Robot Doc is located in the hall, with n computers stand in a line, numbered from left to right from 1 to n. Each computer containsexactly one piece of information, each of which Doc wants to get eventually. The computers are equipped with a security system, so to crack the i-th of them, the robot needs to collect at least ai any pieces of information from the other computers. Doc can hack the computer only if he is right next to it.
The robot is assembled using modern technologies and can move along the line of computers in either of the two possible directions, but the change of direction requires a large amount of resources from Doc. Tell the minimum number of changes of direction, which the robot will have to make to collect all n parts of information if initially it is next to computer with number 1.
It is guaranteed that there exists at least one sequence of the robot's actions, which leads to the collection of all information. Initially Doc doesn't have any pieces of information.
The first line contains number n (1 ≤ n ≤ 1000). The second line contains n non-negative integers a1, a2, ..., an (0 ≤ ai < n), separated by a space. It is guaranteed that there exists a way for robot to collect all pieces of the information.
Print a single number — the minimum number of changes in direction that the robot will have to make in order to collect all n parts of information.
3 0 2 0
1
5 4 2 3 0 1
3
7 0 3 1 0 5 2 6
2
In the first sample you can assemble all the pieces of information in the optimal manner by assembling first the piece of information in the first computer, then in the third one, then change direction and move to the second one, and then, having 2 pieces of information, collect the last piece.
In the second sample to collect all the pieces of information in the optimal manner, Doc can go to the fourth computer and get the piece of information, then go to the fifth computer with one piece and get another one, then go to the second computer in the same manner, then to the third one and finally, to the first one. Changes of direction will take place before moving from the fifth to the second computer, then from the second to the third computer, then from the third to the first computer.
In the third sample the optimal order of collecting parts from computers can look like that: 1->3->4->6->2->5->7.
解题思路：
用一个sum值记录一下他所已知的电脑信息的数量，然后一直判断，从前往后，从后往前
一直判断知道sum = = m，用两个标记值分别标记从前往后是否转向和从后往前是否转向，
然后输出结果，（转向的次数）
上代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
#define MM1(a) memset(a, false, sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e3+5;
const int mod = 1000000007;
const double eps = 1e-7;
const double pi = 3.1415926;
int arr[maxn];
bool flag[maxn];
int main()
{
int m, k;
while(cin>>m)
{
for(int i=0; i<m; i++)
cin>>arr[i];
int sum = 0, ret = 0;
MM1(flag);
///memset(flag, false, sizeof(false));
bool ok = false, yes=false;
while(sum < m)
{
///注意需要在这里判断，
if(ok)
ret++;
///在初始化一次
ok = false;
for(int i=0; i<m; i++)
{
///cout<<9;
if(arr[i]<=sum && !flag[i])
{
k = i;
ok = true;
sum++;
flag[i] = true;
}
}
for(int j=k-1; j>=0; j--)
{
if(arr[j]<=sum && !flag[j])
{
yes = true;
sum++;
flag[j] = true;
}
}
if(yes)
ret++;
///初始化为false
yes = false;
}
cout<<ret<<endl;
}
return 0;
}
