Editorial for VOI 11 Bài 3 - Hàng cây


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Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

Code mẫu của flashmt

#include<iostream>
#include<algorithm>
#define fr(a,b,c) for(a=b;a<=c;a++)
#define frr(a,b,c) for(a=b;a>=c;a--)
#define maxn 100010
#define z 1000000000
using namespace std;

int n,a[maxn],re=2,np,p[maxn],d[maxn*2],f[maxn];

void prime()
{
    int i,j;
    fr(i,2,500)
      if (!d[i])
      {
          j=i*i;
          while (j<=200002)
          {
              d[j]=1;
              j+=i;  
          }
      } 
    fr(i,2,200002)
      if (!d[i]) p[++np]=i;
}

void calc(int n,int y)
{
    int i; 
    fr(i,1,np)
      if (p[i]<=n)
      {
        int x=p[i];
        while (x<=n)
        {
            f[i]+=(n/x)*y;
            if (p[i]<500) x*=p[i];  
            else break;
        }       
      } 
      else break;
}

int main()
{
    int i,j;
    cin >> n >> i;
    fr(i,1,n) scanf("%d",&a[i]);
    frr(i,n-1,1)
      if (a[i]<a[i+1]) re++;
      else break;
    cout << re << endl;
    n++;
    prime();
    calc(n*2,1);
    calc(n,-1);
    calc(n+1,-1);
    long long res=1;
    fr(i,1,np)
      if (f[i])
        fr(j,1,f[i])
          res=(res*p[i])%z;
    cout << res << endl;
    return 0;
}

Code mẫu của happyboy99x

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5, V = 2*(N+1), MOD = 1e9;
bool isp[V+1];
int p[V+1], c[V+1], a[N], n, np;

void eratos() {
    memset(isp, -1, sizeof isp);
    isp[0] = isp[1] = false;
    for(int i = 2; i * i <= V; ++i) if(isp[i])
        for(int j = i+i; j <= V; j += i)
            isp[j] = false;
    for(int i = 2; i <= V; ++i) if(isp[i])
        p[np++] = i;
}

void enter() {
    int h; cin >> n >> h;
    for(int i = 0; i < n; ++i) cin >> a[i];
}

void task1() {
    int res = 2, mn = a[n-1];
    for(int i = n-2; i >= 0; --i)
        if(a[i] <= mn) ++res, mn = a[i];
        else break;
    cout << res << '\n';
}

void analysis(int n, bool dec) {
    for(int i = 0; i < np; ++i) {
        long long x = p[i];
        while(x <= n) {
            if(dec) c[i] -= n / x;
            else c[i] += n / x;
            x *= p[i];
        }
    }
}

long long powmod(long long a, long long n) {
    long long res = 1;
    for(; n != 0; n >>= 1) {
        if(n & 1) res = res * a % MOD;
        a = a * a % MOD;
    }
    return res;
}

void task2() {
    ++n;
    analysis(2*n, false);
    analysis(n, true);
    analysis(n+1, true);
    long long res = 1;
    for(int i = 0; i < np; ++i)
        res = res * powmod(p[i], c[i]) % MOD;
    cout << res << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    eratos();
    enter();
    task1();
    task2();
    return 0;
}

Code mẫu của ladpro98

#include <bits/stdc++.h>
const int N = 222005;
const int B = 1000000000;
using namespace std;
int n, h, lp, Prime[N], Power[N], a[N];
bool isPrime[N];

void initSieve() {
    int m = 2*(n+5), i, j;
    for(i=2; i<=m; i++)
        isPrime[i] = true;
    for(i=2; i<=m; i++)
    if (isPrime[i]) {
        j = i + i;
        while (j <= m) {
            isPrime[j] = false;
            j += i;
        }
    }
    for(i=2; i<=m; i++)
    if (isPrime[i]) {
        Prime[++lp] = i;
    }
}

void addFactorial(int nn, int mul) {
    for(int i=1; i<=lp; i++) {
        long long v = Prime[i];
        while (v <= nn) {
            Power[i] += mul * (nn / v);
            v *= Prime[i];
        }
    }
}

long long powMod(int v, int pw) {
    if (pw == 0) return 1;
    long long t = powMod(v, pw / 2);
    t = (t * t) % B;
    if (pw & 1) t = (t*v) % B;
    return t;
}

int main()
{
    //freopen("TREELINE.in", "r", stdin);
    scanf("%d %d\n", &n, &h);
    int i, res;
    long long day = 1;
    for(i=1; i<=n; i++) scanf("%d", &a[i]);
    for(i=n-1; i>0; i--)
        if (a[i] > a[i+1]) break;
    res = n - i + 1;
    initSieve();
    n++;
    addFactorial(2*n, 1);
    addFactorial(n+1, -1);
    addFactorial(n, -1);
    for(i=1; i<=lp; i++)
        if (Power[i] > 0)
            day = (day * powMod(Prime[i], Power[i])) % B;

    printf("%d\n", res);
    printf("%lld", day);
    return 0;
}

Code mẫu của RR

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <string>
#include <deque>
#include <complex>

