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 (2k*X+2k*X-2k+1)*2k/2/2k=(2k+1*X-2k+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2k+1*X-2k+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
