Codeforces Beta Round #13 E. Holes （分块）-白红宇

Codeforces Beta Round #13 E. Holes （分块）

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发布时间：2019-06-24

本文共 3426 字，大约阅读时间需要 11 分钟。

Little Petya likes to play a lot. Most of all he likes to play a game «Holes». This is a game for one person with following rules:

There are N holes located in a single row and numbered from left to right with numbers from 1 to N. Each hole has it's own power (hole number i has the power a_i). If you throw a ball into hole i it will immediately jump to hole i + a_i, then it will jump out of it and so on. If there is no hole with such number, the ball will just jump out of the row. On each of the M moves the player can perform one of two actions:

Set the power of the hole a to value b.

Throw a ball into the hole a and count the number of jumps of a ball before it jump out of the row and also write down the number of the hole from which it jumped out just before leaving the row.

Petya is not good at math, so, as you have already guessed, you are to perform all computations.

Input

The first line contains two integers N and M (1 ≤ N ≤ 10⁵, 1 ≤ M ≤ 10⁵) — the number of holes in a row and the number of moves. The second line contains N positive integers not exceeding N — initial values of holes power. The following M lines describe moves made by Petya. Each of these line can be one of the two types:

0 a b

Type

0 means that it is required to set the power of hole a to b, and type 1 means that it is required to throw a ball into the a-th hole. Numbers a and b are positive integers do not exceeding N.

Output

For each move of the type 1 output two space-separated numbers on a separate line — the number of the last hole the ball visited before leaving the row and the number of jumps it made.

Examples

Input

8 5 1 1 1 1 1 2 8 2 1 1 0 1 3 1 1 0 3 4 1 2

Output

8 7 8 5 7 3 【分析】数轴上n个点，没个点 i 上有一个值a[i]，表示站在i点上下一步将跳向i+a[i]点，可能跳出区间[1,n]。然后两次操作，第一种是将某个a[i]改成x，另一种就是询问从x节点开始  需要多少步才能跳出去，最后一个经过的点是哪一个。  可以分块做。对于每个块，统计每个点需要多少步跳出此块，并且跳出此块后将跳向哪个点。然后就很简单了，详细看代码

#include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 #include 
                  
                   #include 
                   
                    #include 
                    
                     #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|1typedef long long ll;using namespace std;const int N = 1e5+5;const int M = 4e2+50;int n,m;int l[N],r[N],belong[N];int cnt,num,x,v,ans;int a[N],tot[M],go[N],nxt[N];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=r[i];j>=l[i];j--){ if(nxt[j]>r[i]){ tot[j]=1; go[j]=nxt[j]; } else { tot[j]=tot[nxt[j]]+1; go[j]=go[nxt[j]]; } } }}int main() { int op,ll,rr,x,y; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++)scanf("%d",&a[i]),nxt[i]=min(n+1,a[i]+i); init(); while(m--){ scanf("%d",&op); if(!op){ scanf("%d%d",&x,&y); int b=belong[x]; nxt[x]=min(n+1,x+y); for(int j=r[b];j>=l[b];j--){ if(nxt[j]>r[b]){ tot[j]=1; go[j]=nxt[j]; } else { tot[j]=tot[nxt[j]]+1; go[j]=go[nxt[j]]; } } } else { scanf("%d",&x); int b,y; int ans1,ans2=0; while(x<=n){ ans2+=tot[x]; if(go[x]>n)y=x; x=go[x]; } while(y<=n){ ans1=y; y=nxt[y]; } printf("%d %d\n",ans1,ans2); } } return 0;}

转载于:https://www.cnblogs.com/jianrenfang/p/6674426.html

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