CodeChef--SEPT14小结

这套题目只做了几个相对简单的。其他的做起来比较吃力。

A 找下规律

 /*************************************************************************
     > File Name: A.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月06日 星期六 16时32分54秒
     > Propose:
  ************************************************************************/

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

 const int MOD = 1e9 + ;
 typedef long long LL;
 int t, n;
 char s[];

 int main(void) {
       ios::sync_with_stdio(false);
     cin >> t;
     while (t--) {
           cin >> s;
           n = strlen(s);
         LL res = ;
         for (int i = ; i < n; i++) {
               if ((i + ) %  == ) res = (res * ) % MOD;
             else res = (res *  - ) % MOD;
             if (s[i] == 'r') res = (res + ) % MOD;
         }
         cout << res << endl;
     }
     return ;
 }

B　模拟

 n, m = map(int, raw_input().split());
 a = map(int, raw_input().split());
 now = 1;
 while m:
     m -= 1;
     Q = raw_input().split();
     if Q[0] == 'R':
           print a[(now + int(Q[1]) - 1) % n - 1]
     elif Q[0] == 'C':
         now = (now + int(Q[1])) % n;
     else :
         now = (now - int(Q[1]) + n) % n;

C

这题被我写烦了。。。是想烦了。

我是枚举中间一堆的个数。然后就是组合数学了。。。。。。。

 /*************************************************************************
     > File Name: C.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月06日 星期六 11时08分09秒
     > Propose:
  ************************************************************************/

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

 const int MOD = 1e9 + ;
 typedef long long LL;
 LL fact[], n;

 LL pow_mod(LL a, LL b) {
     LL res = ;
     while (b) {
         if (b & ) res = (res * a) % MOD;
         a = (a * a) % MOD;
         b >>= ;
     }
     return res;
 }

 void init() {
     fact[] = ; fact[] = ;
     int now = ;
     for (int i = ; i <= ; i++, now++) fact[i] = (fact[i - ] * pow_mod(now, MOD - ) % MOD * i) % MOD;
 }

 int main(void) {
     ios::sync_with_stdio(false);
     init();
     while (cin >> n) {
         if (n < ) {
             cout << "" << endl;
             continue;
         }
         LL res = ;
         if (n %  == ){
             for (int i = ; i <= n - ; i += ) {
               　res = (res + fact[(n - i)/ - ]) % MOD;
             }
         } else {
             for (int i = ; i <= n - ; i += ) {
                 res = (res + fact[(n - i)/ - ]) % MOD;
             }
         }
         cout << res << endl;
     }

     return ;
 }

D
E

题解上这题也是个Easy..可是当时愣是没有看出来，知道突破点是ｋ比较小。

正解是枚举数列中第一个和最后一个没有改变的位置,　这个不超过k^2，然后。。。

这里getchar_unlocked是getchar()的线程不安全版本。。好像已经被windows弃用。

还有，n*10可以表达为(n<<3) + (n<<1)

 /*************************************************************************
     > File Name: E.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月15日 星期一 21时04分57秒
     > Propose:
  ************************************************************************/

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

 const int MAX_N = ;
 typedef long long LL;
 #define rep(i, a, b) for (int i = (a); i <= (b); i++)
 int n, k, a[MAX_N];

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

 int main(void) {
     read(n), read(k);
     rep (i, , n) read(a[i]);

     int a1 = 0x3f3f3f3f, d = 0x3f3f3f3f;
     rep (i, , k+) rep (j, n-k+i-, n) {
         int cnt = , tmp_d = (a[j] - a[i]) / (j - i), tmp_a1 = a[i] - (i - ) * tmp_d;
         rep (t, , n) if (a[t] != tmp_a1 + (t - ) * tmp_d) cnt++;
         if (cnt <= k) {
             if (a1 > tmp_a1) a1 = tmp_a1, d = tmp_d;
             else if (a1 == tmp_a1 && d > tmp_d) d = tmp_d;
         }
     }

     rep (i, , n) printf("%d%c", a1 + (i - ) * d, i == n ? '\n' : ' ');

     return ;
 }

F

 /*************************************************************************
     > File Name: F.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月08日 星期一 13时07分43秒
     > Propose:
  ************************************************************************/
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 typedef long long LL;
 LL N, M;

 /*
  *　(a / b) % c 等价于 (a % (b * c)) / b;
  * 　适用于ｂｃ不大而且ｂ在ｃ的剩余系中逆元不存在的情况
  */
 LL sum(LL n) {
       LL m = M * ;
       n %= m;
       return (n * ( * n % m * n % m * n % m * n % m +  * n % m * n % m * n % m +  * n % m * n % m -  + m) % m) % m / ;
 }

 int main(void) {
     ios::sync_with_stdio(false);
     int T;
     cin >> T;
     while (T--) {
         cin >> N >> M;
         if (N == ) {
             cout << N % M << endl;
             continue;
         } else if (N == ) {
             cout << (N + ) % M << endl;
             continue;
         } else if (N == ) {
             cout << (N +  + ) % M << endl;
             continue;
         }

