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。
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 >>>>>> ", (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 ", res); 110 } 111 return 0; 112 }
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 >>>>>> ", (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 }
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 >>>>>> ", (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 }