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

最新推荐文章于 2024-07-17 22:33:54 发布

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最新推荐文章于 2024-07-17 22:33:54 发布

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原文链接：http://www.cnblogs.com/jianrenfang/p/6656819.html

本文介绍了一种通过分块技术优化大规模数组操作的方法。面对数组的区间加操作和查询特定元素最远距离的问题，文章详细阐述了如何通过预处理、二分查找及懒惰标记来提高效率。

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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 a_j = a_i = 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:

First type has form 1 l r x and asks you to increase values of all a_i 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, ... a_n (1 ≤ a_i ≤ 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.

Examples

Input

Output

Input

Output

0
-1
【分析】给出一个数组a，然后q次操作，1 l r x表示将区间[l,r]所有数+1 ； 2 y表示询问这个数组中最左边的y和最右边的y的下标差为多少。
 可分块做。将所有块里面的数排序，（便于查找要查询的y）。对于更新操作，单独处理左端点和右端点的块，改变a[i];然后对于中间跨过的块，用lazy数组标记add值。
 对于查询操作，依次遍历每个块，二分查找y的位置即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 1000000000
#define met(a,b) memset(a,b,sizeof a)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
using namespace std;
const int N = 5e5+5;
const int M = 7e2+50;
int n,m;
int l[N],r[N],belong[N];
int cnt,num,x,v,ans;
int a[N],tot[M],lazy[M];
pair<int,int>p[M][M];
void init(){
    num=sqrt(n);
    cnt=n/num;
    if(n%num)cnt++;
    for(int i=1;i<=n;i++){
        belong[i]=(i-1)/num+1;
    }
    for(int i=1;i<=cnt;i++){
        l[i]=(i-1)*num+1;
        r[i]=min(n,i*num);
        for(int j=l[i];j<=r[i];j++){
            p[i][++tot[i]]=make_pair(a[j],j);
        }
        sort(p[i]+1,p[i]+1+tot[i]);
    }
}
void update1(int b,int x){
    if(lazy[b]+x>inf)lazy[b]=inf+1;
    else lazy[b]+=x;
}
void update2(int b,int ll,int rr,int x){
    if(ll==l[b]&&rr==r[b]){
        update1(b,x);
        return;
    }
    for(int i=1;i<=tot[b];i++){
        int po=p[b][i].second;
        if(po>=ll&&po<=rr){
            p[b][i].first+=x;
            if(p[b][i].first>inf)p[b][i].first=inf+1;
        }
    }
    sort(p[b]+1,p[b]+tot[b]+1);
}
int main() {
    int op,ll,rr,x,y;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    init();
    while(m--){
        scanf("%d",&op);
        if(op==1){
            scanf("%d%d%d",&ll,&rr,&x);
            if(belong[ll]==belong[rr])update2(belong[ll],ll,rr,x);
            else {
                for(int i=belong[ll]+1;i<=belong[rr]-1;i++)update1(i,x);
                update2(belong[ll],ll,r[belong[ll]],x);
                update2(belong[rr],l[belong[rr]],rr,x);
            }
        }
        else {
            scanf("%d",&y);
            int last=0,fir=0;
            for(int i=cnt;i>=1;i--){
                if(!last){
                    int po=upper_bound(p[i]+1,p[i]+tot[i]+1,make_pair(y-lazy[i],1000000))-p[i]-1;
                    if(p[i][po].first+lazy[i]==y){
                        fir=last=p[i][po].second;
                    }
                }
                int po=lower_bound(p[i]+1,p[i]+tot[i]+1,make_pair(y-lazy[i],0))-p[i];
                if(p[i][po].first+lazy[i]==y){
                    fir=p[i][po].second;
                }
            }
            if(!last)puts("-1");
            else printf("%d\n",last-fir);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/jianrenfang/p/6656819.html