A. Roman and Browser
签到.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int n, k, a[110]; 5 6 int get(int b) 7 { 8 int vis[110]; 9 memset(vis, 0, sizeof vis); 10 int c = b; 11 while (c <= n) 12 { 13 vis[c] = 1; 14 c += k; 15 } 16 c = b - k; 17 while (c >= 1) 18 { 19 vis[c] = 1; 20 c -= k; 21 } 22 int l = 0, r = 0; 23 for (int i = 1; i <= n; ++i) if (!vis[i]) 24 { 25 if (a[i] == 1) ++l; 26 else ++r; 27 } 28 return abs(l - r); 29 } 30 31 int main() 32 { 33 while (scanf("%d%d", &n, &k) != EOF) 34 { 35 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 36 int res = 0; 37 for (int i = 1; i <= n; ++i) res = max(res, get(i)); 38 printf("%d ", res); 39 } 40 return 0; 41 }
B. Build a Contest
签到.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 int n, m, a[N]; 6 int cnt[N]; 7 8 int main() 9 { 10 while (scanf("%d%d", &n, &m) != EOF) 11 { 12 for (int i = 1; i <= m; ++i) scanf("%d", a + i); 13 int need = n; 14 memset(cnt, 0, sizeof cnt); 15 for (int i = 1; i <= m; ++i) 16 { 17 if (cnt[a[i]] == 0) 18 --need; 19 ++cnt[a[i]]; 20 if (need) putchar('0'); 21 else 22 { 23 putchar('1'); 24 for (int j = 1; j <= n; ++j) 25 { 26 --cnt[j]; 27 if (!cnt[j]) 28 ++need; 29 } 30 } 31 } 32 puts(""); 33 } 34 return 0; 35 }
C. NN and the Optical Illusion
签到.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const double eps = 1e-8; 5 const double PI = acos(-1.0); 6 7 int main() 8 { 9 int n; double r; 10 while (scanf("%d%lf", &n, &r) != EOF) 11 { 12 double ang1 = 2.0 * PI / n; 13 double ang2 = (PI - ang1) / 2; 14 double R = (r * sin(ang1) * 1.0) / (2 * sin(ang2) - sin(ang1)); 15 printf("%.10f ", R); 16 } 17 return 0; 18 }
D. Dasha and Chess
Upsolved.
题意：
你有一个白王，对方有$666$个黑王
你每次可以八个方向走一步，对方可以任意移动一个黑王的位置
在$2000步以内如果你和某个黑王同行同列，你就赢了$
$给出你必胜的方案$
思路：
先走到中间，再考虑黑王个数最少的那个角落，向它的反方向走就可以了
考虑最差的情况，四个角很平均都有$166个左右$
那么也就是说另外三个角的个数加起来至少是$500个，而从中间走到某一角落只需要499步$
所以一定可以包围一个
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 #define pii pair <int, int> 6 #define x first 7 #define y second 8 int G[N][N], x, y, cnt[4]; 9 pii cor[N]; 10 11 void Move(int edx, int edy) 12 { 13 while (x != edx || y != edy) 14 { 15 if (x < edx && G[x + 1][y] == 0) ++x; 16 else if (x > edx && G[x - 1][y] == 0) --x; 17 if (y < edy && G[x][y + 1] == 0) ++y; 18 else if (y > edy && G[x][y - 1] == 0) --y; 19 printf("%d %d ", x, y); 20 fflush(stdout); 21 int k, a, b; 22 scanf("%d%d%d", &k, &a, &b); 23 if (k == -1) exit(0); 24 G[cor[k].x][cor[k].y] = 0; 25 cor[k].x = a, cor[k].y = b; 26 G[cor[k].x][cor[k].y] = 1; 27 } 28 } 29 30 int main() 31 { 32 while (scanf("%d%d", &x, &y) != EOF) 33 { 34 memset(G, 0, sizeof G); 35 for (int i = 1; i <= 666; ++i) 36 { 37 scanf("%d%d", &cor[i].x, &cor[i].y); 38 G[cor[i].x][cor[i].y] = 1; 39 } 40 Move(500, 500); 41 memset(cnt, 0, sizeof cnt); 42 for (int i = 1; i <= 999; ++i) 43 for (int j = 1; j <= 999; ++j) 44 if (G[i][j]) 45 { 46 if (i < x && j < y) ++cnt[0]; 47 else if (i < x && j > y) ++cnt[1]; 48 else if (i > x && j < y) ++cnt[2]; 49 else ++cnt[3]; 50 } 51 if (cnt[0] <= 166) Move(999, 999); 52 else if (cnt[1] <= 166) Move(999, 1); 53 else if (cnt[2] <= 166) Move(1, 999); 54 else Move(1, 1); 55 } 56 return 0; 57 }
E. Andrew and Taxi
Upsolved.
