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HackerEarth Permutation problem solution

In this HackerEarth Permutation problem solution Given a permutation of 1 to n, you need to perform some operations to make it into increasing order. Each operation is to reverse an interval a1,a2,...,ax(1<=x,+n) (a prefix). Your goal is to minimize the number of operations.

HackerEarth Permutation problem solution

HackerEarth Permutation problem solution.


typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair<int,int> pii;

#define xx first
#define yy second

template<typename T> inline T max(T a,T b){return a>b?a:b;}
template<typename T> inline T min(T a,T b){return a<b?a:b;}
template<typename T> inline T abs(T a){return a>0?a:-a;}
template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;}
template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;}
template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}}
template<typename T> inline T sqr(T x){return x*x;}
#define mp(a,b) std::make_pair(a,b)
#define pb push_back
#define I inline
#define mset(a,b) memset(a,b,sizeof(a))
#define mcpy(a,b) memcpy(a,b,sizeof(a))

#define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
#define fd0(i,n) for(int i=(n)-1;~i;i--)
#define fd1(i,n) for(int i=n;i;i--)
#define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
#define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)

struct Cg{I char operator()(){return getchar();}};
struct Cp{I void operator()(char x){putchar(x);}};
#define OP operator
#define RT return *this;
#define RX x=0;char t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\
#define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x
#define RU x=0;char t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define TR *this,x;return x;
I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x)
{RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x)
{for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS
(t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf()
{llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in;
#define WI(S) if(x){if(x<0)P('-'),x=-x;char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
#define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\
x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5);
#define WU(S) if(x){char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT}
I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP()
(ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT}
I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out;

using std::string;

int n;

int main()
string cur,final;
//if(q.size()==0)out,dis[cur],' ',cur.c_str(),'\n';
return 0;
string nxt=cur;

Second solution

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

const int maxn = 10;
int n;
map<vi, int> hs;

void chemthan() {
cin >> n;
assert(1 <= n && n <= 8);
vi a(n);
FOR(i, 0, n) {
cin >> a[i];
assert(1 <= a[i] && a[i] <= n);
FOR(j, 0, i) assert(a[i] != a[j]);
vi v; FOR(i, 1, n + 1) v.pb(i);
queue<vi> que;
hs[v] = 0, que.push(v);
while (sz(que)) {
vi v = que.front(); que.pop();
FOR(i, 1, n + 1) {
vi nv = v;
reverse(nv.begin(), nv.begin() + i);
if (!hs.count(nv)) {
hs[nv] = hs[v] + 1;
cout << hs[a] << "\n";

int main(int argc, char* argv[]) {
ios_base::sync_with_stdio(0), cin.tie(0);
if (argc > 1) {
assert(freopen(argv[1], "r", stdin));
if (argc > 2) {
assert(freopen(argv[2], "wb", stdout));
cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
return 0;

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