codeforces——875B —— Sorting the Coins
2017-10-23 20:34
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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 hardnes
4000
s 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
output
input
output
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.
一共n+1局游戏,每局游戏互不干扰,一开始,所有硬币处于O状态,每局游戏都将指定某个硬币处于X状态(包括上一局开局所指定的,为X状态的硬币),问从左至右的浏览,每次只能将"XO"状态的两枚硬币交换,输出每局游戏能交换的次数 +1.
注意超时
#include <cstring>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <iostream>
#include <cmath>
using namespace std;
int main() {
bool com[300002];
int n;
while(~scanf("%d",&n)) {
memset(com,false,sizeof com);
printf("1");
int flag=n;
for(int i=0; i<n; i++) {
int temp,ans=i+1;
scanf("%d",&temp);
com[temp]=true;
while(com[flag])
flag--;
printf(" %d",ans+1-n+flag);
}
printf("\n");
}
return 0;
}
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 hardnes
4000
s 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.
一共n+1局游戏,每局游戏互不干扰,一开始,所有硬币处于O状态,每局游戏都将指定某个硬币处于X状态(包括上一局开局所指定的,为X状态的硬币),问从左至右的浏览,每次只能将"XO"状态的两枚硬币交换,输出每局游戏能交换的次数 +1.
注意超时
#include <cstring>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <iostream>
#include <cmath>
using namespace std;
int main() {
bool com[300002];
int n;
while(~scanf("%d",&n)) {
memset(com,false,sizeof com);
printf("1");
int flag=n;
for(int i=0; i<n; i++) {
int temp,ans=i+1;
scanf("%d",&temp);
com[temp]=true;
while(com[flag])
flag--;
printf(" %d",ans+1-n+flag);
}
printf("\n");
}
return 0;
}
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