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

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：

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                   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-9const 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
                   
                    >1; if(b[mid]
                    
                     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\n",ask(p)); } } return 0;}

转载于:https://www.cnblogs.com/qscqesze/p/4572910.html

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