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【LGR-109】洛谷 5 月月赛 II & Windy Round 6
P8344 「Wdoi-6」走在夜晚的莲台野
P8344 「Wdoi-6」走在夜晚的莲台野
这个就是推一个公式，比较简单，但是当时减号写成加号没看出来，无语，画了十多分钟#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <string> #include <cstring> #include <cmath> #include <stack> #include<bits/stdc++.h> using namespace std; #define ll long long #define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define sc(a) scanf("%lld",&a) #define pf(a) printf("%d",a) #define endl "\n" #define int long long #define mem(a,b) memset(a,b,sizeof a) #define ull unsigned long long #define INF 0x3f3f3f3f3f3f3f3f #define inf 0x3f3f3f3f #define rep(i,a,b) for(auto i=a;i<=b;++i) #define bep(i,a,b) for(auto i=a;i>=b;--i) #define LL long long #define lowbit(x) x&(-x) #define PII pair<int,int> #define PLL pair<ll,ll> #define PI acos(-1) #define pb push_back #define x first #define y second const double eps = 1e-6; const int mod = 998244353; const int MOD = 1e9 + 7; int x, y, z; signed main() { int t; cin >> t; while (t--) { cin >> x >> y >> z; int p = x * z - (x + 1) * x / 2 + z - x; if(x > z || p < y ) cout << "Merry" << endl; else cout << "Renko" << endl; } return 0; }- 1
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P8345 「Wdoi-6」华胥之梦
P8345 「Wdoi-6」华胥之梦
有人瞎说图论，差点信了
也是个推式子的题，就直接写
赛时推的#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <string> #include <cstring> #include <cmath> #include <stack> #include<bits/stdc++.h> using namespace std; #define ll long long #define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define sc(a) scanf("%lld",&a) #define pf(a) printf("%d",a) #define endl "\n" #define int long long #define mem(a,b) memset(a,b,sizeof a) #define ull unsigned long long #define INF 0x3f3f3f3f3f3f3f3f #define inf 0x3f3f3f3f #define rep(i,a,b) for(auto i=a;i<=b;++i) #define bep(i,a,b) for(auto i=a;i>=b;--i) #define LL long long #define lowbit(x) x&(-x) #define PII pair<int,int> #define PLL pair<ll,ll> #define PI acos(-1) #define pb push_back #define x first #define y second const double eps = 1e-6; const int mod = 998244353; const int MOD = 1e9 + 7; const int N = 1e6 + 10; int a[N]; int n, c, q; signed main() { fast; cin >> n >> c >> q; for (int i = 1; i <= n; i++) cin >> a[i]; vector<int> v; while(q --) { v.clear(); int x, y; cin >> x; int sum = 0; for(int i=1; i<=x; i++) { cin >> y; v.push_back(a[y]); sum += a[y]; } sort(v.begin(), v.end()); cout << (x - 1) * c - sum - v[x - 1] + 2 * v[0] << endl; } return 0; }- 1
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P8346 「Wdoi-6」最澄澈的空与海
这个题大意就是
这个题目找出来我代码的一些隐患吧
在洛谷，vector不能这么清空
1e6的读入差不多就要用快读了，注意STL，这里用set想去掉重边，给T了，裂开#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <string> #include <cstring> #include <cmath> #include <stack> #include<bitsdc++.h> using namespace std; #define ll long long #define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define sc(a) scanf("%lld",&a) #define pf(a) printf("%d",a) #define endl "\n" #define int long long #define mem(a,b) memset(a,b,sizeof a) #define ull unsigned long long #define INF 0x3f3f3f3f3f3f3f3f #define inf 