The 18th Zhejiang Provincial Collegiate Programming Contest 补题记录（ACFGJLM）
Posted tags: 篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了The 18th Zhejiang Provincial Collegiate Programming Contest 补题记录（ACFGJLM）相关的知识，希望对你有一定的参考价值。 补题链接：Here 签到题，求和判断即可 给定 8 个点(三维XYZ），判断是否是正方体 判断是否有 12 个相等的棱， 12 个相等的面对角线，4 个相等的体对角线，以及棱，面对角线和体对角线可以形成直角三角形（满足勾股定理）就行了。 \\(n\\) 只能减少（并且 \\(n\\) 不能为 \\(0\\)），\\(m\\) 只能增加，求出最终达到状态使 \\(n*k=m\\) 时，\\(n\\) 减少的值与 \\(m\\) 增加的值之和最小。 数论题(分治处理），类似某场ARC的题 \\(n >= m\\) 时，不难发现输出 \\(n-m\\) 即可 \\(n < m\\) 时，我们可以枚举n来求出最终答案，但是枚举并不是一个一个加，因为中间的值在跳跃，所以我们每次确定一个 \\(l\\) ，\\(r = min(n,(m - 1)/((m-1)/l))\\) ，而下个 \\(l = r + 1\\) 此时 \\(k = ⌊\\frac{m-1}i*i⌋\\) 题意待补 题意待补 算是比较套路的题。 根据贪心的原则，我们每次取一个宝珠后应该立马返回到原点。 而我们到这个宝珠的位置走的路和返回时走的路应该是最短路。 转换一下就是我们要取一个价值为 \\(a[i]\\) 的宝珠，需要消耗容量为\\(2∗dis[i]\\) 的时间（\\(dis[i]\\) 表示从 \\(1\\) 到 \\(i\\) 的最短路）。 这样就转换成了经典的完全背包问题。 关于背包问题不懂的可以看经典的背包九讲。 读了半天题，结果题意有问题。。。 给的字符串应该是匹配串。 因此我们判断一下是否存在前缀子串的最长公共前后缀长度不为 \\(0\\)（即 更简单的做法就是判断是否有字符和第一个字符相等即可。 输出 首先我们知道每个学生是不知道 \\(Grammy\\) 的选择，而又因为学生每次都是最优的选择，因此我们可以先算出一个学生在 \\([1 ,20]\\) 选择一个数后赢的分数的期望，每次都选择这\\(20\\) 个数中期望最大的即可。 而学生和 \\(Grammy\\) 的分数是互补的，只有一个输了，另一个赢了的情况 所以 \\(Grammy\\) 胜利的期望也是 \\(0\\) 以上是关于The 18th Zhejiang Provincial Collegiate Programming Contest 补题记录（ACFGJLM）的主要内容，如果未能解决你的问题，请参考以下文章 ZJCPC2021 第18届 浙江省赛The 18th Zhejiang Provincial Collegiate Programming Contest（ACFGJLM 7题） The 17th Zhejiang Provincial Contest B. Bin Packing Problem（线段树+map） The 15th Zhejiang Provincial Collegiate Programming Contest（部分题解） The 19th Zhejiang Provincial Collegiate Programming Contest F - Easy Fix(主席树) The 19th Zhejiang Provincial Collegiate Programming Contest F - Easy Fix(主席树) The 19th Zhejiang Provincial Collegiate Programming Contest F - Easy Fix(主席树)
A. League of Legends
ll suma, sumb;
void solve() {
ll x;
for (int i = 1; i <= 5; ++i)cin >> x, suma += x;
for (int i = 1; i <= 5; ++i)cin >> x, sumb += x;
if (suma < sumb) cout << "Red\\n"; // 注意这里要写suma<sumb
else cout << "Blue\\n";
}
C. Cube
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll inf = 0x3f3f3f3f3f3f3f3f;
struct Point {
int x, y, z;
Point() {}
Point(int x, int y, int z): x(x), y(y), z(z) {}
bool operator<(const Point b) const {
if (x != b.x) return x < b.x;
if (y != b.y) return y < b.y;
if (z != b.z) return z < b.z;
return 0;
}
};
vector<Point>a;
set<Point>a4, ano4;
ll maxDis;
ll getDis(Point a, Point b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) + (a.z - b.z) * (a.z - b.z);
}
void init() {
maxDis = -1;
a4.clear(), ano4.clear();
}
bool sol() {
init();
for (int i = 0; i < 8; ++i)
for (int j = 0; j < i; ++j)