题意：
给出一张有向图，求改变一些边使得它没有环
改变一条边的花费是它的边权，要使得最大花费最小
输出最大花费和需要改变的边数
再输出相应的边
思路：
先二分答案，每次check的时候检查一下边权大于limit的边组成的图是否有环
如果有环，那么一定不行，如果没有环，那么一定可以
对于剩下的边，方向自定，拓扑序小的连向拓扑序大的
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 int n, m; 6 struct Graph 7 { 8 struct node 9 { 10 int to, nx, w, id; 11 node() {} 12 node (int to, int nx, int w, int id) : to(to), nx(nx), w(w), id(id) {} 13 }a[N << 1]; 14 int head[N], pos; 15 void init() 16 { 17 memset(head, 0, sizeof head); 18 pos = 0; 19 } 20 void add(int u, int v, int w, int id) 21 { 22 a[++pos] = node(v, head[u], w, id); head[u] = pos; 23 } 24 }G; 25 #define erp(u) for (int it = G.head[u], v = G.a[it].to, w = G.a[it].w, id = G.a[it].id; it; it = G.a[it].nx, v = G.a[it].to, w = G.a[it].w, id = G.a[it].id) 26 27 int degree[N], ord[N]; 28 bool check(int x) 29 { 30 memset(degree, 0, sizeof degree); 31 for (int i = 1; i <= n; ++i) erp(i) if (w > x) 32 ++degree[v]; 33 int cnt = 0; 34 queue <int> q; 35 for (int i = 1; i <= n; ++i) if (!degree[i]) 36 q.push(i); 37 while (!q.empty()) 38 { 39 ++cnt; 40 int u = q.front(); q.pop(); 41 ord[u] = cnt; 42 erp(u) if (w > x && --degree[v] == 0) 43 q.push(v); 44 } 45 return cnt >= n; 46 } 47 48 int main() 49 { 50 while (scanf("%d%d", &n, &m) != EOF) 51 { 52 G.init(); 53 for (int i = 1, u, v, w; i <= m; ++i) 54 { 55 scanf("%d%d%d", &u, &v, &w); 56 G.add(u, v, w, i); 57 } 58 int l = 0, r = (int)1e9, res = -1; 59 while (r - l >= 0) 60 { 61 int mid = (l + r) >> 1; 62 if (check(mid)) 63 { 64 res = mid; 65 r = mid - 1; 66 } 67 else 68 l = mid + 1; 69 } 70 vector <int> vec; 71 check(res); 72 for (int i = 1; i <= n; ++i) erp(i) if (w <= res && ord[i] > ord[v]) vec.push_back(id); 73 printf("%d %d ", res, (int)vec.size()); 74 for (int i = 0, len = vec.size(); i < len; ++i) printf("%d%c", vec[i], " "[i == len - 1]); 75 } 76 return 0; 77 }
F. Ivan and Burgers
Upsolved.
题意：
询问区间任意数异或最大值
思路：
维护前缀线性基，但是要注意将$位置id大的数更换基底$
$因为这样这些基底被用户更新答案的概率就更高$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 500010 5 int n, q, c[N]; 6 int pos[N][25], p[N][25]; 7 8 void work(int x) 9 { 10 for (int i = 0; i <= 20; ++i) 11 { 12 pos[x][i] = pos[x - 1][i]; 13 p[x][i] = p[x - 1][i]; 14 } 15 int tmp = c[x]; 16 int id = x; 17 for (int i = 20; i >= 0; --i) 18 { 19 if ((tmp >> i) & 1) 20 { 21 if (!p[x][i]) 22 { 23 p[x][i] = tmp; 24 pos[x][i] = id; 25 return; 26 } 27 if (pos[x][i] < id) 28 { 29 swap(tmp, p[x][i]); 30 swap(id, pos[x][i]); 31 } 32 tmp ^= p[x][i]; 33 } 34 } 35 } 36 37 int ask(int l, int r) 38 { 39 int res = 0; 40 for (int i = 20; i >= 0; --i) if (pos[r][i] >= l && (res ^ p[r][i]) > res) 41 res ^= p[r][i]; 42 return res; 43 } 44 45 int main() 46 { 47 while (scanf("%d", &n) != EOF) 48 { 49 for (int i = 1; i <= n; ++i) scanf("%d", c + i); 50 for (int i = 0; i < 25; ++i) pos[0][i] = p[0][i] = 0; 51 for (int i = 1; i <= n; ++i) work(i); 52 scanf("%d", &q); 53 for (int qq = 1, l, r; qq <= q; ++qq) 54 { 55 scanf("%d%d", &l, &r); 56 printf("%d ", ask(l, r)); 57 } 58 59 } 60 return 0; 61 }