ZOJ Monthly, August 2014

135 - ZOJ Monthly, August 2014

A：构造问题，判断序列奇偶性，很容易发现最小值不是1就是0，最大值不是n就是n - 1，注意细节去构造即可

E：dp，dp[i][j]表示长度i，末尾状态为j的最大值，然后每个位置数字取与不取，不断状态转移即可

G：就一个模拟题没什么好说的

H：dfs，每次dfs下去，把子树宽度保存下来，然后找最大值，如果有多个，就是最大值+cnt宽度

I：构造，如果r * 2 > R，肯定无法构造，剩下的就二分底边，按等腰三角形去构造即可

代码：

A:

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int n;

void print(int n) {
    if (n == 3) {
    printf("3 1 2");
    return;
    }
    if (n % 2) {
    int len = (n - 3) / 2;
    printf("%d %d", n, n - len);
    for (int i = n - 1; i > n - len; i--)
        printf(" %d %d", i, i - len);
    printf(" 3 1 2");
    }
    else {
    int len = n / 2;
    printf("%d %d", n, n - len);
    for (int i = n - 1; i > n - len; i--)
        printf(" %d %d", i, i - len);
    }
}

void print2(int n) {
    print(n - 2);
    printf(" %d %d", n - 1, n);
}

void solve() {
    if (n == 1) {
    printf("1 1\n1\n1\n");
    return;
    }
    if (n == 2) {
    printf("1 1\n1 2\n2 1\n");
    return;
    }
    if (n == 3) {
    printf("0 2\n3 1 2\n1 2 3\n");
    return;
    }
    if (n % 2 == 0) {
    if (n / 2 % 2) {
        printf("1 %d\n", n - 1);
        print2(n); printf("\n");
        print2(n - 1);
        printf(" %d\n", n);
    }
    else {
        printf("0 %d\n", n);
        print(n); printf("\n");
        print(n - 1); printf(" %d\n", n);
    }
    }
    else {
    if ((n + 1) / 2 % 2) {
        printf("1 %d\n", n);
        print(n - 2); printf(" %d %d\n", n - 1, n);
        print(n - 1); printf(" %d\n", n);
    }
    else {
        printf("0 %d\n", n - 1);
        print(n); printf("\n");
        print2(n - 1); printf(" %d\n", n);
    }
    }
}

int main() {
    while (~scanf("%d", &n)) {
    solve();
    }
    return 0;
}

E:

 

 

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <map>
 using namespace std;

 const int INF = 0x3f3f3f3f;
 int t, n;

 map<int, int> dp[2];
 map<int, int>::iterator it;

 int lowbit(int x) {
     return (x&(-x));
 }

 int solve() {
     dp[0].clear();
     int pre = 1, now = 0;
     int num;
     dp[0][0] = 0;
     for (int i = 0; i < n; i++) {
     scanf("%d", &num);
     num /= 2;
     swap(pre, now);
     dp[now].clear();
     for (it = dp[pre].begin(); it != dp[pre].end(); it++) {
         int s = it->first;
         if (dp[now].count(s) == 0) dp[now][s] = dp[pre][s];
         else dp[now][s] = max(dp[now][s], dp[pre][s]);
         int next;
         if (s % num) {
         next = num;
         if (dp[now].count(next) == 0) dp[now][next] = dp[pre][s] + num * 2;
         else dp[now][next] = max(dp[now][next], dp[pre][s] + num * 2);
         }
         else {
         next = s + num;
         int add = (s % lowbit(next) * 2 + num) * 2;
         if (dp[now].count(next) == 0) dp[now][next] = dp[pre][s] + add;
         else dp[now][next] = max(dp[now][next], dp[pre][s] + add);
         }
     }
     }
     int ans = 0;
     for (it = dp[now].begin(); it != dp[now].end(); it++)
     ans = max(ans, it->second);
     return ans;
 }

 int main() {
     scanf("%d", &t);
     while (t--) {
     scanf("%d", &n);
     printf("%d\n", solve());
     }
     return 0;
 }

G:

 

 