if (!(a[i] < a[j]) and !(a[j] < a[i]))return false;
vector<ll>diss;
for (int i = 0; i < 8; ++i)
for (int j = 0; j < i; ++j)
diss.push_back(getDis(a[i], a[j]));
sort(diss.begin(), diss.end());
for (int i = 0; i < 12; ++i)
for (int j = 0; j < j; ++j)
if (diss[i] != diss[j])return false;
for (int i = 0; i < 12; ++i)
for (int j = 0; j < j; ++j)
if (diss[12 + i] != diss[12 + j])return false;
for (int i = 0; i < 4; ++i)
for (int j = 0; j < i; ++j)
if (diss[24 + i] != diss[24 + j])return false;
if (diss[0] + diss[12] != diss[24]) return 0;
return true;
}
void solve() {
a.resize(8);
for (int i = 0; i < 8; ++i) {
cin >> a[i].x >> a[i].y >> a[i].z;
}
cout << (sol() ? "YES\\n" : "NO\\n");
}
int main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int _;
for (cin >> _; _--;) solve();
return 0;
}
F. Fair Distribution
using ll = long long;
void solve() {
ll n, m; cin >> n >> m;
if (n > m)cout << n - m << "\\n";
else {
ll cnt = 0x3f3f3f3f;
for (ll l = 1, r; l <= n; l = r + 1) {
r = min(n, (m - 1) / ((m - 1) / l));
cnt = min(cnt, (m - 1) / l * l);
}
cout << cnt + n - m << \'\\n\';
}
}
G. Wall Game
// 待补#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (register int i = (s); i <= (t); i++)
#define RP(i,t,s) for (register int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define pY puts("YES")
#define pN puts("NO")
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\\n";
#define outval2(a,b) cout << "Debuging...|" << #a << ": " << a <<"\\t"<< #b << ": " << b << "\\n";
#define outval3(a,b,c) cout << "Debuging...|" << #a << ": " << a <<"\\t"<< #b << ": " << b <<"\\t"<< #c << ": " << c << "\\n";
using namespace std;
int debug = 0;
ll gcd(ll a, ll b) {
return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
return a / gcd(a, b) * b;
}
inline int read() {
int s = 0, f = 1;
char ch = getchar();
while (ch < \'0\' || ch > \'9\') {
if (ch == \'-\') f = -1;
ch = getchar();
}
while (ch >= \'0\' && ch <= \'9\') {
s = s * 10 + ch - \'0\';
ch = getchar();
}
return s * f;
}
const int N = 5e5 + 7;
int fa[N];
int w[N];
map<pii, int> mp;
int dir[6][2] = {0, 1, 1, 0, 1, -1, 0, -1, -1, 0, -1, 1};
int find(int x) {
return x == fa[x] ? fa[x] : fa[x] = find(fa[x]);
}
void solve() {
int n = read();
int tot = 0;
while (n--) {
int op = read(), x = read(), y = read();
if (op == 1) {
mp[m_p(x, y)] = ++tot;
fa[tot] = tot;
w[tot] = 6;
int sum = 0;
int minus = 0;
int cnt = 0;
set<int> ss;
vector<pii> vv;
rp(k, 0, 5) {
int nx = x + dir[k][0];
int ny = y + dir[k][1];
if (mp.find(m_p(nx, ny)) == mp.end()) continue;
int id = find(mp[m_p(nx, ny)]);
vv.push_back(m_p(k, id));
if (ss.find(id) != ss.end()) cnt++;
