Codeforces Round #354 (Div. 2)

贪心 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;
}

模拟＋DFSB Pyramid of Glasses

设酒杯满了值为1.0，每一次暴力传递下去

#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

从左到右维护一段连续的区间，改变次数不大于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，旋转) D Theseus and labyrinth

多加一维表示旋转次数(0~3)，dis[x][y][z]表示走到(x, y)旋转z次后的最小步数．

#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;
}

数学 E The Last Fight Between Human and AI

原题转化为求P(k)==0.如果k==0，判断a[0]是否能被玩家设置成０．否则判断剩余数字个数的奇偶以及现在是谁出手，判断最后一步是否为玩家走，最后一步总能使得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;
}