标签：规律 题解 平面 ++ ace std color 找规律 rectangle
中规中矩的一场。
题目链接：http://acm.hdu.edu.cn/contests/contest_show.php?cid=855
C：
定义函数f(d,k)为数字d在数字k中出现的次数。给定d和x，找到尽量大的k使得k<=x且f(d,k)==k。
很诡异的一题，最好的做法仍然是打表找规律。题解给了一个很神奇的结论：满足条件的k<1011且k的分布非常稀疏。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const ll m = 1e5, maxn = 1e6 + 10; 21 ll a[maxn][10], ans[10][maxn], cnt[10]; 22 23 bool check(ll x, ll y) { 24 if (x < 0 && y >= 0) return 1; 25 if (x > y && x >= 0 && x < m) return 1; 26 return 0; 27 } 28 29 int main() { 30 for (int i = 1; i < m * 10; i++) { 31 for (int j = 0; j < 10; j++) a[i][j] = a[i / 10][j]; 32 a[i][i % 10]++; 33 } 34 for (int d = 1; d < 10; d++) { 35 int r = 1; 36 for (int i = 0; i < m * d; i++) 37 if (check(r, r - m / 2 + m * a[i][d])) { 38 for (int j = 0; j < m; j++) { 39 r += a[i][d] + a[j][d] - 1; 40 if (!r) ans[d][cnt[d]++] = m * i + j; 41 } 42 } else r += m * a[i][d] - m / 2; 43 } 44 int query; scanf("%d", &query); 45 while (query--) { 46 int d; ll k; scanf("%d%lld", &d, &k); 47 int r = cnt[d] - 1; 48 while (r >= 0 && ans[d][r] > k) r--; 49 if (r >= 0) printf("%lld\n", ans[d][r]); 50 else puts("0"); 51 } 52 return 0; 53 }
I：
给定两个矩形的左上角和右下角坐标，问这两个矩形把平面分成几个区域。
离散化是重点。
/* Codeforces Contest 2019_mutc_08 * Problem I * Au: SJoshua */ #include <set> #include <unordered_map> #include <queue> #include <cstdio> #include <vector> #include <string> #include <cstring> #include <iostream> #include <algorithm> using namespace std; bool board[20][20]; const int movement[4][2] = { {0, -1}, {0, 1}, {1, 0}, {-1, 0} }; struct node { int x, y; }; struct Rectangle { int x1, y1, x2, y2; void fill(void) { for (int x = x1; x <= x2; x++) { board[x][y1] = board[x][y2] = true; } for (int y = y1; y <= y2; y++) { board[x1][y] = board[x2][y] = true; } } }; void init(void) { memset(board, 0, sizeof(board)); } int calc(void) { int ans = 0; for (int i = 0; i < 20; i++) { for (int j = 0; j < 20; j++) { if (!board[i][j]) { queue <node> info; info.push({i, j}); while (!info.empty()) { auto t = info.front(); info.pop(); if (board[t.x][t.y]) { continue; } board[t.x][t.y] = true; for (int k = 0; k < 4; k++) { int nx = t.x + movement[k][0], ny = t.y + movement[k][1]; if (0 <= nx && nx < 20 && 0 <= ny && ny < 20 && !board[nx][ny]) { info.push({nx, ny}); } } } ans++; } } } return ans; } int main(void) { std::ios::sync_with_stdio(false); std::cin.tie(0); int T; cin >> T; while (T--) { vector <int> vec(8); set <int> hash; unordered_map <int, int> mp; for (int i = 0; i < 8; i++) { cin >> vec[i]; hash.insert(vec[i]); } int cur = 1; for (auto e : hash) { mp[e] = cur; cur += 2; } Rectangle A = {mp[vec[0]], mp[vec[1]], mp[vec[2]], mp[vec[3]]}; Rectangle B = {mp[vec[4]], mp[vec[5]], mp[vec[6]], mp[vec[7]]}; init(); A.fill(); B.fill(); cout << calc() << endl; } return 0; }
J：
签到题，水题。
1 /* Codeforces Contest 2019_mutc_08 2 * Problem J 3 * Au: SJoshua 4 */ 5 #include <cstdio> 6 #include <vector> 7 #include <string> 8 #include <iostream> 9 #include <algorithm> 10 11 using namespace std; 12 13 struct info { 14 string name; 15 int score, penalty; 16 }; 17 18 int main(void) { 19 std::ios::sync_with_stdio(false); 20 std::cin.tie(0); 21 int T; 22 cin >> T; 23 while (T--) { 24 int n, d; 25 cin >> n >> d; 26 vector <info> board(n); 27 for (int i = 0; i < n; i++) { 28 cin >> board[i].name >> board[i].score >> board[i].penalty; 29 } 30 sort(board.begin(), board.end(), [](info & a, info & b)->bool { 31 return a.score == b.score ? a.penalty < b.penalty : a.score > b.score; 32 }); 33 d *= 10; 34 if ((n * d) % 100 == 50) { 35 cout << board[(n * d) / 100].name << endl; 36 } else { 37 cout << "Quailty is very great" << endl; 38 } 39 } 40 return 0; 41 }
K：
有n个班级，一个班有ai个人并准备了bi杯奶茶。每个人只能喝一杯奶茶，而且不能喝本班的奶茶。问最多有多少人能喝奶茶。
按班级人数从大到小排序，然后对于第i个班，从第i+1个班开始绕一圈喝奶茶即可。
1 /* Codeforces Contest 2019_mutc_08 2 * Problem K 3 * Au: SJoshua 4 */ 5 #include <queue> 6 #include <cstdio> 7 #include <vector> 8 #include <string> 9 #include <list> 10 #include <iostream> 11 #include <algorithm> 12 13 using namespace std; 14 15 struct _Class { 16 int num, tea; 17 }; 18 19 int main(void) { 20 std::ios::sync_with_stdio(false); 21 std::cin.tie(0); 22 int T; cin >> T; 23 while (T--) { 24 int n; cin >> n; 25 vector <_Class> _class(n); 26 for (int i = 0; i < n; i++) { 27 cin >> _class[i].num >> _class[i].tea; 28 } 29 sort(_class.begin(), _class.end(), [](_Class & a, _Class & b) { 30 return a.tea == b.tea ? a.num > b.num : a.tea > b.tea; 31 }); 32 long long int ans = 0; 33 vector <int> nxt(n); 34 for (int i = 0; i < n - 1; i++) nxt[i] = i + 1; 35 nxt[n - 1] = 0; 36 for (int i = 0; i < n; i++) { 37 int nxt = i; 38 long long int cnt = 0, round = 0; 39 while (_class[i].num) { 40 round++; 41 int last = nxt; nxt = nxt[nxt]; 42 if (nxt == i) break; 43 if (_class[nxt].tea) { 44 int drink = min(_class[nxt].tea, _class[i].num); 45 _class[i].num -= drink, _class[nxt].tea -= drink; 46 ans += drink, cnt += drink; 47 } else nxt[last] = nxt[nxt]; 48 if (round > n) break; 49 } 50 if (!cnt && _class[i].num) break; 51 } 52 cout << ans << endl; 53 } 54 return 0; 55 }
2019 HDOJ Multi-University Training Contest Stage 8(杭电多校)
标签：规律 题解 平面 ++ ace std color 找规律 rectangle
原文地址：https://www.cnblogs.com/JHSeng/p/11354513.html