0x3f3f3f3f #define rep(i,a,b) for(auto i=a;i<=b;++i) #define bep(i,a,b) for(auto i=a;i>=b;--i) #define LL long long #define lowbit(x) x&(-x) #define PII pair<int,int> #define PLL pair<ll,ll> #define PI acos(-1) #define pb push_back #define x first #define y second const double eps = 1e-6; const int mod = 998244353; const int MOD = 1e9 + 7; const int N = 2e6 + 10; int n, m, t; vector<int> mp[N]; int in[N]; int u, v; set<PII> s; bool vis[N]; signed main() { fast; cin >> t; while(t --) { cin >> n >> m; for(int i=1; i<=m; i++) { cin >> u >> v; v += n; // if(s.count({u, v}) || s.count({v, u})) continue; mp[u].push_back(v); mp[v].push_back(u); in[u] ++ ; in[v] ++; // s.insert({u, v}); } n *= 2; int res = 0; queue<int> q; for(int i=1; i<=n; i++) if(in[i] == 1) q.push(i); while(q.size()) { u = q.front(); q.pop(); if(vis[u] || in[u] != 1) continue; vis[u] = 1; in[u] --; for(auto i: mp[u]) { if(in[i] == 1 && !vis[i]) { res ++; in[i] --; vis[i] = 1; } else if(!vis[i]) { res ++; vis[i] = 1; for(auto j : mp[i]) { -- in[i]; -- in[j]; if(in[j] == 1 && !vis[j]) q.push(j); } } } } if(res == n / 2) cout << "Renko" << endl; else cout << "Merry" << endl; // s.clear(); for(int i=1; i<=n; i++) { in[i] = 0; vis[i] = 0; mp[i].clear(); } } return 0; }- 1
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P8347 「Wdoi-6」另一侧的月
疑问和注意点在代码里面
大致方法就是，如果能找到
圈起来的这种结构，就一定是先手必胜，此外还有别的情况，不一一列举了，想看的自己看吧!((n + 1) & n)讲一下这个判断，这是判断n+1是不是2的幂次方的，也就是能判断满二叉树，感觉参考1不应该过，有bug，我再想想
#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <string> #include <cstring> #include <cmath> #include <stack> #include<bits/stdc++.h> using namespace std; #define ll long long #define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define sc(a) scanf("%lld",&a) #define pf(a) printf("%d",a) #define endl "\n" #define int long long #define mem(a,b) memset(a,b,sizeof a) #define ull unsigned long long #define INF 0x3f3f3f3f3f3f3f3f #define inf 0x3f3f3f3f #define rep(i,a,b) for(auto i=a;i<=b;++i) #define bep(i,a,b) for(auto i=a;i>=b;--i) #define LL long long #define lowbit(x) x&(-x) #define PII pair<int,int> #define PLL pair<ll,ll> #define PI acos(-1) #define pb push_back #define x first #define y second const double eps = 1e-6; const int mod = 998244353; const int MOD = 1e9 + 7; const int N = 1e5 + 10; int in[N]; vector<int> v[N]; bool dfs(int u, int fa) { for(auto i : v[u]) { if(i == fa) continue; if(!(in[u] & 1) && in[i] == 1) return true; if(dfs(i, u)) return true; } return false; } signed main() { int t; cin >> t; while (t--) { int n; cin >> n; int a, b; for(int i=1; i<n; i++) { cin >> a >> b; v[a].push_back(b); v[b].push_back(a); in[a] ++; in[b] ++; } bool flag = 0; for(int i=1; i<=n; i++) if(in[i] != 2) flag = 0; if(dfs(1, -1) || !((n + 1) & n)) cout << "Hifuu" << endl; //!((n + 1) & n)这个判断的是是否为满二叉树，但是这个判断方法感觉不严谨，不知道为什么要特判 //同学有种写法，思路一样，但是他直接判断有没有度是2的点，也能过，就不用特判了 else cout << "Luna" << endl; for(int i=1; i<=n; i++) { in[i] = 0; v[i].clear(); } } return 0; }- 1
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- 原文地址：https://blog.csdn.net/weixin_51176105/article/details/125611681
