A、Chino with Geometry
思路：简单数学。过点A作直线BC的垂线交于点F，然后根据勾股定理就可以化简出 $ |BD| imes |BE| = |AB|^2 - r^2$ 。注意要开long long！
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 1e5+5; 27 const int inf = 0x3f3f3f3f; 28 29 LL x_0, y_0, r, x_1, y_1, y_2; 30 31 int main() { 32 while(cin >> x_0 >> y_0 >> r >> x_1 >> y_1 >> y_2) { 33 cout << ((x_0 - x_1) * (x_0 - x_1) + (y_0 - y_1) * (y_0 - y_1)) - r * r << endl; 34 } 35 return 0; 36 }
B、Chino with Repeater
思路：简单贪心。每次翻倍复读即可。
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 1e5+5; 27 const int inf = 0x3f3f3f3f; 28 29 LL n, sum, ans; 30 31 int main() { 32 while(cin >> n) { 33 sum = 1LL; 34 ans = 0LL; 35 while(sum < n) { 36 sum <<= 1; 37 ++ans; 38 } 39 cout << ans << endl; 40 } 41 return 0; 42 }
D、Chino with Equation
思路：经典的隔板法。
咋们来举2个简单的栗子：
①求方程 $ x + y + z = 10 $ 的正整数解的个数。
解：相当于先将10个1摆好，然后在9个空隙中插入2块隔板，则将10个1分成3部分，每部分至少有1个1，取法有 $ C_{10-1}^{3-1} = C_{9}^{2} $。
②求方程 $ x + y + z = 10 $ 的非负整数解的个数。
解：相当于将10个1和2块隔板随便摆（一共12个位置），最终总能分成3部分，且至少有1部分有1~n个1，其余部分不要求（可以为0也可以不为0）。于是先从12个位置中挑2个位置放隔板，剩下的位置放1，取法有$ C_{10+3-1}^{3-1} = C_{12}^{2} $。
了解了上面2个栗子之后，这道题就轻而易举了，该方程的正整数解有 $ C_{n-1}^{m-1} $ 个，非负整数解有 $ C_{n+m-1}^{m-1} $ 个。因为 m和n的值不超过 $ 10^6 $，且是组合数取模大质数，所以用卢卡斯定理求解即可。
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 using namespace std; 18 typedef long long LL; 19 typedef unsigned long long ULL; 20 const double eps = 1e-6; 21 const double PI = acos(-1.0); 22 const int maxn = 1e5+5; 23 const int inf = 0x3f3f3f3f; 24 const int p = 1e9+7; 25 26 LL m, n; 27 28 LL quick_mul(LL a, LL b, LL mod){ 29 LL res = 0LL; 30 while(b) { 31 if(b & 1) res = (res + a) % mod; 32 a = (a + a) % mod; 33 b >>= 1; 34 } 35 return res; 36 } 37 38 LL quick_mod(LL a, LL b, LL mod) { 39 LL res = 1LL; 40 while(b) { 41 if(b & 1) res = quick_mul(res, a, mod); 42 a = quick_mul(a, a, mod); 43 b >>= 1; 44 } 45 return res; 46 } 47 48 LL comb(LL n, LL m, LL mod) { 49 if(n < m) return 0LL; 50 if(n - m < m) m = n - m; 51 LL x = 1LL, y = 1LL; 52 for(LL i = 1LL; i <= m; ++i) x = x * (n - i + 1LL) % mod, y = y * i % mod; 53 return x * quick_mod(y, mod - 2LL, mod) % mod; 54 } 55 56 LL Lucas(LL n, LL m, LL mod) { 57 if(n < mod && m < mod) return comb(n, m, mod); 58 return comb(n % mod, m % mod, mod) * Lucas(n / mod, m / mod, mod) % mod; 59 } 60 61 62 int main() { 63 while(cin >> m >> n) { 64 cout << Lucas(n - 1, m - 1, p) << ' ' << Lucas(n + m - 1, m - 1, p) << endl; 65 } 66 return 0; 67 }