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 using namespace std;

 const int N = 55;
 const int d[8][2] = {{1, 0}, {1, 1}, {1, -1}, {0, 1}, {0, -1}, {-1, 0}, {-1, 1}, {-1, -1}};

 typedef pair<int, int> pii;

 int t;
 int n, m, f, k;
 int g[N][N];
 int gg[N][N];
 char str[55];
 vector<pii> go[1005];

 void solve() {
     for (int ti = 1; ti <= f; ti++) {
     memset(gg, 0, sizeof(gg));
     for (int i = 1; i <= n; i++) {
         for (int j = 1; j <= m; j++) {
         if (g[i][j] == 1) {
             for (int k = 0; k < 8; k++) {
             int xx = i + d[k][0];
             int yy = j + d[k][1];
             if (xx <= 0 || xx > n || yy <= 0 || yy > m) continue;
             gg[xx][yy]++;
             }
         }
         }
     }

     for (int i = 1; i <= n; i++)
         for (int j = 1; j <= m; j++) {
         if (g[i][j] == 2) continue;
         else if (g[i][j] == 0) {
             if (gg[i][j] == 3) g[i][j] = 1;
         }
         else {
             if (gg[i][j] < 2 || gg[i][j] > 3) g[i][j] = 0;
         }
         }
     for (int i = 0; i < go[ti].size(); i++) {
         g[go[ti][i].first][go[ti][i].second] = 2;
     }
     }
     for (int i = 1; i <= n; i++) {
     for (int j = 1; j <= m; j++) {
         if (g[i][j] == 2) printf("X");
         else printf("%d", g[i][j]);
     }
     printf("\n");
     }
 }

 int main() {
     scanf("%d", &t);
     while (t--) {
     scanf("%d%d%d%d", &n, &m, &f, &k);
     for (int i = 1; i <= f; i++)
         go[i].clear();
     for (int i = 1; i <= n; i++) {
         scanf("%s", str + 1);
         for (int j = 1; j <= m; j++) {
         g[i][j] = str[j] - '0';
         }
     }
     int ti, x, y;
     while (k--) {
         scanf("%d%d%d", &ti, &x, &y);
         go[ti].push_back(make_pair(x, y));
     }
     solve();
     }
     return 0;
 }

 

H:

 

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 using namespace std;

 const int N = 10005;

 int n;
 vector<int> g[N];

 int dfs(int u) {
     int sz = g[u].size();
     vector<int> save;
     for (int i = 0; i < sz; i++)
     save.push_back(dfs(g[u][i]));
     sort(save.begin(), save.end());
     sz = save.size();
     int cnt = 0;
     int ans = 1;
     for (int i = sz - 1; i >= 0; i--) {
     if (i != sz - 1 && save[i] != save[i + 1]) break;
     ans = save[i] + cnt;
     cnt++;
     }
     return ans;
 }

 int main() {
     while (~scanf("%d", &n)) {
     for (int i = 1; i <= n; i++)
         g[i].clear();
     int v;
     for (int i = 2; i <= n; i++) {
         scanf("%d", &v);
         g[v].push_back(i);
     }
     printf("%d\n", dfs(1));
     }
     return 0;
 }

I:

 

 

 #include <cstdio>
 #include <cstring>
 #include <cmath>

 double r, R;

 double h, x;

 double cal(double a) {
     double d = a / 2;
     h = sqrt(R * R - d * d) + R;
     x = sqrt(h * h + d * d);
     return a * x * x / (2 * R * (a + x + x));
 }

 void solve() {
     double lx = 0, rx = sqrt(3.0) * R;
     double mid;
     for (int i = 0; i < 1000; i++) {
     mid = (lx + rx) / 2;
     double tmp = cal(mid);
     if (tmp > r) rx = mid;
     else lx = mid;
     }
     cal((lx + rx) / 2);
     printf("%.10lf %.10lf %.10lf\n", mid, x, x);
 }

 int main() {
     while (~scanf("%lf%lf", &r, &R)) {
     if (r * 2 > R) printf("NO Solution!\n");
     else solve();
     }
     return 0;
 }