And I've got a TLE on test 13.Anyone help me?

/*
ID: jerrywcy
LINK: http://codeforces.me/contest/13/problem/E
LANG: C++
STATUS:
*/
#include <bits/stdc++.h>

#define init(array,x) memset(array,x,sizeof(array))

using namespace std;

const int inf=0x3f3f3f3f;
const int maxn=100010;

int n,q;
int a[maxn];
int m,len;
int l[1010],r[1010];
int pos[maxn],nxt[maxn],step[maxn];

void build(){
	m=len=sqrt(n);
	for (int cur=1;cur<=m;cur++){
		l[cur]=len*(cur-1)+1;
		r[cur]=len*cur;
		for (int i=l[cur];i<=r[cur];i++){
			pos[i]=cur;
			int j=i;
			while (j<=r[cur] && j<=n){
				j=min(j+a[j],n+1);
				step[i]++;
			}
			nxt[i]=j;
		}
	}
	if (r[m]!=n){
		m++;
		l[m]=r[m-1]+1;
		r[m]=n;
		for (int i=l[m];i<=r[m];i++){
			pos[i]=m;
			int j=i;
			while (j<=r[m] && j<=n){
				j=min(j+a[j],n+1);
				step[i]++;
			}
			nxt[i]=j;
		}
	}
	return ;
}

void add(int x,int y){
	int bl=pos[x];
	a[x]=y;
	for (int i=x;i>=l[bl];i--){
		nxt[i]=max(nxt[i+a[i]],n+1),step[i]=step[i+a[i]]+1;
	}
	return ;
}

void query(int x){
	int cnt=0;
	while (nxt[x]<=n){
		cnt+=step[x];
		x=nxt[x];
	}
	while (x+a[x]<=n)x+=a[x],cnt++;
	printf("%d %d\n",x,cnt+1);
	return ;
}

void solve(){
	int opt;
	scanf("%d",&opt);
	if (!opt){
		int x,y;
		scanf("%d %d",&x,&y);
		add(x,y);
	}
	else{
		int x;
		scanf("%d",&x);
		query(x);
	}
}

int main()
{
	scanf("%d %d",&n,&q);
	for (int i=1;i<=n;i++)scanf("%d",&a[i]);
	build();
	while (q--)solve();

	return 0;
}

Anyone noticed that Problem E also required the number of the last hole the ball visited?

So we'll have to jump to the block before we jump to the fictious hole and jump to the last hole before we jump to the fictious hole. So I had to add another while loop.

But it seems that the complexity is the same? Well, I'm just here to say that the type 1 query is not that easy(probably still easy for great coder,but not for this noob)