codeforces contest 13 problem E(分块)

E. Holes

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

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 toN. Each hole has it's own power (hole numberi has the power a_i). If you throw a ball into holei 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 theM 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 andM (1 ≤ N ≤ 10⁵,1 ≤ M ≤ 10⁵) — the number of holes in a row and the number of moves. The second line containsN 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
• 1 a

Type 0 means that it is required to set the power of holea to b, and type1 means that it is required to throw a ball into thea-th hole. Numbers a and b are positive integers do not exceedingN.

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 51 1 1 1 1 2 8 21 10 1 31 10 3 41 2

Output

8 78 57 3

这题的操作是真的骚

如果当前点跳的下个点在块内 直接延伸到下一个点的后续结点相当于 把整个块都联系起来了 如果不在一个块内 就将后续结点设置为 后续结点的编号

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>#include <bits/stdc++.h>using namespace std;const int N = 300000+10;typedef  long long LL;const LL mod = 998244353;int a[N],next1[N],cnt[N];int lx[N],rx[N],id[N],end1[N];int n, m;void add(int i,int pos){    if(pos>n)next1[i]=n+1, cnt[i]=1, end1[i]=i;    else    {        if(id[i]==id[pos])        {            next1[i]=next1[pos];            cnt[i]=cnt[pos]+1;            end1[i]=end1[pos];        }        else        {            next1[i]=pos;            cnt[i]=1;            end1[i]=i;        }    }    return ;}int ans;int solve(int x){    int c=cnt[x];    ans=end1[x];    while(x<=n)    {        x=next1[x];        if(x>n)break;        c+=cnt[x];        ans=end1[x];    }    return c;}int main(){    scanf("%d %d", &n, &m);    for(int i=1;i<=n;i++) scanf("%d", &a[i]);    int k=(int)sqrt(1.0*n),i;    for(i=1;i*k<n;i++)    {        lx[i]=(i-1)*k+1,rx[i]=i*k;        for(int j=lx[i];j<=rx[i];j++) id[j]=i;    }    lx[i]=(i-1)*k+1,rx[i]=n;    for(int j=lx[i];j<=rx[i];j++) id[j]=i;    for(int i=n;i>=1;i--) add(i,i+a[i]);    while(m--)    {        int x, u, v;        scanf("%d", &x);        if(x)        {            scanf("%d", &u);            printf("%d %d\n",ans,solve(u));        }        else        {            scanf("%d %d", &u, &v);            add(u,u+v);            for(int i=u-1;i>=lx[id[u]];i--) add(i,i+a[i]);            a[u]=v;        }    }    return 0;}