codeforces 551 E. GukiZ and GukiZiana （分块）

最新推荐文章于 2019-10-07 16:05:30 发布

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最新推荐文章于 2019-10-07 16:05:30 发布

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分类专栏：分块

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本文介绍了一个名为 GukiZiana 的函数，该函数用于处理数组查询问题，包括增加指定范围内元素的值及查找特定值的最大跨度。通过分块算法优化了查询效率。

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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 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:

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.

Examples

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

题解：分块

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define N 500003
#define LL long long
using namespace std;
int belong[N],l[N],r[N];
LL val[N],delta[N];
struct point
{
	LL x; int pos;
};
struct data{
	point a[720];
}c[720];
int n,m;
int cmp(point a,point b)
{
	return a.x<b.x||a.x==b.x&&a.pos<b.pos;
}
void build(int x)
{
	int cnt=0;
	for (int i=l[x];i<=r[x];i++) c[x].a[++cnt].x=val[i],c[x].a[cnt].pos=i;
	sort(c[x].a+1,c[x].a+cnt+1,cmp);
}
void calc(int &x,int &y,int k,LL v)
{
	LL vl=v-delta[k];
	if (vl<=0) return;
	int ls=1; int rs=r[k]-l[k]+1; int ans=n+1;
	while (ls<=rs) {
		int mid=(ls+rs)/2;
		if (c[k].a[mid].x>=vl) ans=min(ans,mid),rs=mid-1;
		else ls=mid+1; 
	}
	if (c[k].a[ans].x==vl&&ans!=n+1) 
	 x=min(x,c[k].a[ans].pos),y=max(y,c[k].a[ans].pos);
	ls=1; rs=r[k]-l[k]+1;  ans=0;
	while (ls<=rs) {
		int mid=(ls+rs)/2;
		if (c[k].a[mid].x<=vl) ans=max(ans,mid),ls=mid+1;
		else rs=mid-1;
	}
	if (c[k].a[ans].x==vl&&ans!=0) 
	  x=min(x,c[k].a[ans].pos),y=max(y,c[k].a[ans].pos);
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%I64d",&val[i]);
	int block=sqrt(n);
	int t=ceil(n*1.0/block);
	for (int i=1;i<=n;i++) belong[i]=(i-1)/block+1;
    for (int i=1;i<=t;i++) l[i]=n,r[i]=0;
    for (int i=1;i<=n;i++) {
    	int k=belong[i];
    	l[k]=min(i,l[k]);
    	r[k]=max(r[k],i);
	}
	for (int i=1;i<=t;i++) build(i);
	for (int i=1;i<=m;i++) {
		int opt,x,y; LL k;
		scanf("%d",&opt);
		if (opt==1) {
			scanf("%d%d%I64d",&x,&y,&k);
			if (belong[x]==belong[y]) {
				for (int i=x;i<=y;i++) val[i]+=k;
				build(belong[x]); continue;
			}
			for (int i=x;i<=r[belong[x]];i++) val[i]+=k;
			build(belong[x]);
			for (int i=l[belong[y]];i<=y;i++) val[i]+=k;
			build(belong[y]);
			for (int i=belong[x]+1;i<=belong[y]-1;i++) delta[i]+=k;
		}
		else {
			scanf("%I64d",&k);
			int x=n+1; int y=0;
			for (int i=1;i<=t;i++) 
			 calc(x,y,i,k);
			printf("%d\n",max(-1,y-x));
		}
	}
}