Codeforces Round #354 (Div. 2)

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貪心 A Nicholas and Permutation

#include <bits/stdc++.h>typedef long long ll;
const int N = 1e5 + 5;
int a[105];
int pos[105];int main() {int n;scanf ("%d", &n);for (int i=1; i<=n; ++i) {scanf ("%d", a+i);pos[a[i]] = i;}int ans = abs (pos[1] - pos[n]);if (pos[n] != 1) {ans = std::max (ans, abs (pos[n] - 1));}if (pos[n] != n) {ans = std::max (ans, abs (pos[n] - n));}if (pos[1] != 1) {ans = std::max (ans, abs (pos[1] - 1));}if (pos[1] != n) {ans = std::max (ans, abs (pos[1] - n));}printf ("%d\n", ans);return 0;
}

模擬＋DFSB Pyramid of Glasses

設酒杯滿了值為1.0，每一次暴力傳遞下去

#include <bits/stdc++.h>typedef long long ll;
const int N = 1e5 + 5;
double a[15][15];
int n;void DFS(int x, int y, double v) {if (x == n + 1 || v == 0) {return ;}if (a[x][y] == 1.0) {DFS (x + 1, y, v / 2);DFS (x + 1, y + 1, v / 2);} else {double sub = 1.0 - a[x][y];if (sub <= v) {a[x][y] = 1.0;DFS (x + 1, y, (v - sub) / 2);DFS (x + 1, y + 1, (v - sub) / 2);} else {a[x][y] += v;}}
}int main() {int t;scanf ("%d%d", &n, &t);for (int i=1; i<=t; ++i) {DFS (1, 1, 1.0);}int ans = 0;for (int i=1; i<=n; ++i) {for (int j=1; j<=i+1; ++j) {if (a[i][j] == 1.0) {ans++;}}}printf ("%d\n", ans);return 0;
}

尺取法(two points) C Vasya and String

從左到右維護一段連續的區間，改變次數不大于ｋ，取最大值．

#include <bits/stdc++.h>const int N = 1e5 + 5;
char str[N];
int n, m;void solve() {int c[2] = {0};int ans = 0;for (int i=0, j=0; i<n; ++i) {while (j < n) {c[str[j] == 'a' ? 0 : 1]++;if (std::min (c[0], c[1]) <= m) {++j;} else {c[str[j] == 'a' ? 0 : 1]--;break;}}ans = std::max (ans, j - i);c[str[i] == 'a' ? 0 : 1]--;}printf ("%d\n", ans);
}int main() {scanf ("%d%d", &n, &m);scanf ("%s", str);solve ();return 0;
}

BFS(方向，旋轉)?D Theseus and labyrinth

多加一維表示旋轉次數(0~3)，dis[x][y][z]表示走到(x, y)旋轉z次后的最小步數．

#include <bits/stdc++.h>const int N = 1e3 + 5;
const int INF = 0x3f3f3f3f;
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
bool dir[N][N][4];
int dis[N][N][4];
char str[N];
struct Point {int x, y, d;
};
int n, m;
int sx, sy, ex, ey;bool judge(int x, int y) {if (x < 0 || x >= n || y < 0 || y >= m) {return false;} else {return true;}
}int BFS() {memset (dis, INF, sizeof (dis));dis[sx][sy][0] = 0;std::queue<Point> que;que.push ((Point) {sx, sy, 0});while (!que.empty ()) {Point p = que.front (); que.pop ();int &pd = dis[p.x][p.y][p.d];int td = (p.d + 1) % 4;if (dis[p.x][p.y][td] > pd + 1) {dis[p.x][p.y][td] = pd + 1;que.push ((Point) {p.x, p.y, td});}for (int i=0; i<4; ++i) {int tx = p.x + dx[i];int ty = p.y + dy[i];if (!judge (tx, ty)) {continue;}if (dir[p.x][p.y][(p.d+i)%4] && dir[tx][ty][(p.d+i+2)%4]) {if (dis[tx][ty][p.d] > pd + 1) {dis[tx][ty][p.d] = pd + 1;que.push ((Point) {tx, ty, p.d});}}}}int ret = INF;for (int i=0; i<4; ++i) {ret = std::min (ret, dis[ex][ey][i]);}return (ret != INF ? ret : -1);
}int main() {scanf ("%d%d", &n, &m);for (int i=0; i<n; ++i) {scanf ("%s", str);for (int j=0; j<m; ++j) {char ch = str[j];if (ch=='+' || ch=='-' || ch=='>' || ch=='U' || ch=='L' || ch=='D') dir[i][j][0] = true;if (ch=='+' || ch=='|' || ch=='^' || ch=='R' || ch=='L' || ch=='D') dir[i][j][1] = true;if (ch=='+' || ch=='-' || ch=='<' || ch=='R' || ch=='U' || ch=='D') dir[i][j][2] = true;if (ch=='+' || ch=='|' || ch=='v' || ch=='R' || ch=='U' || ch=='L') dir[i][j][3] = true;}}scanf ("%d%d%d%d", &sx, &sy, &ex, &ey);sx--; sy--; ex--; ey--;printf ("%d\n", BFS ());return 0;
}

數學 E The Last Fight Between Human and AI

原題轉化為求P(k)==0.如果k==0，判斷a[0]是否能被玩家設置成０．否則判斷剩余數字個數的奇偶以及現在是誰出手，判斷最后一步是否為玩家走，最后一步總能使得P(k)==0．

#include <bits/stdc++.h>typedef long long ll;
const int N = 1e5 + 5;int read() {int ret = 0, f = 1;char ch = getchar ();while (ch < '0' || ch > '9') {if (ch == '?') {return -11111;}if (ch == '-') {f = -1;}ch = getchar ();}while (ch >= '0' && ch <= '9') {ret = ret * 10 + (ch - '0');ch = getchar ();}return ret * f;
}int a[N];int main() {int n, k;scanf ("%d%d", &n, &k);int who = 0, m = 0;for (int i=0; i<=n; ++i) {a[i] = read ();if (a[i] != -11111) {who = 1 ^ who; //0: Computer, 1: player} else {m++;}}if (k == 0) {if (a[0] == -11111) {if (!who) {puts ("No");} else {puts ("Yes");}} else {if (a[0] == 0) {puts ("Yes");} else {puts ("No");}}} else {if (m & 1) {if (!who) {puts ("No");} else {puts ("Yes");}} else {if (m > 0) {if (!who) {puts ("Yes");} else {puts ("No");}} else {double sum = 0;for (int i=n; i>=0; --i) {sum = sum * k + a[i];}if (fabs (sum - 0) < 1e-8) {puts ("Yes");} else {puts ("No");}}}}return 0;
}

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