2020-2021 ICPC - Gran Premio de Mexico

A. Atsa’s Checkers Board

给定一个 n ∗ m n * m n∗m大小的矩阵，在该矩阵内放置黑色白色的石头，要保证每个 2 ∗ 2 2 * 2 2∗2的矩阵都有两个黑石头和两个白石头，问有多少种摆放方式
可以发现当有两块颜色相同的石头摆放在一起的时候，整个矩阵的摆放方式就确定了
可以发现第一行至少有一对颜色相同石子相连的方案数是 1 < < m − 2 1 << m - 2 1<<m−2
而剩下两种没有颜色相同的石子相连的情况，黑白…和白黑…
取其中一种情况作计算。可以发现当第一行确定为黑白…的时候，第一列的排列方式数就为 1 < < ( n − 1 ) 1 << (n - 1) 1<<(n−1)。同理可得第一行确定为白黑…的时候也为 1 < < ( n − 1 ) 1 << (n - 1) 1<<(n−1)，两者总和即为 1 < < n 1 << n 1<<n
因此总情况数就是 ( 1 < < m ) + ( 1 < < n ) − 2 (1 << m) + (1 << n) - 2 (1<<m)+(1<<n)−2

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 3200;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline ll rd() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
int main() {ll n, m;n = rd();m = rd();if (n < 2 || m < 2) {puts("0");return 0;}printf("%lld\n", (1LL << n) + (1LL << m) - 2);return 0;
}

C. CLETS Patrols

假设一个1*n的矩阵 ( a 1 , a 2 , a 3 . . . a n ) (a_1,a_2,a_3...a_n) (a1,a2,a3...an) ， a i a_i ai 表示当前守卫在第 i i i 个点的概率
而输入的 n*n 矩阵表示，明显得到以下关系：
( a 1 ′ , a 2 ′ , a 3 ′ , . . . , a n ′ ) = ( a 1 , a 2 , a 3 , . . . , a n ) ∗ 输入的 n ∗ n 矩阵 (a_1',a_2',a_3',...,a_n')=(a_1,a_2,a_3,...,a_n)* 输入的n*n矩阵 (a1′,a2′,a3′,...,an′)=(a1,a2,a3,...,an)∗输入的n∗n矩阵
所以直接从原始矩阵 ( 1 , 0 , 0 , 0... , 0 ) ∗ ( n ∗ n ) 矩阵 m (1,0,0,0...,0)*(n*n)矩阵^m (1,0,0,0...,0)∗(n∗n)矩阵m
跑一个矩阵快速幂即可。

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e7 + 10;
const int INF = 0x3f3f3f3f;
inline ll rd() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
ll gcd(ll x, ll y) {if (y == 0) return x;return gcd(y, x % y);
}
int n, m;
struct matrix {double a[202][202];matrix operator *(const matrix& b) {matrix c;/* for (int k = 1;k <= n;++k) {c.a[1][k] = 0;for (int i = 1;i <= n;++i) {c.a[1][k] += this->a[1][i] * b.a[i][k];}}return c;*/for (int i = 1;i <= n;++i) {for (int j = 1;j <= n;++j) {c.a[i][j] = 0;for (int k = 1;k <= n;++k) {c.a[i][j] += this->a[i][k] * b.a[k][j];}}}return c;}
}mp;
matrix fp(matrix a, int y) {matrix ans = a;while (y) {if (y & 1) ans = ans * a;a = a * a;y >>= 1;}return ans;
}
int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin >> n >> m;for (int i = 1;i <= n;++i) {for (int j = 1;j <= n;++j) {cin >> mp.a[i][j];}}matrix ans = fp(mp, m - 1);for (int i = 1;i <= n;++i) {cout << setprecision(10) << ans.a[1][i] << '\n';}return 0;
}