else {
cnt++;
sum += w[id];
ss.insert(id);
}
}
int len = vv.size();
rp(k, 0, len - 1) {
rp(j, 0, len - 1) {
if (vv[k].first == 0 && vv[j].first == 1 && vv[k].second != vv[j].second) minus++;
if (vv[k].first == 1 && vv[j].first == 2 && vv[k].second != vv[j].second) minus++;
if (vv[k].first == 2 && vv[j].first == 3 && vv[k].second != vv[j].second) minus++;
if (vv[k].first == 3 && vv[j].first == 4 && vv[k].second != vv[j].second) minus++;
if (vv[k].first == 4 && vv[j].first == 5 && vv[k].second != vv[j].second) minus++;
if (vv[k].first == 5 && vv[j].first == 0 && vv[k].second != vv[j].second) minus++;
}
}
for (auto val : ss) {
// cout<<val<<" ";
fa[val] = tot;
}
// cout<<endl;
w[tot] = sum + w[tot] - 2 * cnt - 2 * minus;
// cout<<w[tot]<<endl;
// outval3(sum,cnt,minus);
} else {
cout << w[find(mp[m_p(x, y)])] << endl;
}
}
// for(auto val:mp){
// pii t=val.first;
// outval3(t.first,t.second,find(mp[t]));
// }
}
int main() {
//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
//debug = 1;
#endif
//time_t beg, end;
//if(debug) beg = clock();
solve();
/*
if(debug) {
end = clock();
printf("time:%.2fs\\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);
}
*/
return 0;
}
J. Grammy and Jewelry
const int N = 3010, inf = 0x3f3f3f3f;
vector<int>e[N];
struct node {
int u, w;
node() {}
node(int u, int w): u(u), w(w) {}
friend bool operator <(node a, node b) {
return a.w > b.w;
}
};
int a[N], dis[N], vis[N], dp[N];
int n, m, t;
void Dijlstra() {
priority_queue<node> q;
for (int i = 1; i <= n; ++i)dis[i] = inf;
q.push(node{1, 0});
dis[1] = 0;
memset(vis, 0, sizeof vis);
while (!q.empty()) {
node t = q.top();
q.pop();
vis[t.u] = 1;
int u = t.u;
int w = t.w;
// cout << u << " " << w << endl;
for (auto v : e[u]) {
// cout << v << " " << dis[v] << endl;
if (!vis[v] && dis[v] > w + 1) {
dis[v] = w + 1;
q.push(node{v, dis[v]});
}
}
}
}
void solve() {
cin >> n >> m >> t;
for (int i = 2; i <= n; ++i)cin >> a[i];
for (int i = 1; i <= m; ++i) {
int u, v; cin >> u >> v;
if (u == v)continue;
e[u].push_back(v);
e[v].push_back(u);
}
Dijlstra();
for (int i = 1; i <= n; ++i)
for (int j = dis[i] * 2; j <= t; ++j)
dp[j] = max(dp[j], dp[j - dis[i] * 2] + a[i]);
for (int i = 1; i <= t; ++i)cout << dp[i] << (i == t ? "\\n" : " ");
}
L. String Freshman
next 数组不为 \\(0\\) ）。void solve() {
int n; string s;
cin >> n >> s;
for (int i = 1; i <= n - 1; ++i)
if (s[i] == s[0]) {
cout << "Wrong Answer\\n";
return ;
}
cout << "Correct\\n";
}
M. Game Theory
0.0000 即可。
我们写出计算期望的代码可以发现，每个数赢的分数的期望都是 \\(0\\) 。void solve() {
int n; cin >> n;
cout << fixed << setprecision(4) << 0.0;
}