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.

`#include<bits/stdc++.h>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 secondtemplate<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;\if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'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;std::map<string,int>dis;std::queue<string>q;int main(){    in,n;    string cur,final;    fo0(i,n)cur.pb(char(47+(int)in)),final.pb(char(48+i));    dis[cur]=0;    q.push(cur);    while(!q.empty())    {        cur=q.front();q.pop();        //if(dis[cur]==5)out,dis.size(),'\n';        //if(q.size()==0)out,dis[cur],' ',cur.c_str(),'\n';        if(cur==final)        {            out,dis[cur],'\n';            return 0;        }        fo(i,2,n)        {            string nxt=cur;            std::reverse(nxt.begin(),nxt.begin()+i);            if(dis.find(nxt)==dis.end())            {                dis[nxt]=dis[cur]+1;                q.push(nxt);            }        }    }    out,dis.size(),'\n';}`

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;                que.push(nv);            }        }    }    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));    }    chemthan();    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";    return 0;} `