D. Determine the Winner Marshaland

有题推得，求的是以下条件：
x ≡ p i − 2 ( m o d p i ) x\equiv p_i-2\ (mod\ p_i) x≡pi−2 (mod pi)
换过来 x + 2 ≡ p i ≡ 0 ( m o d p i ) x+2\equiv p_i\equiv0\ (mod\ p_i) x+2≡pi≡0 (mod pi)
所以就对 x + 2 x+2 x+2 进行质因数分解即可。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> pii;
typedef pair<ll, ll>pll;
template<typename T>
inline void rd(T& x) {int tmp = 1;char c = getchar();x = 0;while (c > '9' || c < '0') {if (c == '-')tmp = -1;c = getchar();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}x *= tmp;
}
const int N = 1e3 + 10;
const int M = 2e7 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ull seed = 31;
ll gcd(ll x, ll y) {if (y == 0) return x;return gcd(y, x % y);
}int head[N], cntE = 0;
struct edge {int next, to;
}e[M];
void add(int u, int v) {e[cntE].to = v;e[cntE].next = head[u];head[u] = cntE++;
}
int n, m, q;
int prime[N], cnt = 0;
bool isprime[N];
void init() {for (int i = 2; i <= N - 5; ++i) {if (!isprime[i]) prime[++cnt] = i;for (int j = 1; j <= cnt && prime[j] * i <= N - 5; ++j) {isprime[i * prime[j]] = 1;if (i % prime[j] == 0) break;}}
}
int main() {#ifdef _DEBUGFILE* _INPUT = freopen("input.txt", "r", stdin);//    FILE* _OUTPUT = freopen("output.txt", "w", stdout);
#endifint T = 1, cas = 0;rd(T);init();while (T--) {rd(n);n += 2;vector<int>ans;while (!(n & 1)) n >>= 1;for (int i = 2; i <= cnt && prime[i] * prime[i] <= n; ++i) {if (n % prime[i] == 0) {ans.push_back(prime[i]);while (n % prime[i] == 0) n /= prime[i];}}if (n > 1) ans.push_back(n);if (ans.empty()) puts("-1");else {for (int i = 0; i < ans.size(); ++i) {printf("%d%s", ans[i], i == ans.size() - 1 ? "\n" : " ");}}}return 0;
}

E. Enterprise Recognition Program

给定一个树形图代表一个公司的主从结构，并且公司中有 n n n个员工，老板节点编号为 0 0 0，老板不属于员工
操作 m m m次，并且有两种操作
第一种操作是给员工 e e e发 v v v元的奖金
第二种操作是给员工 e e e及其所有上司发 v v v元的奖金
询问 q q q次，并且有两种询问
第一种询问是询问员工 x x x有多少奖金
第二种询问是询问员工 x x x的下属和员工 x x x共有多少奖金
设置一个 e a r n earn earn数组， e a r n [ i ] [ 0 ] earn[i][0] earn[i][0]表示公司用操作一发给员工 i i i的奖金， e a r n [ i ] [ 1 ] earn[i][1] earn[i][1]表示公司用操作二发给员工 i i i的奖金， e a r n [ i ] [ 2 ] earn[i][2] earn[i][2]表示员工 i i i的下属们的总奖金
从老板节点开始往下 d f s dfs dfs，从叶子节点往上累加奖金即可
以下为两个递推公式， u u u为当前员工， v v v为 u u u的下属
e a r n [ u ] [ 1 ] + = e a r n [ v ] [ 1 ] ; earn[u][1] += earn[v][1]; earn[u][1]+=earn[v][1];
e a r n [ u ] [ 2 ] + = e a r n [ v ] [ 0 ] + e a r n [ v ] [ 1 ] + e a r n [ v ] [ 2 ] ; earn[u][2] += earn[v][0] + earn[v][1] + earn[v][2]; earn[u][2]+=earn[v][0]+earn[v][1]+earn[v][2];
具体逻辑在代码中体现

