Codeforces 992 范围内GCD,LCM要求找一对数 衣柜裙子期望

A

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v)
{
        to[++tot] = v;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
inline void read(int &v)
{
        v = 0;
        char c = 0;
        int p = 1;
        while (c < '0' || c > '9')
        {
                if (c == '-')
                {
                        p = -1;
                }
                c = getchar();
        }
        while (c >= '0' && c <= '9')
        {
                v = (v << 3) + (v << 1) + c - '0';
                c = getchar();
        }
        v *= p;
}
map<int, int> mp;
int main()
{
        int n;
        read(n);
        int ans = 0;
        mp[0] = 1;
        for (int i = 1; i <= n; i++)
        {
                int now;
                read(now);
                if (!mp[now])
                {
                        mp[now] = 1;
                        ans++;
                }
        }
        cout << ans << endl;
        return 0;
}

B

题意:

给你L,R,A,B四个数 要你找出一对在L,R范围内的数使得他们的GCD为A,LCM为B 问你有几对

解:

假设我们找到的数是X,Y 那么X*Y肯定为LCM*GCD 因为LCM=X*Y/GCD

所以我们只要在1~sqrt(LCM*GCD)范围内枚举数即可 但是LCM*GCD的最大值是1e18 平方根下来是1e9还是不行

我们观察到X,Y的GCD是X 那么就说明X,Y都是GCD的倍数 所以我们不用++枚举 直接+X枚举即可

这样的复杂度是sqrt(LCM*GCD)/GCD=sqrt(LCM/GCD)  LCM/GCD最大是1e9 可以接受

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v)
{
        to[++tot] = v;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
inline void read(int &v)
{
        v = 0;
        char c = 0;
        int p = 1;
        while (c < '0' || c > '9')
        {
                if (c == '-')
                {
                        p = -1;
                }
                c = getchar();
        }
        while (c >= '0' && c <= '9')
        {
                v = (v << 3) + (v << 1) + c - '0';
                c = getchar();
        }
        v *= p;
}
ll gcd(ll a, ll b)
{
        ll t;
        while (b)
        {
                t = b;
                b = a % b;
                a = t;
        }
        return a;
}
int main()
{
        ll l, r, x, y;
        cin >> l >> r >> x >> y;
        ll rl = l;
        ll sum = x * y;
        ll ans = 0;
        ll a, b;
        if (l % x != 0)
        {
                l = (l / x + 1) * x;
        }
        for (ll i = l; i <= sqrt(sum) && i <= r; i += x)
        {
                if (sum % i == 0)
                {
                        a = i, b = sum / i;
                        if (b >= rl && b <= r)
                        {
                                if (gcd(a, b) == x)
                                {
                                        if (a == b)
                                        {
                                                ans++;
                                        }
                                        else
                                        {
                                                ans += 2;
                                        }
                                        //cout << a << " " << b << endl;
                                }
                        }

                }
        }
        cout << ans << endl;
        return 0;
}

C

题意:

你开始有X个裙子 你有K+1次增长机会 前K次会100%的增长一倍 但是增长后有50%的机会会减少一个

给你X,K(1e18) 问你最后裙子数量的期望值是多少(mod 1e9+7)

解:

纯推公式找规律题

我们很容易知道其实在K月之后(总共有K+1月)最后可能得到的裙子数是连续的

即如果刚开始有X个裙子且K=2时 他经过K月(没经过最后特殊的那一月)后可能得到的为 4*X,4*X-1,4*X-2,4*X-3 这四种答案

所以可以得到公式经过K月后的期望值为 (2^k*X+2^k*X-2^k+1)*2^k/2/2^k=(2^k+1*X-2^k+1)/2

这是K月后的期望值 还有最后一月要*2 所以直接*2 最后的答案即为2^k+1*X-2^k+1

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1;
inline void addedge(int u, int v)
{
        to[++tot] = v;
        nxt[tot] = Head[u];
        Head[u] = tot;
}*/
inline void read(int &v)
{
        v = 0;
        char c = 0;
        int p = 1;
        while (c < '0' || c > '9')
        {
                if (c == '-')
                {
                        p = -1;
                }
                c = getchar();
        }
        while (c >= '0' && c <= '9')
        {
                v = (v << 3) + (v << 1) + c - '0';
                c = getchar();
        }
        v *= p;
}
ll Qpow(ll a, ll b)
{
        ll ans = 1, base = a;
        while (b != 0)
        {
                if (b & 1 != 0)
                {
                        ans *= base;
                        ans %= mod;
                }
                base *= base;
                base %= mod;
                b >>= 1LL;
        }
        return ans;
}
int main()
{
        ll x, k;
        cin >> x >> k;
        if (x == 0)
        {
                cout << 0 << endl;
                return 0;
        }
        ll ans1 = Qpow(2, k + 1);
        x %= mod;
        ans1 = (ans1 * x) % mod;
        ll ans2 = (Qpow(2, k) - 1 + mod) % mod;
        cout << (ans1 - ans2 + mod) % mod << endl;
        return 0;
}

D