最大的K-偏差排列
每次取可选范围里的最大的数字,如果最左侧的数字还没有使用就直接使用最左侧的数字
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#pragma comment(linker, "/STACK:102400000,102400000") #include<stdio.h> #include<string.h> #include<stdlib.h> #include<vector> #include<algorithm> #include<iostream> #include<map> #include<queue> #include<stack> #include<string> #include<functional> #include<math.h> //#include<bits/stdc++.h> using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); int n, k; bool f[200]; cin >> n >> k; memset(f, false, sizeof(f)); for (int i = 1; i <= n; i++) { if (i - k > 0 && !f[i - k]) { f[i - k] = true; cout << i - k << ' '; continue; } for (int j = i + k; j > i - k; j--) { if (j > n || j < 0) continue; if (!f[j]) { f[j] = true; cout << j << ' '; break; } } } return 0; }
孤独的字符
枚举每个字符,计算它是一个孤独字符时包含它的区间有多少个
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#pragma comment(linker, "/STACK:102400000,102400000") #include<stdio.h> #include<string.h> #include<stdlib.h> #include<vector> #include<algorithm> #include<iostream> #include<map> #include<queue> #include<stack> #include<string> #include<functional> #include<math.h> //#include<bits/stdc++.h> using namespace std; typedef long long lint; typedef vector<int> VI; typedef pair<int, int> PII; typedef queue<int> QI; void makedata() { freopen("input.txt", "w", stdout); fclose(stdout); } string s; int pre[110000], nex[110000], last[256]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); cin >> s; int n = s.length(); memset(last, -1, sizeof(last)); for (int i = 0; i < n; i++) { pre[i] = last[s[i]]; nex[i] = n; if (last[s[i]] >= 0) nex[pre[i]] = i; last[s[i]] = i; } lint ans = 0; for (int i = 0; i < n; i++) { ans += 1LL * (nex[i] - i) * (i - pre[i]); } cout << ans << endl; return 0; }
秋天来了
这题有问题,都说了l是唯一最高完了样例好几个最高???最高的还能看到别人???
Nim森林
这题跟这棵树没啥关系。两个回合以后,是一个典型的反nim游戏,先手获胜的条件为:1)所有堆的石子个数为1,且NIM_sum=0或2)至少有一堆的石子个数大于1,且 NIM_sum≠0。第一种获胜情况是所有堆狮子个数为1且有偶数堆,这种情况是不可能达到的。经过第一回合操作后,只要有石子数为1的石堆,B都可以拿完多于1的石堆且剩下奇数个石子数为1的石堆,所以只需考虑第二种情况(且所有石堆石子数大于1)。按代价从大到小的顺序添加各个石堆的石子数到线性基中,成功添加进去的就是第二回合给B留下的。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; using LL = int64_t; using LD = long double; const LL INF = 0x3f3f3f3f; const LL mod = 1e9 + 7; template <typename T> struct LnBase{ int sz, szc; T *x; int *y; LnBase (){x = 0; sz = sizeof(T) << 3; szc = -1; resize(sz);} void resize(int size){ sz = size; if(!x) delete(x); x = new T[sz + 2]; y = new int[sz + 2]; memset(x, 0, sz * sizeof(T)); memset(y, 0, sz << 2); } T operator[](int h){return x[h];} int add(T v){ for(int i = sz - 1; i >= 0; i--) if(v & (T)1 << i){if(!x[i]){x[i] = v; szc = -1; return i;} v ^= x[i];} return -1; } int find(T v){ for(int i = sz - 1; i >= 0; i--){ if(v & (T)1 << i && x[i]) v ^= x[i]; if(!v) return 1;} return 0; } T Max(){ T s = 0; for(int i = sz - 1; i >= 0; i--){if((s ^ x[i]) > s) s ^= x[i];} return s; } T Min(){ for(int i = 0; i < sz; i++) if(x[i]) return x[i]; return -1; } void Canonicity(){ int i, j; for(i = sz - 1; i > 0; i--) for(j = i - 1; j >= 0; j--) if(x[i] & (T)1 << j) x[i] ^= x[j]; for(szc = i = 0; i < sz; i++) if(x[i]) y[szc++] = i; } T Kth(long long K){ if(szc < 0) Canonicity(); if(K >= 1ll << szc) return -1; T s = 0; for(int i = szc - 1; i >= 0; i--) if(K & 1ll << i) s ^= x[y[i]]; return s; } }; struct node { __int128 a, t; bool operator<(const node &e) const { return t > e.t; } }; int main() { ios::sync_with_stdio(0); cin.tie(0); int n; cin >> n; n--; vector<node> a(n); for (auto &i : a) { LL u, v, a, t; cin >> u >> v >> a >> t; i.a = a; i.t = u ^ v ^ t; } sort(a.begin(), a.end()); int flag = 1; __int128 ans = 0, sum = 0; LnBase<__int128> b; for (auto& s : a) { sum += s.t; if (s.a == 1) continue; flag = 0; if (b.add(s.a) != -1) ans += s.t; } if (flag) cout << "No "; else { ans = sum - ans; string s = ""; while (ans) { s += '0' + ans % 10; ans /= 10; } if (s == "") s = "0"; reverse(s.begin(), s.end()); cout << s << ' '; } }