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#define lowbit(x) ((x) & -(x))
#define endl "\n"
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6 + 10, NN = 2e3 + 10, INF = 0x3f3f3f3f, LEN = 20;
const ll MOD = 1e9 + 7;
const ull seed = 31;
const double pi = acos(-1.0), eps = 1e-8;
inline int read() {int x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
struct Edge {int next, to;
} edge[N];
int num_edge, n, m, q, boss;
int head[N], cha[N];
ll earn[N][3];
void add_edge(int from, int to) {edge[num_edge].next = head[from];edge[num_edge].to = to;head[from] = num_edge++;
}
void init() {memset(head, -1, sizeof head);
}
void dfs(int u) {for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].to;dfs(v);earn[u][1] += earn[v][1];earn[u][2] += earn[v][0] + earn[v][1] + earn[v][2];}
}
int main() {// ios::sync_with_stdio(false);// cin.tie(0);// cout.tie(0);// freopen("input.txt", "r", stdin);// freopen("output.txt", "w", stdout);n = read(), m = read(), q = read();init();for (int i = 1; i <= n; ++i) {int x = read();add_edge(x, i);if (x == 0)boss = i;}for (int i = 1; i <= m; ++i) {int m = read(), e = read(), v = read();earn[e][m - 1] += v;}dfs(boss);while (q--) {int op = read(), x = read();if (op == 1)printf("%lld\n", earn[x][0] + earn[x][1]);elseprintf("%lld\n", earn[x][2] + earn[x][1] + earn[x][0]);}return 0;
}

G. Goombas Colliding

无论怎么撞，最终都会有一辆车子走的是他自己的路程或者别人的路程

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e7 + 10;
const int INF = 0x3f3f3f3f;
inline ll rd() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
ll n, m;
void solve() {scanf("%lld %lld", &m, &n);ll ans = 0;for (int i = 1;i <= n;++i) {ll p, x;scanf("%lld %lld", &p, &x);if (x) ans = max(ans, m - p);else ans = max(ans, p);}printf("%lld\n", ans);
}
int main() {int T = 1;while (T--)solve();return 0;
}

H. Hamsters Training

最后出来的公式是 C 2 n − 1 n C_{2n-1}^n C2n−1n, 怎么来的真不知道

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> pii;
typedef pair<ll, ll>pll;
template<typename T>
inline void rd(T& x) {int tmp = 1;char c = getchar();x = 0;while (c > '9' || c < '0') {if (c == '-')tmp = -1;c = getchar();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}x *= tmp;
}
const int N = 2e5 + 10;
const int M = 2e7 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ull seed = 31;
ll gcd(ll x, ll y) {if (y == 0) return x;return gcd(y, x % y);
}
int n, m, q;
ll fac[N], ifac[N];
ll fp(ll x, ll y) {ll ans = 1;while (y) {if (y & 1) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}
ll C(int x, int y) {ll ans = fac[x];ll res = ifac[x - y] * ifac[y] % mod;ans = ans * res % mod;return ans;
}
int main() {#ifdef _DEBUGFILE* _INPUT = freopen("input.txt", "r", stdin);//    FILE* _OUTPUT = freopen("output.txt", "w", stdout);
#endifint T = 1, cas = 0;rd(T);fac[0] = 1;for (int i = 1; i <= N - 10; ++i) fac[i] = fac[i - 1] * i % mod;ifac[N - 10] = fp(fac[N - 10], mod - 2);for (int i = N - 10; i > 0; --i) ifac[i - 1] = ifac[i] * i % mod;while (T--) {rd(n);ll ans = C(2 * n - 1, n);printf("%lld\n", ans);}return 0;
}

J. Just Turn the Wheels!

根据题目意思，首先车轮滚动的距离要大于 s s s ,所以一个首先满足的 t u r n s turns turns 是 s c ∗ 360 30 \frac{s}{c}*\frac{360}{30} cs∗30360
剩下的 s 1 = s % c s1=s\%c s1=s%c 的路程：需要满足在 t u r n s turns turns 的情况下还要让两个车轮的边同时平行于地面
首先去求共同转动的角度 α \alpha α 能够两车轮的边同时平行于地面： α = 360 ° / g c d ( f , b ) \alpha=360°/gcd(f,b) α=360°/gcd(f,b) , f , b f,b f,b 为输入的车轮顶点数
随后求 α \alpha α 和 t u r n s turns turns 的最小公倍数，即 θ = l c m ( α , 30 ° ) \theta=lcm(\alpha,30°) θ=lcm(α,30°)
对于 s % c s\%c s%c 需要的角度是 β = ⌈ ( s % c ) / c 360 ° ⌉ \beta=\lceil(s\%c)/\ \frac{c}{360°}\rceil β=⌈(s%c)/ 360°c⌉
所以最后 s % c s\%c s%c 需要的 t u r n s turns turns 是 ⌈ β θ ⌉ / 30 ° \lceil \frac{\beta}{\theta} \rceil /30° ⌈θβ⌉/30°