#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; i--)
#define REP(i,a) for(int i=0,_a=(a); i<_a; i++)
#define ll long long
#define F first
#define S second
#define PB push_back
#define MP make_pair
using namespace std;

//Buffer reading
int INP,AM;
#define BUFSIZE (1<<10)
char BUF[BUFSIZE+1], *inp=BUF;
#define GETCHAR(INP) { \
    if(!*inp) { \
        fread(BUF,1,BUFSIZE,stdin); \
        inp=BUF; \
    } \
    INP=*inp++; \
}
#define DIG(a) (((a)>='0')&&((a)<='9'))
#define GN(j) { \
    AM=0;\
    GETCHAR(INP); while(!DIG(INP) && INP!='-') GETCHAR(INP);\
    if (INP=='-') {AM=1;GETCHAR(INP);} \
    j=INP-'0'; GETCHAR(INP); \
    while(DIG(INP)){j=10*j+(INP-'0');GETCHAR(INP);} \
    if (AM) j=-j;\
}
//End of buffer reading

const double PI = acos(-1.0);

int n, a[1000111];
int nprime, prime[100111];
bool sieve[200111];
const int BASE = 1000000000;

void init() {
    FOR(i,2,450)
    if (!sieve[i]) {
        int j = i*i;
        while (j <= 200000) {
            sieve[j] = true;
            j += i;
        }
    }

    FOR(i,2,200000)
        if (!sieve[i]) prime[nprime++] = i;
}

int get(int n, int p) {
    if (n < p) return 0;
    return n / p + get(n/p, p);
}

ll power(ll x, int k) {
    if (!k) return 1;
    if (k == 1) return x;
    ll mid = power(x, k >> 1);
    mid = (mid * mid) % BASE;
    if (k & 1) return (x * mid) % BASE;
    else return mid;
}

int main() {
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    int tmp;
    GN(n); GN(tmp); FOR(i,1,n) GN(a[i]);

    int nn = a[n], u = 0;
    FORD(i,n-1,1) {
        if (a[i] > nn) {
            u = i;
            break;
        }
        nn = min(nn, a[i]);
    }
    printf("%d\n", n-u+1);
    init();

    n++;

    ll res = 1;
    ll now = n + 1;
    REP(x,nprime) {
        int k = get(n<<1, prime[x]) - (get(n,prime[x]) << 1);
        while (now % prime[x] == 0) {
            now /= prime[x];
            k--;
        }
        res = (res * power(prime[x], k)) % BASE;
    }
    cout << res << endl;
    return 0;
}

Code mẫu của hieult

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <deque>
#include <bitset>
#include <cctype>
#include <utility>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;

#define Rep(i,n) for(int i = 0; i < (n); ++i)
#define Repd(i,n) for(int i = (n)-1; i >= 0; --i)
#define For(i,a,b) for(int i = (a); i <= (b); ++i)
#define Ford(i,a,b) for(int i = (a); i >= (b); --i)
#define Fit(i,v) for(__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i)
#define Fitd(i,v) for(__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sz(a) ((int)(a).size())
#define all(a) (a).begin(), (a).end()
#define ms(a,x) memset(a, x, sizeof(a))

template<class F, class T> T convert(F a, int p = -1) {
    stringstream ss;
    if (p >= 0)
        ss << fixed << setprecision(p);
    ss << a;
    T r;
    ss >> r;
    return r;
}
template<class T> T gcd(T a, T b) {
    T r;
    while (b != 0) {
        r = a % b;
        a = b;
        b = r;
    }
    return a;
}
template<class T> T lcm(T a, T b) {
    return a / gcd(a, b) * b;
}
template<class T> T sqr(T x) {
    return x * x;
}
template<class T> T cube(T x) {
    return x * x * x;
}
template<class T> int getbit(T s, int i) {
    return (s >> i) & 1;
}
template<class T> T onbit(T s, int i) {
    return s | (T(1) << i);
}
template<class T> T offbit(T s, int i) {
    return s & (~(T(1) << i));
}
template<class T> int cntbit(T s) {
    return s == 0 ? 0 : cntbit(s >> 1) + (s & 1);
}

