原题链接
A. Palindrome Dance
A group of n dancers rehearses a performance for the closing ceremony. The dancers are arranged in a row, they’ve studied their dancing moves and can’t change positions. For some of them, a white dancing suit is already bought, for some of them — a black one, and for the rest the suit will be bought in the future.

On the day when the suits were to be bought, the director was told that the participants of the olympiad will be happy if the colors of the suits on the scene will form a palindrome. A palindrome is a sequence that is the same when read from left to right and when read from right to left. The director liked the idea, and she wants to buy suits so that the color of the leftmost dancer’s suit is the same as the color of the rightmost dancer’s suit, the 2nd left is the same as 2nd right, and so on.

The director knows how many burls it costs to buy a white suit, and how many burls to buy a black suit. You need to find out whether it is possible to buy suits to form a palindrome, and if it’s possible, what’s the minimal cost of doing so. Remember that dancers can not change positions, and due to bureaucratic reasons it is not allowed to buy new suits for the dancers who already have suits, even if it reduces the overall spending.
题目大意：这个题目本意就是回文串的应用，在这之间还加入了价值的运算，思维量不大，注意中间的一些细节就好了。。
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

int f[21];
int n,a,b;

int main()
{
    while(cin>>n>>a>>b)
    {
        for(int i=0;i<n;i++) cin>>f[i];
        int sum=0;
        int flag=0;
        for(int i=0;i<n/2;i++)
        {
            if(f[i]!=f[n-i-1])
            {
                if(!(f[i]==2||f[n-i-1]==2))//1 1 2 1 0这样的就不可以了
                {
                    flag=1;
                    break;
                }
                else 
                {
                    if(f[i]!=2) sum+=(f[i]==0)?a:b;
                    if(f[n-i-1]!=2) sum+=(f[n-i-1]==0)?a:b;
                }
            }
            else if(f[i]==f[n-i-1]&&f[i]==2) sum+=2*min(a,b);
        }
        if(n&1)/n为奇数的时候需要特判一下
        {
            if(f[n/2]==2) sum+=min(a,b);
        }
        if(flag) cout<<"-1"<<endl;
        else cout<<sum<<endl;
    }
    return 0;
}