135-ZOJ Monthly, August 2014

135-ZOJ Monthly, August 2014
135-ZOJ Monthly, August 2014

A: Construct the problem and judge the parity of the sequence. It is easy to find that the minimum value is either 1 or 0, and the maximum value is n or n-1. Pay attention to the details to construct it.

E: dp, dp [I] [j] indicates the length I, and the end state is the maximum value of j. Then, the numbers at each position are obtained and not retrieved, and the status can be transferred continuously.

G: There is nothing to say about a simulation question.

H: dfs. Store the subtree width and find the maximum value. If there are multiple values, the maximum value + cnt width is used.

I: constructor. If r * 2> R, it cannot be constructed. The rest is the two-point bottom edge, which can be constructed according to the isosceles triangle.

Code:

A:

#include 
 
  #include 
  
   #include 
   
    #include 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 
 
  #include 
  
   #include #include 
   using namespace std;const int INF = 0x3f3f3f3f;int t, n;map
    
      dp[2];map
     
      ::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 
 
  #include 
  
   #include 
   
    #include 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
    
      pii;int t;int n, m, f, k;int g[N][N];int gg[N][N];char str[55];vector
     
       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 
 
  #include 
  
   #include 
   
    #include using namespace std;const int N = 10005;int n;vector
    
      g[N];int dfs(int u) {    int sz = g[u].size();    vector
     
       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 
 
  #include 
  
   #include 
   
    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;}