ZOJ Monthly, November 2014

做了一次月赛，没想到这么难，加上后来补上的题目也只有3个题。第一名也只有4个题啊啊啊啊~.其中两道还是水题。留坑慢慢补上来。

 

 3832 Tilt Cylinder

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

给定如图所示有盖圆柱体，R，H,水面高度h，倾角a，求水得体积。

分析：明显的数值积分题，这样考虑。圆下底面即A点与地面高度lim1, 圆上底面一点B与地面高度lim2,h所处的范围进行讨论从而确定积分几何体的两边的高度。我们积分的几何体应该是一个圆柱体被削掉一部分了。

h>lim1时，几何体左半部分可以减掉一个圆柱，对剩下部分积分，剩下部分左边截面的高度2*R；否则高度为2*R/cos(a);

h>lim2时，几何体右半部分截面的高度需要计算，为(h-lim2)/cos(a);否则为0.

注意：这里所说的结合体截面的高度是相对与下面的母线而言，并不是对地高度。

然后对上面给出的高度范围结合夹角a进行积分，需要推导截面的面积。最后答案需要加上截掉的那部分完整的圆柱体体积（如果是那种情况的话）。

代码：

  1 #include <bits/stdc++.h>
  2 #define pb push_back
  3 #define mp make_pair
  4 #define esp 1e-12
  5 #define lowbit(x) ((x)&(-x))
  6 #define lson   l, m, rt<<1
  7 #define rson   m+1, r, rt<<1|1
  8 #define sz(x) ((int)((x).size()))
  9 #define pb push_back
 10 #define pf(x) ((x)*(x))
 11 
 12 #define pi acos(-1.0)
 13 
 14 #define in freopen("solve_in.txt", "r", stdin);
 15 #define out freopen("solve_out.txt", "w", stdout);
 16 
 17 #define bug(x) printf("Line : %u >>>>>>\n", (x));
 18 #define inf 0x0f0f0f0f
 19 using namespace std;
 20 
 21 
 22 int dblcmp(double x) {
 23     if(fabs(x) < esp) return 0;
 24     return x > 0 ? 1 : -1;
 25 }
 26 const int N = 200;
 27 double R, H, a, h;
 28 double f1(double x) {
 29     double cth = (R-x)/R;
 30     return acos(cth)*pf(R)-(R-x)*sqrt(pf(R)-pf(R-x));
 31 }
 32 double f2(double x) {
 33     double cth = (x-R)/R;
 34     return (pi-acos(cth))*pf(R)+(x-R)*sqrt(pf(R)-pf(x-R));
 35 }
 36 double s1(double l, double r) {
 37     double x = 0, y = (l-r)/tan(a);
 38     double h0 = (y-x)/N;
 39 
 40     double res = 0.0;
 41     res = f1(l)+f1(r);
 42     for(int i = 1; i <= N; i++) {
 43         if(i < N)
 44             res += 2*f1(l-(x+i*h0)*tan(a));
 45         res += 4*f1(l-(x+(2*i-1)*h0/2)*tan(a));
 46     }
 47     return res*h0/6;
 48 
 49 }
 50 double s2(double l, double r) {
 51     double x = 0, y = (l-r)/tan(a);
 52     double h0 = (y-x)/N;
 53 
 54     double res = 0.0;
 55     res = f2(l)+f2(r);
 56     for(int i = 1; i <= N; i++) {
 57         if(i < N)
 58             res += 2*f2(l-(x+i*h0)*tan(a));
 59         res += 4*f2(l-(x+(2*i-1)*h0/2)*tan(a));
 60     }
 61     return res*h0/6;
 62 }
 63 
 64 double getAns(double h0, double h1) {
 65     if(h1 > R) {
 66         return s2(h0, h1);
 67     } else if(h0 < R) {
 68         return s1(h0, h1);
 69     } else {
 70         return s2(h0, R)+s1(R, h1);
 71     }
 72 }
 73 int main() {
 74     
 75     while(scanf("%lf%lf%lf%lf", &R, &H, &h, &a) == 4) {
 76         double res = 0.0;
 77         if(fabs(a) > esp && fabs(a-90.0) > esp) {
 78             a = a*pi/180.0;
 79             double lim2 = sin(a)*H;
 80             double lim1 = 2*R*cos(a);
 81 
 82             double h1, h2;
 83             if(h > lim1) {
 84                 h1 = 2*R;
 85                 res += pi*pf(R)*(h/sin(a)-2*R/tan(a));
 86             } else {
 87                 h1 = h/cos(a);
 88             }
 89             if(h > lim2) {
 90                 h2 = (h-lim2)/cos(a);
 91             } else {
 92                 h2 = 0.0;
 93             }
 94             res += getAns(h1, h2);
 95 
 96         } else {
 97             if(fabs(a) < esp) {
 98                 double Sr;
 99                 if(h > R){
100                     Sr = f2(h);
101                 }else{
102                     Sr = f1(h);
103                 }
104                 res = Sr*H;
105             } else {
106                 res = pi*pf(R)*h;
107             }
108         }
109         printf("%.12f\n", res);
110     }
111     return 0;
112 }