         LL res = , i = , fst = , last = N, j;
         while (true) {
             LL r = N / (i + ), l = i;
             j = N / i;
             res = (res + (sum(last) - sum(r) + M) % M * (i % M) % M) % M;
             if (r != fst || last != l)
                 res = (res + (sum(l) - sum(fst) + M) % M * (j % M) % M) % M;
             if (l >= r) break;
             i++;
             last = r;
             fst = l;
         }

         cout << res << endl;
     }

     return ;
 }

G

需要维护每个区间内比右短点小的个数和比左端点小的个数，BIT系列。。是个不错的题。

 /*************************************************************************
     > File Name: E.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月08日 星期一 17时18分23秒
     > Propose:
  ************************************************************************/

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

 #define rep(i, n) for (int i = (1); i <= (n); i++)
 #define per(i, n) for (int i = (n); i >= (1); i--)
 const int MAX_N = ;
 typedef long long LL;
 int N, M, w, A[MAX_N], c[MAX_N];
 struct node {
     int id, l, r, ltr, ltl, eql, eqr;
     node () {}
     node (int id, int l, int r):id(id), l(l), r(r),ltr(), ltl() {}
     friend bool operator < (const node &x, const node &y) {
           return x.id < y.id;
     }
 }Q[MAX_N];

 int lowbit(int x) {
       return x & -x;
 }

 void add(int x) {
       while (x <= w) {
         c[x]++;
         x += lowbit(x);
     }
 }

 int sum(int x) {
       int res = ;
       while (x > ) {
         res += c[x];
         x -= lowbit(x);
     }
     return res;
 }

 int main(void) {
     ios::sync_with_stdio(false);
     while (cin >> N >> M) {
         vector<int> var;
         rep (i, N) {
             cin >> A[i];
             var.push_back(A[i]);
         }
         sort(var.begin(), var.end());
         var.resize(unique(var.begin(), var.end()) - var.begin());
         w = var.size();

         memset(c, , sizeof(c));
         LL res = ;
         rep (i, N) {
             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + ;
             res += i -  - sum(pos - ) - (sum(pos) - sum(pos - ));
             add(pos);
         }

 //        cerr << res << endl;

         vector<int> lft[MAX_N], rgt[MAX_N];
         int x, y;
         rep (i, M) {
             cin >> x >> y;
             if (x > y) swap(x, y);
             Q[i] = node(i, x, y);
             lft[x].push_back(i);
             rgt[y].push_back(i);
         }

         memset(c, , sizeof(c));
         rep (i, N) {
               int szl = lft[i].size(), szr = rgt[i].size();
             for (int j = ; j < szl; j++) {
                 int id = lft[i][j];
                 int pos = lower_bound(var.begin(), var.end(), A[Q[id].r]) - var.begin() + ;
                 Q[id].eqr = sum(pos) - sum(pos - );
                 Q[id].ltr = sum(pos - );
             }

             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + ;
             for (int j = ; j < szr; j++) {
                 int id = rgt[i][j];
                 Q[id].ltr = sum(pos - ) - Q[id].ltr;
                 Q[id].eqr = sum(pos) - sum(pos - ) - Q[id].eqr;
             }
             add(pos);
         }

         memset(c, , sizeof(c));
         per (i, N) {
             int szl = lft[i].size(), szr = rgt[i].size();
             for (int j = ; j < szr; j++) {
                 int id = rgt[i][j];
                 int pos = lower_bound(var.begin(), var.end(), A[Q[id].l]) - var.begin() + ;
                 Q[id].ltl = sum(pos - );
                 Q[id].eql = sum(pos) - sum(pos - );
             }
             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + ;
             for (int j = ; j < szl; j++) {
                 int id = lft[i][j];
                 Q[id].ltl = sum(pos - ) - Q[id].ltl;
                 Q[id].eql = sum(pos) - sum(pos - ) - Q[id].eql;
             }
             add(pos);
         }
     //    cerr << "say hello\n";

         sort (Q + , Q + M + );
         rep (i, M) {
             int l = Q[i].l, r = Q[i].r;
             LL tmp = Q[i].ltr - Q[i].ltl - (r - l +  - Q[i].ltr - Q[i].eqr) + (r - l +  - Q[i].ltl - Q[i].eql);
             if (A[r] > A[l]) tmp--;
             else if (A[r] < A[l]) tmp++;
             cout << res + tmp << endl;
         }
     }

     return ;
 }

H

I

J