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

E. GukiZ and GukiZiana

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/551/problem/E

Description

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.

Sample Input

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

Sample Output

2

HINT

题意

一堆数，俩操作

[l,r]区间的数都加 v

查询最大的j-i,要求满足a[j]=a[i]=v

题解：

时间10s，这就是明摆着告诉我们这道题是分块，然后我们就分块乱搞就好了……

二分写挂的都是傻逼（没错就是我！

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

ll n,q,m,block;
ll a[1000001],b[1000001],pos[1000001],add[1000001];
void reset(ll x)
{
    ll l=(x-1)*block+1,r=min(x*block,n);
    for(ll i=l;i<=r;i++)
        b[i]=a[i];
    sort(b+l,b+r+1);
}
ll find(ll x,ll v)
{
    ll l=(x-1)*block+1,r=min(x*block,n);
    ll last=r;
    while(l<r)
    {
        ll mid=(l+r)>>1;
        if(b[mid]<v)l=mid+1;
        else
            r=mid;
        if(b[mid]==v)
            return 1;
    }
    if(b[l]==v||b[r]==v)
        return 1;
    return 0;
}
void update(ll x,ll y,ll v)
{
    if(pos[x]==pos[y])
    {
        for(int i=x;i<=y;i++)a[i]=a[i]+v;
    }
    else 
    {
        for(int i=x;i<=pos[x]*block;i++)a[i]=a[i]+v;
        for(int i=(pos[y]-1)*block+1;i<=y;i++)a[i]=a[i]+v;
    }
    reset(pos[x]);reset(pos[y]);
    for(int i=pos[x]+1;i<pos[y];i++)
       add[i]+=v;
}
ll query(ll x,ll y,ll v)
{
    ll sum=0;
    if(pos[x]==pos[y])
    {
        for(int i=x;i<=y;i++)if(a[i]+add[pos[i]]==v)sum++;
    }
    else 
    {
        for(int i=x;i<=pos[x]*block;i++)
            if(a[i]+add[pos[i]]==v)sum++;
        for(int i=(pos[y]-1)*block+1;i<=y;i++)
            if(a[i]+add[pos[i]]==v)sum++;
    }
    for(int i=pos[x]+1;i<pos[y];i++)
        sum+=find(i,v-add[i]);
    
    //cout<<sum<<endl;
    return sum;
}
ll deall(ll x,ll y,ll v)
{
    ll sum=0;
    if(pos[x]==pos[y])
    {
        for(int i=x;i<=y;i++)if(a[i]+add[pos[i]]==v)return i;
    }
    else 
    {
        for(int i=x;i<=pos[x]*block;i++)
            if(a[i]+add[pos[i]]==v)return i;
        for(int i=pos[x]+1;i<pos[y];i++)
        {
            int d=find(i,v-add[i]);
            if(d>0)
            {
                for(int j=i*block-block+1;j<=i*block;j++)
                    if(a[j]+add[pos[j]]==v)
                        return j;
            }
        }
        for(int i=(pos[y]-1)*block+1;i<=y;i++)
            if(a[i]+add[pos[i]]==v)return i;
    }
}
ll dealr(ll x,ll y,ll v)
{
    ll sum=0;
    if(pos[x]==pos[y])
    {
        for(int i=y;i>=x;i--)if(a[i]+add[pos[i]]==v)return i;
    }
    else 
    {
        for(int i=y;i>=(pos[y]-1)*block+1;i--)
            if(a[i]+add[pos[i]]==v)
                return i;

        for(int i=pos[y];i>=pos[x]+1;i--)
        {
            int d=find(i,v-add[i]);
            if(d>0)
            {
                for(int j=i*block;j>=i*block-block;j--)
                    if(a[j]+add[pos[j]]==v)
                        return j;
            }
        }
        for(int i=pos[x]*block;i>=x;i--)
            if(a[i]+add[pos[i]]==v)return i;

    }
}
int ask(ll cc)
{
    int aa=query(1,n,cc);
    if(aa==0)
        return -1;
    return dealr(1,n,cc)-deall(1,n,cc);
}
int main()
{
    //test;
    n=read(),q=read();
    block=int(sqrt(n));
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        pos[i]=(i-1)/block+1;
        b[i]=a[i];
    }
    int x,y,v,ch;
    if(n%block)m=n/block+1;
    else m=n/block;
    for(int i=1;i<=m;i++)reset(i);
    for(int i=1;i<=q;i++)
    {
        ch=read();
        if(ch==1)
        {
            x=read(),y=read(),v=read();
            update(x,y,v);
        }
        else
        {
            int p;
            p=read();
            printf("%d
",ask(p));
        }
    }
    return 0;
}