a[i] += x;        }        sort( s[t].begin(), s[t].end(), cmp );    }}inline ll query( ll x ){    int l = -1, r = -1, i;    for( i = 1; i <= N; ++i ) {        a[0] = x - add[i];        auto it = lower_bound( s[i].begin(), s[i].end(), 0, cmp );        if( it == s[i].end() ) continue;        if( a[*it] + add[i] == x ) {            l = *it;            break;        }    }    if( l == -1 ) return l;    for( int  j = N; j >= i; --j ) {        a[n + 1] = x - add[j];        auto it = lower_bound( s[j].begin(), s[j].end(), n + 1, cmp );        if( it == s[j].begin() ) continue;        it--;        if( a[*it] + add[j] == x ) {            r = *it;            break;        }    }    return r - l;}int main(){    ios::sync_with_stdio( 0 );    cin >> n >> q;    bk = ceil( sqrt( 1.*n ) + 0.005 );    for( int i = 1; i <= n; ++i ) {        cin >> a[i];        pos[i] = ( i - 1 ) / bk + 1;        s[pos[i]].push_back( i );    }    N = ( n - 1 ) / bk + 1;    for( int i = 1; i <= N; ++i ) {        sort( s[i].begin(), s[i].end(), cmp );    }    ll cmd, l, r, x;    for( int i = 1; i <= q; ++i ) {        cin >> cmd;        if( cmd == 1 ) {            cin >> l >> r >> x;            modify( l, r, x );        } else {            ++ti;            cin >> x;            cout << query( x ) << "\n";        }    }}

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