AEIOU
选出的子串中由AEI构成的子串和由OU构成的子串之间并没有什么关系，分别算出最长的加起来。
#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 dp[1110000]; int p[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.size(); int max1 = 0, max2 = 0; memset(dp, 0, sizeof(dp)); for (int i = 0; i < 256; i++) p[i] = -1; for (int i = 0; i < n; i++) { if (s[i] == 'a') { dp[i] = 1; if (p['a'] != -1) dp[i] = max(dp[i], dp[p['a']] + 1); if (dp[i] > dp[p['a']]) p['a'] = i; } if (s[i] == 'e') { dp[i] = 1; if (p['a'] != -1) dp[i] = max(dp[i], dp[p['a']] + 1); if (p['e'] != -1) dp[i] = max(dp[i], dp[p['e']] + 1); if (dp[i] > dp[p['e']]) p['e'] = i; } if (s[i] == 'i') { dp[i] = 1; if (p['a'] != -1) dp[i] = max(dp[i], dp[p['a']] + 1); if (p['e'] != -1) dp[i] = max(dp[i], dp[p['e']] + 1); if (p['i'] != -1) dp[i] = max(dp[i], dp[p['i']] + 1); if (dp[i] > dp[p['i']]) p['i'] = i; } max1 = max(max1, dp[i]); } memset(dp, 0, sizeof(dp)); for (int i = 0; i < 256; i++) p[i] = -1; for (int i = 0; i < n; i++) { if (s[i] == 'o') { dp[i] = 1; if (p['o'] != -1) dp[i] = max(dp[i], dp[p['o']] + 1); if (dp[i] > dp[p['o']]) p['o'] = i; } if (s[i] == 'u') { dp[i] = 1; if (p['o'] != -1) dp[i] = max(dp[i], dp[p['o']] + 1); if (p['u'] != -1) dp[i] = max(dp[i], dp[p['u']] + 1); if (dp[i] > dp[p['u']]) p['u'] = i; } max2 = max(max2, dp[i]); } cout << max1 + max2 << endl; return 0; }
数字游戏
这题看别人的代码貌似跟全排列有关，但是没想通关系在哪。
我的解法是这样的：首先加减和交换的顺序并不会影响操作的步数，所以可以看做是先进行若干步交换，然后依次把每一位加减为B的对应位数字。以A为初始状态进行bfs，然后卡时，就过了、
#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); } char A[15], B[15]; int cost[10][10]; lint p[15]; int n; struct state { int step; char num[15]; bool operator < (state s) const { if (step == s.step) return false; else return step > s.step; } }; lint calc(state s) { lint rtn = 0; for (int i = 0; i < n; i++) rtn += s.num[i] * p[n - i - 1]; return rtn; } map<lint, int> mp; priority_queue<state> q; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); scanf("%d%s%s", &n, A, B); for (int i = 0; i < n; i++) { A[i] -= '0'; B[i] -= '0'; } p[0] = 1; for (int i = 1; i < 15; i++) p[i] = p[i - 1] * 10; for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (i == j) cost[i][j] = 0; if (i > j) cost[i][j] = cost[j][i]; if (i < j) cost[i][j] = min(j - i, i + 10 - j); } } lint b = 0; for (int i = 0; i < n; i++) b += B[i] * p[n - i - 1]; state s, u; s.step = 0; memcpy(s.num, A, sizeof(A)); q.push(s); mp[calc(s)] = 1; int ans = 1000, iter = 0; while (!q.empty()) { s = q.top(); q.pop(); int step = s.step; for (int i = 0; i < n; i++) step += cost[s.num[i]][B[i]]; ans = min(ans, step); iter++; if (iter > 50000) break; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { u = s; char t = u.num[i]; u.num[i] = u.num[j]; u.num[j] = t; u.step++; lint c = calc(u); if (mp.find(c) != mp.end()) continue; else { mp[c] = 1; q.push(u); } } } } cout << ans << endl; return 0; }
第K小分数
前K个分数显然由每个质数构成的分数的前连续若干项构成，由于最多有1000个质数，所以最多有1000个最后一项，所以第K个分数一定是这最多1000个最后一项中的一个。可以用二分法来确定每种分母的最后一项并计算其排名，若其排名为K，则找到了题目要求的分数。
#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); } lint p[2000]; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif //makedata(); std::ios::sync_with_stdio(0), cin.tie(0); int n; lint k; cin >> n >> k; for (int i = 0; i < n; i++) cin >> p[i]; for (int i = 0; i < n; i++) { lint l = 0, r = p[i], mid; while (l + 1 < r) { mid = (l + r) >> 1; lint tmp = 0; for (int j = 0; j < n; j++) tmp += mid * p[j] / p[i]; if (tmp == k) cout << mid << '/' << p[i] << endl; if (tmp <= k) l = mid; else r = mid; } } return 0; }
逆序异或和
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 100010; int n, a[N]; struct BIT { int c[N]; void init() { memset(c, 0, sizeof(c)); } void insert(int x) { for (; x < N; x += x & -x) c[x] ++; } int count(int x, int res=0) { for (; x; x ^= x & -x) res += c[x]; return res; } }bit[20], num; int main() { scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", a + i); num.init(); for (int i = 0; i < 20; ++i) bit[i].init(); ll ans = 0; for (int i = n; i >= 1; --i) { int tar = num.count(a[i] - 1); num.insert(a[i]); for (int j = 0; j < 20; ++j) { int _num = 1 << j; int _bit = a[i] & _num; int _bit_tar = bit[j].count(a[i] - 1); if (_bit) { ans += 1LL * _num * (tar - _bit_tar); bit[j].insert(a[i]); } else { ans += 1LL * _num * _bit_tar; } } } cout << ans << endl; return 0; }