DP:

According to the meaning of problems,if we check n to m, assume x and y are both solvable,then we only should:

(1). check  xy;

(2). check AxA

(3). check AxAyA

if any one of the three is solvable ,  n to m is solvable

#include<cstdio>

#include<string>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int MAXN = 200;

char str[MAXN], dp[MAXN][MAXN];

bool dfs(int n, int m){

    if(n > m) return true;

    if(n == m) return false;

    if(dp[n][m] == 1) return true;

    if(dp[n][m] == -1) return false;

    for(int i = n+1;i < m-1;i ++){

        if(dfs(n, i) && dfs(i+1, m)){               //check xy;

            dp[n][m] = 1;

            return true;

        }

    }

    if(str[n] == str[m]){

        if(dfs(n+1, m-1)){                          //check AxA

            dp[n][m] = 1;

            return true;;

        }

        for(int i = n+1;i < m;i ++){

            if(str[i] != str[n]) continue;

            if(dfs(n+1, i-1) && dfs(i+1, m-1)){     //check AxAyA

                dp[n][m] = 1;

                return true;

            }

        }

    }

    dp[n][m] = -1;

    return false;

}

int main(){

    memset(str, 0, sizeof(str));

    freopen("in.c", "r", stdin);

    while(~scanf("%s", str)){

        memset(dp, 0, sizeof(dp));

        int len = strlen(str);

        if(dfs(0, len-1)) printf("solvable\n");

        else printf("unsolvable\n");

    }

    return 0;

}