COOK50小结

题目链接

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。

 /*************************************************************************
     > File Name: A.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月23日 星期二 13时57分22秒
     > Propose:
  ************************************************************************/
 #include <map>
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 const int MAX_N = ;
 int N, A[MAX_N];
 #define rep(i, n) for (int i = (1); i <= (n); i++)

 int gcd(int a, int b) {
       if (!b) return a;
     return gcd(b, a % b);
 }

 int main(void) {
     ios::sync_with_stdio(false);
     int T;
     cin >> T;
     while (T--) {
         cin >> N;
         bool flag = false;
         rep (i, N) {
             cin >> A[i];
             if (i >  && gcd(A[i], A[i-]) == ) flag = true;
         }
         if (flag) cout << N << endl;
         else cout << - << endl;
     }

     return ;
 }

Ｂ题

统计一个每个位置向右和向下是否都没有障碍。

 /*************************************************************************
     > File Name: B.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月23日 星期二 14时30分39秒
     > Propose:
  ************************************************************************/

 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 char a[][];
 int N, T;
 bool col[][], row[][];
 #define rep(i, n) for (int i = (1); i <= n; i++)
 #define per(i, n) for (int i = (n); i >= 1; i--)

 int main(void) {
     ios::sync_with_stdio(false);
     cin >> T;
     while (T--) {
         cin >> N;
         rep (i, N) rep (j, N) cin >> a[i][j], row[i][j] = col[i][j] = true;

         rep (i, N) per (j, N) if ((j < N && !row[i][j+]) || (a[i][j] != '.')) row[i][j] = false;
         rep (j, N) per (i, N) if ((i < N && !col[i+][j]) || (a[i][j] != '.')) col[i][j] = false;

         int res = ;
         rep (i, N) rep (j, N) if (row[i][j] && col[i][j]) res++;
         cout << res << endl;
     }

     return ;
 }

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。

 /*************************************************************************
     > File Name: C.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月23日 星期二 17时24分56秒
     > Propose:
  ************************************************************************/
 #include <set>
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 const int MAX_N = ;
 int A[MAX_N], p[MAX_N], cnt, last[MAX_N*];
 bool vis[MAX_N*];
 #define rep(i, n) for (int i = (1); i <= n; i++)

 void prime() {
     cnt = ;
     memset(vis, false, sizeof(vis));
     for (int i = ; i < ; i++) if (!vis[i]) {
         p[cnt++] = i;
         for (int j = *i; j < ; j += i) vis[j] = true;
     }
 }

 void read(int &res) {
     res = ;
     char c = ' ';
     while (c < '' || c > '') c = getchar();
     while (c >= '' && c <= '') res = res *  + c - '', c = getchar();
 }

 vector<int> factor[];
 vector<int>::iterator it;
 int main(void) {
     prime();
     rep (i, ) {
         int x = i;
         for (int j = ; p[j]*p[j] <= x; j++) {
             if (x % p[j] == ) {
                 factor[i].push_back(p[j]);
                 while (x % p[j] == ) x /= p[j];
             }
         }
         if (x > ) factor[i].push_back(x);
     }

     int T, N;
     read(T);
     while (T--) {
         read(N);
         rep (i, N) read(A[i]);

         int l = , r = , res = ;
         memset(last, , sizeof(last));
         for (it = factor[A[l]].begin(); it != factor[A[l]].end(); ++it) last[*it] = l;
         while (l <= r && r <= N) {
               int m = ;
             for (it = factor[A[r]].begin(); it != factor[A[r]].end(); ++it) {
                 if (last[*it] >= l) m = max(m, last[*it]);
                 last[*it] = r;
             }
             if ( == m) res = max(res, r - l + );
             else l = m + ;
             r++;
         }

         if (res == ) puts("-1");
         else printf("%d\n", res);
     }

     return ;
 }

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。

 /*************************************************************************
     > File Name: D.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月24日 星期三 13时20分04秒
     > Propose:
  ************************************************************************/

 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 typedef long long LL;
 const int MOD = 1e9 + ;
 #define rep(i, n) for (int i = (1); i <= (n); i++)

 struct Matrix {
      LL mat[][];
     int n, m;
     Matrix() {}
     Matrix(int n, int m):n(n), m(m) {
         memset(mat, , sizeof(mat));
     }
     Matrix operator * (const Matrix &b) const {
         Matrix c(n, b.m);
         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;
         return c;
     }
 };

 Matrix pow_mod(Matrix a, int b) {
     Matrix res(a.n, a.m);
     rep (i, a.n) res.mat[i][i] = ;
     while (b) {
         if (b & ) res = res * a;
         a = a * a;
         b >>= ;
     }
     return res;
 }

 int main(void) {
     ios::sync_with_stdio(false);
     int T, N, M;
     cin >> T;
     while (T--) {
         cin >> N >> M;
         Matrix a(M, M), b(M, M);
         rep (i, M) {
             if (i -  >= ) a.mat[i][i-] = b.mat[i][i-] = ;
             if (i +  <= M) a.mat[i][i+] = b.mat[i][i+] = ;
             a.mat[i][i] = ;
         }
         Matrix res = pow_mod(a*b, (N% ==  ? N/- : N/));
         if (N %  == ) res = b * res;
         Matrix f1(M, );
         rep (i, M) f1.mat[i][] = ;
         res = res * f1;

         LL ans = ;
         rep (i, M) ans = (ans + res.mat[i][]) % MOD;
         cout << ans << endl;
     }

     return ;
 }

Ｅ题