题目肯定会有解，使用IDA*算法。。。枚举转的次数，对于确定的次数，dfs搜索，搜索剪枝是当前8个格子最大的相同的数量+当前还剩余的步数还到不了8个，就剪枝。。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 400005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head
int dir[8][7] = {
{0, 2, 6, 11, 15, 20, 22},
{1, 3, 8, 12, 17, 21, 23},
{10, 9, 8, 7, 6, 5, 4},
{19, 18, 17, 16, 15, 14, 13},
{23, 21, 17, 12, 8, 3, 1},
{22, 20, 15, 11, 6, 2, 0},
{13, 14, 15, 16, 17, 18, 19},
{4, 5, 6, 7, 8, 9, 10}};
int f[8] = {5, 4, 7, 6, 1, 0, 3, 2};
vector<int> ans;
int a[maxn];
int flag, aa, bb, cc;
void fun(int x)
{
for(int i = 0; i < 6; i++) swap(a[dir[x][i]], a[dir[x][i+1]]);
}
int calc(int x)
{
int res = 0;
for(int i = 6; i <= 8; i++) res += a[i] == x;
for(int i = 15; i <= 17; i++) res += a[i] == x;
res += a[11] == x;
res += a[12] == x;
return res;
}
void debug()
{
for(int i = 0; i < 24; i++) printf("PPP %d\n", a[i]);
}
void dfs(int step)
{
if(flag) return;
int ok = 0;
int a2 = calc(1), b2 = calc(2), c2 = calc(3);
if(a2 == 8 || b2 == 8 || c2 == 8) {
flag = 1;
for(int i = 0; i < ans.size(); i++) printf("%c", ans[i] + 'A');
if(ans.size() == 0) printf("No moves needed");
printf("\n");
if(a2 == 8) printf("1\n");
else if(b2 == 8) printf("2\n");
else printf("3\n");
return;
}
if(aa >= 8 && a2 + step >= 8) ok = 1;
if(bb >= 8 && b2 + step >= 8) ok = 1;
if(cc >= 8 && c2 + step >= 8) ok = 1;
if(!ok) return;
if(step == 0) return;
for(int i = 0; i < 8; i++) {
fun(i);
ans.push_back(i);
dfs(step-1);
fun(f[i]);
ans.pop_back();
if(flag) return;
}
}
void work()
{
for(int i = 1; i < 24; i++) scanf("%d", &a[i]);
aa = bb = cc = flag = 0;
for(int i = 0; i < 24; i++) {
if(a[i] == 1) aa++;
if(a[i] == 2) bb++;
if(a[i] == 3) cc++;
}
ans.clear();
for(int i = 0; i <= 100; i++) {
if(!flag) dfs(i);
}
}
int main()
{
while(scanf("%d", &a[0]), a[0]) {
work();
}
return 0;
}
