比赛链接：http://hihocoder.com/contest/hihointerview28/problems
A. 固定一个方向，两两相邻的点顺时针或逆时针构造三个向量，判断这个点在这个向量的左侧还是右侧，看看是否在同一侧。trick就是点在向量上，对应的情况就是值为0.
1 def do(p1x, p1y, p2x, p2y, p3x, p3y): 2 return (p3x - p1x) * (p2y - p1y) - (p2x - p1x) * (p3y - p1y); 3 4 T = map(int, raw_input().split(' '))[0] 5 while T > 0: 6 T -= 1 7 px, py, ax, ay, bx, by, cx, cy = map(int, raw_input().split(' ')) 8 x, y, z = do(ax, ay, px, py, bx, by), do(bx, by, px, py, cx, cy), do(cx, cy, px, py, ax, ay) 9 if (x >= 0 and y >= 0 and z >= 0) or (x <= 0 and y <= 0 and z <= 0): 10 print 'YES' 11 else: 12 print 'NO'
B.打表规律，发现<=16的时候可以暴搜，>16的时候f(n)=4*f(n-5)（？如果没记错的话），矩阵加速一下就行了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long LL; 5 const LL mod = 1000000007LL; 6 const int maxn = 10; 7 LL n; 8 LL ret; 9 10 typedef struct Matrix { 11 LL m[maxn][maxn]; 12 int r; 13 int c; 14 Matrix() { 15 r = c = 0; 16 memset(m, 0, sizeof(m)); 17 } 18 } Matrix; 19 20 Matrix mul(Matrix m1, Matrix m2, LL mod) { 21 Matrix ans = Matrix(); 22 ans.r = m1.r; 23 ans.c = m2.c; 24 for(int i = 1; i <= m1.r; i++) { 25 for(int j = 1; j <= m2.r; j++) { 26 for(int k = 1; k <= m2.c; k++) { 27 if(m2.m[j][k] == 0) continue; 28 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod; 29 } 30 } 31 } 32 return ans; 33 } 34 35 Matrix quickmul(Matrix m, LL n, LL mod) { 36 Matrix ans = Matrix(); 37 for(int i = 1; i <= m.r; i++) { 38 ans.m[i][i] = 1; 39 } 40 ans.r = m.r; 41 ans.c = m.c; 42 while(n) { 43 if(n & 1) { 44 ans = mul(m, ans, mod); 45 } 46 m = mul(m, m, mod); 47 n >>= 1; 48 } 49 return ans; 50 } 51 52 void dfs(LL n, LL cur, LL sz) { 53 if(n == 0) { 54 ret = max(ret, cur); 55 return; 56 } 57 if(n >= 3) dfs(n-3,cur*2%mod, cur); 58 if(n >= 1 && sz) dfs(n-1, (cur+sz)%mod, sz); 59 dfs(n-1,(cur+1)%mod, sz); 60 } 61 62 int main() { 63 // freopen("in", "r", stdin); 64 while(cin >> n) { 65 if(n < 16) { 66 ret = 0; 67 dfs(n, 0, 0); 68 cout << ret << endl; 69 continue; 70 } 71 Matrix x; x.r = 5, x.c = 1; 72 x.m[1][1] = 81, x.m[2][1] = 64, x.m[3][1] = 48, x.m[4][1] = 36, x.m[5][1] = 27; 73 Matrix p; p.r = p.c = 5; 74 memset(p.m, 0, sizeof(p.m)); 75 p.m[1][5] = 4; 76 for(int i = 2; i <= 5; i++) p.m[i][i-1] = 1; 77 p = quickmul(p, n-15, mod); 78 p = mul(p, x, mod); 79 cout << p.m[1][1] << endl; 80 } 81 return 0; 82 }
C.对于trie上的每个节点u，求最小的整数x满足 节点u对应的字符串(trie上root->u的路径) 是 S[1..x]的子序列。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 10100; 5 const int maxm = 100100; 6 const int maxc = 30; 7 8 typedef pair<int, int> pii; 9 typedef struct Trie { 10 int rt, sz; 11 int id[maxm][maxc], val[maxm]; 12 int dp[maxm]; 13 string s; 14 void build() { 15 rt = sz = 0; 16 memset(dp, 0, sizeof(dp)); 17 memset(id, -1, sizeof(id)); 18 memset(val, 0, sizeof(val)); 19 } 20 void insert(const char* str) { 21 int u = rt; 22 int n = strlen(str); 23 for(int i = 0; i < n; i++) { 24 int v = str[i] - 'a'; 25 if(id[u][v] == -1) id[u][v] = ++sz; 26 u = id[u][v]; 27 } 28 val[u] = max(val[u], n); 29 } 30 void dfs(int x, int u) { 31 if(x == s.length()) return; 32 for(int i = 0; i < 26; i++) { 33 int &v = id[u][i]; 34 if(v == -1) continue; 35 int y = x; 36 while(s[y] != i + 'a' && y < s.length()) y++; 37 if(y == s.length()) break; 38 if(i + 'a' == s[y]) { 39 dp[v] = dp[u] + 1; 40 dfs(y+1, v); 41 } 42 } 43 } 44 }Trie; 45 46 int n; 47 char tmp[maxm]; 48 Trie trie; 49 50 int main() { 51 // freopen("in", "r", stdin); 52 while(~scanf("%d", &n)) { 53 trie.build(); 54 for(int i = 0; i < n; i++) { 55 scanf("%s", tmp); 56 trie.insert(tmp); 57 } 58 scanf("%s", tmp); 59 trie.s = tmp; 60 trie.dfs(0, 0); 61 int ret = 0; 62 for(int i = 0; i <= trie.sz; i++) { 63 // printf("%d %d ", trie.dp[i], trie.val[i]); 64 if(trie.dp[i] != 0) { 65 ret = max(ret, trie.val[i]); 66 } 67 } 68 printf("%d ", ret); 69 } 70 return 0; 71 }