hdoj.1295 Move Move Look【水题】 2015/08/06

Move Move Look

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 853 Accepted Submission(s): 542

[align=left]Problem Description[/align]
There are two color (white and black) chessmen in a row. The number of white chessman is equal to the number of black chessman. The left half of the row is black chessmen and the right half of the row is white chessmen at first. There
are two rules for the game.

1. Only two neighbor chessmen (double black, double white or one black and one white) can be moved to vacancy at one time. The new vacancy can be filled by the two new neighbor chessmen. Repeat moving chessmen until all chessmen are interphase with black chessman
and white chessman.

2. The number of move must be the minimum.

The case n=3 is shown in Fig. 1. Your task is to get the number of the minimum steps.



[align=left]Input[/align]
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer means the number of black chessmen (the same as white chessmen) n(3<=n<=1000).

[align=left]Output[/align]
For each test case, there is only one integer means the minimum steps.

[align=left]Sample Input[/align]

3

[align=left]Sample Output[/align]

3

[align=left]Author[/align]
SmallBeer(CML)

[align=left]Source[/align]
杭电ACM集训队训练赛（VIII）

n = 4 时：

0 0 0 0 1 1 1 1

0 0 1 1 1 1 0 0

0 1 1 0 1 1 0 0

0 1 1 0 1 0 1 0

1 0 1 0 1 0 1 0

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
int n;
while( ~scanf("%d",&n) ){
printf("%d\n",n);
}
return 0;
}