View Code

3839 Poker Face

分析：第n个是将第n/2个倒插进去然后加上眼睛等部分。

代码：

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define esp 1e-8
 5 #define lowbit(x) ((x)&(-x))
 6 #define lson   l, m, rt<<1
 7 #define rson   m+1, r, rt<<1|1
 8 #define sz(x) ((int)((x).size()))
 9 #define pb push_back
10 #define in freopen("solve_in.txt", "r", stdin);
11 #define out freopen("solve_out.txt", "w", stdout);
12 
13 #define bug(x) printf("Line : %u >>>>>>\n", (x));
14 #define inf 0x0f0f0f0f
15 #define Fill(x, b1, b2, l, r) {\
16     for(int i = 0; i < l; i++)\
17         maze[x][i+b1][b2] = maze[x][i+b1][b2+r-1] = '*';\
18     for(int i = 0; i < r; i++)\
19         maze[x][b1][i+b2] = maze[x][b1+l-1][i+b2] = '*';\
20 }\
21 
22 using namespace std;
23 typedef long long LL;
24 typedef pair<int, int> PII;
25 typedef map<string, int> MPS;
26 
27 using namespace std;
28 const int maxn = 1100;
29 char s[maxn][maxn] = {
30         {"********"}, {"***  ***"}, {"***  ***"}, {"***  ***"}, {"* **** *"}, {"* *  * *"},
31         {"* *  * *"}, {"********"}
32     };
33 
34 char maze[12][maxn][maxn];
35 
36 int popcount(int x){
37     int ans = 0;
38     while(1){
39         if(x&1) break;
40         ans++;
41         x >>= 1;
42     }
43     return ans;
44 }
45 void dfs(int n){
46     if(n <= 8) return;
47     int x = popcount(n);
48 //    cout << x << endl;
49     for(int i = 0; i < n; i++) for(int j = 0; j < n; j++)
50         maze[x][i][j] = ' ';
51     Fill(x, 0, 0, n, n)
52     int st1 = n/8, st2 = st1+n/2;
53     Fill(x,  n/8, st1, n/4+1, n/4)
54     Fill(x, n/8, st2, n/4+1, n/4)
55     dfs(n>>1);
56     int b1 = n/2, b2 = n/4;
57     int nn = n>>1;
58     for(int i = 0; i < (n>>1); i++)
59     for(int j = 0; j < (n>>1); j++){
60         maze[x][b1+i][b2+j] = maze[x-1][nn-1-i][nn-1-j];
61     }
62 }
63 int main() {
64     
65     int n;
66     for(int i = 0; i < 8; i++)
67         strcpy(maze[3][i], s[i]);
68     dfs(1024);
69 
70     while(scanf("%d", &n), n >= 8) {
71 //            cout << n <<endl;
72         int x = popcount(n);
73 //        cout << x << endl;
74         for(int i = 0; i < n; i++)
75             puts(maze[x][i]);
76         puts("");
77     }
78     return 0;
79 }

View Code

 

3838 Infusion Altar

分析：对每个点和其对称点访问一遍，将数目最多的那种保留，其他全部替换成这种。

代码：

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define esp 1e-8
 5 #define lowbit(x) ((x)&(-x))
 6 #define lson   l, m, rt<<1
 7 #define rson   m+1, r, rt<<1|1
 8 #define sz(x) ((int)((x).size()))
 9 #define pb push_back
10 #define in freopen("solve_in.txt", "r", stdin);
11 #define out freopen("solve_out.txt", "w", stdout);
12 
13 #define bug(x) printf("Line : %u >>>>>>\n", (x));
14 #define inf 0x0f0f0f0f
15 using namespace std;
16 const int maxn = 110;
17 char maze[maxn][maxn];
18 int vis[maxn][maxn];
19 vector<char> tmp;
20 int n;
21 
22 void dfs(int x, int y){
23     if(vis[x][y]) return;
24     vis[x][y] = 1;
25     tmp.pb(maze[x][y]);
26     dfs(y, x);
27     dfs(n-1-y, n-1-x);
28     dfs(n-1-x, y);
29     dfs(x, n-1-y);
30 }
31 int main(){
32 
33     int T;
34     for(int t = scanf("%d", &T); t <= T; t++){
35         scanf("%d", &n);
36         int ans = 0;
37         for(int i = 0; i < n; i++)
38             scanf("%s", maze[i]);
39 //        for(int i = 0; i < n; i++)
40 //            cout << maze[i];
41         memset(vis, 0, sizeof vis);
42         for(int i = 0; i < n; i++)for(int j = 0; j < n; j++){
43             if(vis[i][j]) continue;
44             tmp.clear();
45             dfs(i, j);
46             sort(tmp.begin(), tmp.end());
47             int jj;
48             int mx = 0;
49             for(int ii = 0; ii < sz(tmp); ii = jj){
50                 int ok = 0;
51                 for(jj = ii; jj < sz(tmp) && tmp[jj] == tmp[ii]; jj++)
52                     ok++;
53                 mx = max(ok, mx);
54             }
55             ans += sz(tmp)-mx;
56         }
57         cout << ans << endl;
58     }
59     return 0;
60 }

View Code