F、Chino with Expectation
思路：简单数学求期望值。对于询问"1 2 3"，不难想到有以下5种情况：(2,2,3,4,5)，(1,3,3,4,5)，(1,2,4,4,5)，(1,2,3,5,5)，(1,2,3,4,6)。显然，每种情况等概率出现即 $ frac{1}{n} $，于是将红色部分求和（用前缀和的差求子区间和）除以 $ (rt-lt+1) $ 再乘上其概率 $frac{1}{n} $ 即可。
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 1e5+5; 27 const int inf = 0x3f3f3f3f; 28 29 int n, q, lt, rt, len; 30 double x, a, now, per[maxn], sum[maxn]; 31 32 int main() { 33 while(~scanf("%d%d", &n, &q)) { 34 per[0] = 0; 35 for(int i = 1; i <= n; ++i) { 36 scanf("%lf", &a); 37 per[i] = per[i - 1] + a; 38 } 39 while(q--) { 40 scanf("%lf%d%d", &x, <, &rt); 41 len = rt - lt + 1; 42 now = per[rt] - per[lt - 1]; 43 printf("%.6f ", ((n - len) * now + len * (now + x)) / len / n); 44 } 45 } 46 return 0; 47 }
H、Chino with Ciste
思路：简单的bfs。先将起点出发的4个方向加入队列中，然后用优先队列每次维护当前转弯次数最少的坐标点，注意走过的坐标点可能再次走到，因为我们的目标是转弯次数最少而不是步数最少，这点应该不难想到。对于是否转弯只需简单判断一下上一个方向dir与当前方向i的和值是奇数还是偶数即可。
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 2005; 27 const int inf = 0x3f3f3f3f; 28 29 int n, m, sx, sy, ex, ey, ans, carry, vis[maxn][maxn][4]; 30 char mp[maxn][maxn]; 31 32 struct node{int x, y, dir, cnt;} per, now; 33 deque<node> que; 34 35 bool check(int x, int y) { 36 if(x < 0 || y < 0 || x >= n || y >= m) return false; 37 return true; 38 } 39 40 int bfs(int x, int y) { 41 while(!que.empty()) que.pop_front(); 42 now.x = x, now.y = y, now.cnt = 0; 43 for(int i = 0; i < 4; ++i) { 44 now.dir = i; 45 vis[x][y][i] = 0; 46 que.push_back(now); 47 } 48 while(!que.empty()) { 49 per = que.front(), que.pop_front(); 50 if(per.x == ex && per.y == ey) return per.cnt; 51 for(int i = 0; i < 4; ++i) { 52 now.x = per.x + dir[i][0], now.y = per.y + dir[i][1], now.cnt = per.cnt, now.dir = i, carry = (per.dir + i) & 1; 53 if(carry) ++now.cnt; 54 if(check(now.x, now.y) && mp[now.x][now.y] != '*' && vis[now.x][now.y][i] > now.cnt) { 55 // cout << now.x << ' ' << now.y << ' ' << now.cnt << endl; 56 carry ? que.push_back(now) : que.push_front(now); 57 vis[now.x][now.y][i] = now.cnt; 58 } 59 } 60 } 61 return -1; 62 } 63 64 int main() { 65 while(~scanf("%d%d", &n, &m)) { 66 for(int i = 0; i < n; ++i) scanf("%s", mp[i]); 67 for(int i = 0; i < n; ++i) { 68 for(int j = 0; j < m; ++j) { 69 if(mp[i][j] == 'S') sx = i, sy = j; 70 else if(mp[i][j] == 'T') ex = i, ey = j; 71 } 72 } 73 memset(vis, 0x3f, sizeof(vis)); 74 ans = bfs(sx, sy); 75 if(ans != -1) printf("%d ", ans); 76 else puts("troil"); 77 } 78 return 0; 79 }