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 3200;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline ll rd() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
int sgn(double x) {if (fabs(x) < eps)return 0;return x < 0 ? -1 : 1;
}
ll gcd(ll x, ll y) {if (!y) return x;return gcd(y, x % y);
}
ll c, f, b, s;
int main() {int T;scanf("%d", &T);while (T--) {scanf("%lld %lld %lld %lld", &c, &f, &b, &s);ll ans = 12LL * (s / c);s %= c;ll tmp1 = gcd(f, b);ll t = gcd(360, tmp1);ll x = 1LL * 360 / t;ll y = 30 ;ll tmp2 = x * y / gcd(x, y);ll res = (s * 360 / c);if (s * 360 % c != 0) ++res;res = res / tmp2 + (res % tmp2 != 0);res = res * tmp2 / 30;ans += res;printf("%lld\n", ans);}
}

K. Kitchen Waste

有 n n n个面包， m m m口锅，每口锅中都装了一定体积的汤，要将汤淋在面包上，面包也有体积，一体积汤能淋一体积面包
一次淋面包的操作必须要将整个面包都淋上汤，若锅中剩余汤的体积不足以将整个面包都淋上，则将这个锅中的汤都倒掉
签到题，根据题意直接模拟即可

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e7 + 10;
const int INF = 0x3f3f3f3f;
inline ll read() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
int n, m;
int a[N], b[N];
int main() {n = read();m = read();for (int i = 1; i <= n; ++i) a[i] = read();for (int i = 1; i <= m; ++i)b[i] = read();int pos = 1;for (int i = 1; i <= m; ++i) {while (b[i] >= a[pos]) {b[i] -= a[pos];++pos;if (pos > n)break;}if (pos > n)break;}int ans = 0;for (int i = 1; i <= m; ++i)ans += b[i];printf("%d\n", ans);return 0;
}

L. Lets Count Factors

题意很明确，求 A , B A,B A,B 的共同质因数
这题稍有不慎就会超时，下欧拉筛的范围控制到 1 0 7 \sqrt{10^7} 107 就可以了
对于共同质因数的标记，使用 u n o r d e r e d _ m a p unordered\_map unordered_map 就万无一失了

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 3200;
const int INF = 0x3f3f3f3f;
inline ll rd() {ll x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}return x * f;
}
int cnt;
int vis[N], pri[N];
void Euler(int n) {for (int i = 2; i <= n; i++) {if (!vis[i])pri[++cnt] = i;for (int j = 1; j <= cnt; j++) {if (i * pri[j] > n)break;vis[i * pri[j]] = true;if (i % pri[j] == 0)break;}}
}
unordered_map<int, bool>visn;
void cal(int x) {for (int i = 1;i <= cnt;++i) {if (x % pri[i] == 0) {visn[pri[i]] = 1;while (x % pri[i] == 0) x /= pri[i];}}if (x > 1) visn[x] = 1;
}
void solve() {int a = (int)rd(), b = (int)rd();visn.clear();cal(a);cal(b);printf("%d\n", visn.size());
}
int main() {int T = (int)rd();Euler(N);while (T--)solve();return 0;
}

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2020-2021 ICPC - Gran Premio de Mexico - Repechaje

A. Atsa’s Checkers Board

C. CLETS Patrols

D. Determine the Winner Marshaland

E. Enterprise Recognition Program

G. Goombas Colliding

H. Hamsters Training

J. Just Turn the Wheels!

K. Kitchen Waste

L. Lets Count Factors

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