CodeForces 875B ( Sorting the Coins) 难度蛮大的一道题目 题目说了一堆主要还是找规律

                B. Sorting the Coins
            time limit per test： 1 second
            memory limit per test： 512 megabytes
            input： standard input
            output： standard output

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  →  OXXX

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.

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**题意：**题目说了一大堆，有n个硬币，刚开始所有的硬币均为不流通，然后分开n次使得n个硬币流通，小明每次只能从左到右看，若看的位置是流通硬币且下一个硬币是不流通的，则小明则在心中将这两个硬币交换位置，若前后两个硬币均为流通或不流通则不交换，问每次将此时流通的硬币全部排到最后的难度是多少（因为每次看完要记住当前的位置，所以看的次数越多越难，不用看时的难度为1多看一次加一）。其实就是把它移动到最右边
例如 0x0x 意思就是将第一个x右移1位 在1的基础上+1 输出2
**思路：**已经在最后的相连的流通硬币是不会交换的，然后每次将一个流通硬币换到下一个流通硬币时这个流通硬币就不能再换的，要开始换下一个流通硬币，等同于就是每次看把最后的没有在尾部的流通硬币移到尾部，即每次看的次数就是没有连接在尾部的流通硬币个数。难度+1
你只需要把总的x减去最右边连续的x的个数再加1就为答案 n-x+1;

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#include <stdio.h>
#include <string.h>
int flag[300001];
int main()
{
    int n,k,m;
    scanf("%d",&n);
    k=n;
    memset(flag,0,sizeof(flag));
    printf("1");
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&m);
        flag[m]=1;
        while(flag[k])
            k--;
        printf(" %d",i+k-n+1);
    }
    puts("");
    return 0;
}