http://codeforces.com/problemset/problem/875/B
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
He looks through all the coins from left to right;
If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input
The first line contains single integer n (1 ≤ n ≤ 300 000) — number of coins that Sasha puts behind Dima.
Second line contains n distinct integers p1, p2, …, pn (1 ≤ pi ≤ n) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.
Output
Print n + 1 numbers a0, a1, …, an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.
Examples
input
4
1 3 4 2
output
1 2 3 2 1
input
8
6 8 3 4 7 2 1 5
output
1 2 2 3 4 3 4 5 1
Note
Let’s denote as O coin out of circulation, and as X — coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won’t make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO → OOOX
After replacement of the third coin, Dima’s actions look this way:
XOXO → OXOX → OOXX
After replacement of the fourth coin, Dima’s actions look this way:
XOXX → OXXXRound #
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.
题意
往ｎ个位置放硬币，第ｉ次放到从a[1]到第a[i]的位置，通过移动，把所有的硬币放到最右边，问一共移动多少次；
规则：从左往右看，如果硬币的右边没有硬币，则可以移动，一次可移动多位直到有硬币的地方。
输出第ｉ个次放硬币需要移动的次数。
样例一：
一、１０００－》０００１ 一次
二、１０１０－》０１０１－》００１１ 二次
三、１０１１－》０１１１一次
四、１１１１－》０次
因为初始次数是１，所以所有次数都加上１
先输出初始的次数（难度）：
所以输出１ １ ２ ３ ２ １
题解
ａｎｓ＝放硬币的数量－最右边连续的硬币数＋１
我们可以看做从需要移动的最右边的硬币开始往右移动到最右边，直到最左边的硬币移动到最右边，那么每个硬币往右移动一次，所以总移动次数就是除了最右边硬币，其他硬币个数的总和，比如：
样例一：
二、１０１０－＞１００１－＞００１１
CODE:
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int a,vis[300001];
int main()
{
int n;
scanf("%d",&n);
cout << '1';
int k=n;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
vis[a]=1;
while(vis[k])
k--;
printf(" %d",i+k-n+1);
}
cout <<endl;
return 0;
}