const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
    if (rsz <= 0) {
        ptr = 0;
        rsz = (int) fread(bf, 1, bfsz, stdin);
        if (rsz <= 0)
            return EOF;
    }
    --rsz;
    return bf[ptr++];
}
void ga(char &c) {
    c = EOF;
    while (!isalpha(c))
        c = gc();
}
int gs(char s[]) {
    int l = 0;
    char c = gc();
    while (isspace(c))
        c = gc();
    while (c != EOF && !isspace(c)) {
        s[l++] = c;
        c = gc();
    }
    s[l] = '\0';
    return l;
}
template<class T> bool gi(T &v) {
    v = 0;
    char c = gc();
    while (c != EOF && c != '-' && !isdigit(c))
        c = gc();
    if (c == EOF)
        return false;
    bool neg = c == '-';
    if (neg)
        c = gc();
    while (isdigit(c)) {
        v = v * 10 + c - '0';
        c = gc();
    }
    if (neg)
        v = -v;
    return true;
}

typedef pair<int, int> II;
const ld PI = acos(-1.0);
const ld eps = 1e-12;
const int dr[] = { -1, 0, +1, 0};
const int dc[] = { 0, +1, 0, -1};
const int inf = (int) 1e9 + 5;
const ll linf = (ll) 1e16 + 5;
const int mod = (ll) (1e9 + eps);

using namespace std;
template<int MAXP> struct BitSieve {
    #define isOn(x) ( p[x >> 6] & ( 1 << ( (x & 63) >> 1) ) )
    #define turnOn(x) ( p[x >> 6] |= ( 1 << ( (x & 63) >> 1 ) ) )
    int p[(MAXP >> 6) + 1];
    BitSieve() {
        for (int i = 3; i * i <= MAXP; i += 2) {
            if (!isOn(i)) {
                int ii = i * i, i2 = i << 1;
                for (int j = ii; j <= MAXP; j += i2) turnOn(j);
            }
        }
    }
    bool operator [] (const int x) {
        return x > 1 && (x == 2 || ( (x & 1) && !isOn(x) ) );
    }
};

int n,x,a[100002],m,b[100002];
BitSieve<200005> B;

int GCD (int a,int b)
{
    int r;
    while(b!=0)
    {
        r = a%b;
        a = b;
        b = r;
    }
    return a;
}

int cal(int x, int k){
    int res = 0;
    while(x){
        x /= k;
        res += x;
    }
    return res;
}

int main()
{
//  freopen("in.txt", "r", stdin);
    long long KQ = 1;
    scanf("%d %d",&n,&x);
    for(int i = 1;i<=n;i++)
        scanf("%d",&a[i]);

    int chay ; if(n!=0) chay = n-1; else chay = 0;
    while(chay>0)
    {
        if(a[chay]<a[chay+1])
            chay--;
        else break;
    }
    printf("%d\n",n-chay+1);
    n++;
    int num;


    For(i, 2, n + n) if(B[i]){
        num = cal(n + n, i) - cal(n + 1, i) - cal(n, i);
        Rep(run, num) KQ = KQ * i % mod;
    }

    cout << KQ;

    //getch();
}

Code mẫu của ll931110

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <utility>
#include <vector>
using namespace std;

int n,h,a[100010];
int module = 1000000000,phi = module / 10 * 4;
long long ret = 1,p2 = 0,p5 = 0;

long long power(int x,int p,int mod)
{
    if (!p) return 1;
    long long q = power(x,p/2,mod);
    q = (q * q) % mod;
    if (p & 1) q = (q * x) % mod;
    return q;
}

void multiply(int x)
{
    while (x % 2 == 0)
    {
        p2++;  x /= 2;
    }
    while (x % 5 == 0)
    {
        p5++;  x /= 5;
    }
    ret = (1LL * ret * x) % module;
}

void divide(int x)
{
    while (x % 2 == 0)
    {
        p2--;  x /= 2;
    }
    while (x % 5 == 0)
    {
        p5--;  x /= 5;
    }
    ret = (1LL * ret * power(x,phi - 1,module)) % module;
}

int main()
{
//  freopen("TREELINE.INP","r",stdin);
//  freopen("TREELINE.OUT","w",stdout);
    scanf("%d %d", &n, &h);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int way = n + 1,minh = a[n - 1];
    for (int i = n - 2; i >= 0; i--) if (a[i] > minh)
    {
        way = n - i;  break;
    }
    else minh = a[i];
    printf("%d\n", way);

    n++;
    for (int i = n + 2; i <= 2 * n; i++) multiply(i);
    for (int i = 1; i <= n; i++) divide(i);
    while (p2--) ret = (ret * 2LL) % module;
    while (p5--) ret = (ret * 5LL) % module;
    cout << ret << endl;
}

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