Lock Puzzle CodeForces - 936C (构造)

大意: 给定字符串$s$,$t$, 每次操作可以将$S=AB$变为$S=B^RA$, 要求$3n$次操作内将$s$变为$t$.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n;
char s[N], t[N];
vector<int> g;
void shift(int x) {
	if (!x) return;
	reverse(s+1,s+1+n);
	reverse(s+x+1,s+1+n);
	g.pb(x);
}
int main() {
	scanf("%d%s%s", &n, s+1, t+1);
	REP(i,1,n) {
		int pos = i;
		for (; s[pos]!=t[n-i+1]; ++pos) ;
		if (pos>n) return puts("-1"),0;
		shift(n),shift(pos-1),shift(1);
	}
	printf("%d\n",int(g.size()));
	for (auto &&t:g) printf("%d ", t);hr;
}