很遗憾。看到第五题的通过人数就不敢做了。待日后补上。
A题
求最长的连续子序列，使得他们满足gcd为１。
如果有相邻的两个数的gcd为１，那么整个序列的gcd值也就是１，
否则就是该序列不存在。
1 /************************************************************************* 2 > File Name: A.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月23日 星期二 13时57分22秒 6 > Propose: 7 ************************************************************************/ 8 #include <map> 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <vector> 13 #include <fstream> 14 #include <cstring> 15 #include <iostream> 16 #include <algorithm> 17 using namespace std; 18 /*Let's fight!!!*/ 19 20 const int MAX_N = 100005; 21 int N, A[MAX_N]; 22 #define rep(i, n) for (int i = (1); i <= (n); i++) 23 24 int gcd(int a, int b) { 25 if (!b) return a; 26 return gcd(b, a % b); 27 } 28 29 int main(void) { 30 ios::sync_with_stdio(false); 31 int T; 32 cin >> T; 33 while (T--) { 34 cin >> N; 35 bool flag = false; 36 rep (i, N) { 37 cin >> A[i]; 38 if (i > 1 && gcd(A[i], A[i-1]) == 1) flag = true; 39 } 40 if (flag) cout << N << endl; 41 else cout << -1 << endl; 42 } 43 44 return 0; 45 }
Ｂ题
统计一个每个位置向右和向下是否都没有障碍。
1 /************************************************************************* 2 > File Name: B.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月23日 星期二 14时30分39秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 char a[1005][1005]; 20 int N, T; 21 bool col[1005][1005], row[1005][1005]; 22 #define rep(i, n) for (int i = (1); i <= n; i++) 23 #define per(i, n) for (int i = (n); i >= 1; i--) 24 25 int main(void) { 26 ios::sync_with_stdio(false); 27 cin >> T; 28 while (T--) { 29 cin >> N; 30 rep (i, N) rep (j, N) cin >> a[i][j], row[i][j] = col[i][j] = true; 31 32 rep (i, N) per (j, N) if ((j < N && !row[i][j+1]) || (a[i][j] != '.')) row[i][j] = false; 33 rep (j, N) per (i, N) if ((i < N && !col[i+1][j]) || (a[i][j] != '.')) col[i][j] = false; 34 35 int res = 0; 36 rep (i, N) rep (j, N) if (row[i][j] && col[i][j]) res++; 37 cout << res << endl; 38 } 39 40 return 0; 41 }
Ｃ题
先预处理出每个数的所有的质因子。 并记录下每个质因子在序列中最后一次出现的位置即可。
1 /************************************************************************* 2 > File Name: C.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月23日 星期二 17时24分56秒 6 > Propose: 7 ************************************************************************/ 8 #include <set> 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <vector> 13 #include <fstream> 14 #include <cstring> 15 #include <iostream> 16 #include <algorithm> 17 using namespace std; 18 /*Let's fight!!!*/ 19 20 const int MAX_N = 100050; 21 int A[MAX_N], p[MAX_N], cnt, last[MAX_N*10]; 22 bool vis[MAX_N*10]; 23 #define rep(i, n) for (int i = (1); i <= n; i++) 24 25 void prime() { 26 cnt = 0; 27 memset(vis, false, sizeof(vis)); 28 for (int i = 2; i < 1000005; i++) if (!vis[i]) { 29 p[cnt++] = i; 30 for (int j = 2*i; j < 1000005; j += i) vis[j] = true; 31 } 32 } 33 34 void read(int &res) { 35 res = 0; 36 char c = ' '; 37 while (c < '0' || c > '9') c = getchar(); 38 while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar(); 39 } 40 41 vector<int> factor[1000005]; 42 vector<int>::iterator it; 43 int main(void) { 44 prime(); 45 rep (i, 1000000) { 46 int x = i; 47 for (int j = 0; p[j]*p[j] <= x; j++) { 48 if (x % p[j] == 0) { 49 factor[i].push_back(p[j]); 50 while (x % p[j] == 0) x /= p[j]; 51 } 52 } 53 if (x > 1) factor[i].push_back(x); 54 } 55 56 int T, N; 57 read(T); 58 while (T--) { 59 read(N); 60 rep (i, N) read(A[i]); 61 62 int l = 1, r = 2, res = 1; 63 memset(last, 0, sizeof(last)); 64 for (it = factor[A[l]].begin(); it != factor[A[l]].end(); ++it) last[*it] = l; 65 while (l <= r && r <= N) { 66 int m = 0; 67 for (it = factor[A[r]].begin(); it != factor[A[r]].end(); ++it) { 68 if (last[*it] >= l) m = max(m, last[*it]); 69 last[*it] = r; 70 } 71 if (0 == m) res = max(res, r - l + 1); 72 else l = m + 1; 73 r++; 74 } 75 76 if (res == 1) puts("-1"); 77 else printf("%d ", res); 78 } 79 80 return 0; 81 }
Ｄ题
分奇数行和偶数行讨论，并得出递推式，即可用矩阵加速。
1 /************************************************************************* 2 > File Name: D.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 5 > Created Time: 2014年09月24日 星期三 13时20分04秒 6 > Propose: 7 ************************************************************************/ 8 9 #include <cmath> 10 #include <string> 11 #include <cstdio> 12 #include <fstream> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 using namespace std; 17 /*Let's fight!!!*/ 18 19 typedef long long LL; 20 const int MOD = 1e9 + 7; 21 #define rep(i, n) for (int i = (1); i <= (n); i++) 22 23 struct Matrix { 2 LL mat[32][32]; 25 int n, m; 26 Matrix() {} 27 Matrix(int n, int m):n(n), m(m) { 28 memset(mat, 0, sizeof(mat)); 29 } 30 Matrix operator * (const Matrix &b) const { 31 Matrix c(n, b.m); 32 rep (i, n) rep (k, n) rep (j, b.m) c.mat[i][j] = (c.mat[i][j] + mat[i][k] * b.mat[k][j]) % MOD; 33 return c; 34 } 35 }; 36 37 Matrix pow_mod(Matrix a, int b) { 38 Matrix res(a.n, a.m); 39 rep (i, a.n) res.mat[i][i] = 1; 40 while (b) { 41 if (b & 1) res = res * a; 42 a = a * a; 43 b >>= 1; 44 } 45 return res; 46 } 47 48 int main(void) { 49 ios::sync_with_stdio(false); 50 int T, N, M; 51 cin >> T; 52 while (T--) { 53 cin >> N >> M; 54 Matrix a(M, M), b(M, M); 55 rep (i, M) { 56 if (i - 1 >= 1) a.mat[i][i-1] = b.mat[i][i-1] = 1; 57 if (i + 1 <= M) a.mat[i][i+1] = b.mat[i][i+1] = 1; 58 a.mat[i][i] = 1; 59 } 60 Matrix res = pow_mod(a*b, (N%2 == 0 ? N/2-1 : N/2)); 61 if (N % 2 == 0) res = b * res; 62 Matrix f1(M, 1); 63 rep (i, M) f1.mat[i][1] = 1; 64 res = res * f1; 65 66 LL ans = 0; 67 rep (i, M) ans = (ans + res.mat[i][1]) % MOD; 68 cout << ans << endl; 69 } 70 71 return 0; 72 }
Ｅ题