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

 1 /*************************************************************************

 2     > File Name: A.cpp

 3     > Author: Stomach_ache

 4     > Mail: [email protected]

 5     > Created Time: 2014年09月06日 星期六 16时32分54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 const int MOD = 1e9 + 7;

20 typedef long long LL;

21 int t, n;

22 char s[100005];

23 

24 int main(void) {

25       ios::sync_with_stdio(false);

26     cin >> t;

27     while (t--) {

28           cin >> s;

29           n = strlen(s);

30         LL res = 1;

31         for (int i = 0; i < n; i++) {

32               if ((i + 1) % 2 == 1) res = (res * 2) % MOD;

33             else res = (res * 2 - 1) % MOD;

34             if (s[i] == 'r') res = (res + 2) % MOD;

35         }

36         cout << res << endl;

37     }

38     return 0;

39 }

B　模拟

 1 n, m = map(int, raw_input().split());

 2 a = map(int, raw_input().split());

 3 now = 1;

 4 while m:

 5     m -= 1;

 6     Q = raw_input().split();

 7     if Q[0] == 'R':

 8           print a[(now + int(Q[1]) - 1) % n - 1]

 9     elif Q[0] == 'C':

10         now = (now + int(Q[1])) % n;

11     else :

12         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。。。。。。。

 1 /*************************************************************************

 2     > File Name: C.cpp

 3     > Author: Stomach_ache

 4     > Mail: [email protected]

 5     > Created Time: 2014年09月06日 星期六 11时08分09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 const int MOD = 1e9 + 7;

20 typedef long long LL;

21 LL fact[1000005], n;

22 

23 LL pow_mod(LL a, LL b) {

24     LL res = 1;

25     while (b) {

26         if (b & 1) res = (res * a) % MOD;

27         a = (a * a) % MOD;

28         b >>= 1;

29     }

30     return res;

31 }

32 

33 void init() {

34     fact[5] = 1; fact[6] = 6;

35     int now = 2;

36     for (int i = 7; i <= 1000000; i++, now++) fact[i] = (fact[i - 1] * pow_mod(now, MOD - 2) % MOD * i) % MOD;

37 }

38 

39 int main(void) {

40     ios::sync_with_stdio(false);

41     init();

42     while (cin >> n) {

43         if (n < 13) {

44             cout << "0" << endl;

45             continue;

46         }

47         LL res = 0;

48         if (n % 2 == 0){

49             for (int i = 2; i <= n - 12; i += 2) {

50               　res = (res + fact[(n - i)/2 - 1]) % MOD;

51             }

52         } else {

53             for (int i = 1; i <= n - 12; i += 2) {

54                 res = (res + fact[(n - i)/2 - 1]) % MOD;

55             }

56         }

57         cout << res << endl;

58     }

59 

60     return 0;

61 }

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)

 1 /*************************************************************************

 2     > File Name: E.cpp

 3     > Author: Stomach_ache

 4     > Mail: [email protected]

 5     > Created Time: 2014年09月15日 星期一 21时04分57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 const int MAX_N = 100050;

20 typedef long long LL;

21 #define rep(i, a, b) for (int i = (a); i <= (b); i++)

22 int n, k, a[MAX_N];

23 

24 void read(int &res) {

25     char c = ' ';

26     res = 0;

27     int sign = 1;

28     while (c < '0' || c > '9') {if (c == '-') sign = -1; c = getchar_unlocked();}

29     while (c >= '0' && c <= '9') res = (res<<3) + (res<<1) + c - '0', c= getchar_unlocked();

30     res *= sign;

31 }

32 

33 int main(void) {

34     read(n), read(k);

35     rep (i, 1, n) read(a[i]);

36 

37     int a1 = 0x3f3f3f3f, d = 0x3f3f3f3f;

38     rep (i, 1, k+1) rep (j, n-k+i-1, n) {

39         int cnt = 0, tmp_d = (a[j] - a[i]) / (j - i), tmp_a1 = a[i] - (i - 1) * tmp_d;

40         rep (t, 1, n) if (a[t] != tmp_a1 + (t - 1) * tmp_d) cnt++;

41         if (cnt <= k) {

42             if (a1 > tmp_a1) a1 = tmp_a1, d = tmp_d;

43             else if (a1 == tmp_a1 && d > tmp_d) d = tmp_d;

44         }

45     }

46 

47     rep (i, 1, n) printf("%d%c", a1 + (i - 1) * d, i == n ? '\n' : ' ');

48 

49     return 0;

50 }

F

 1 /*************************************************************************

 2     > File Name: F.cpp

 3     > Author: Stomach_ache

 4     > Mail: [email protected]

 5     > Created Time: 2014年09月08日 星期一 13时07分43秒

 6     > Propose: 

 7  ************************************************************************/

 8 #include <cmath>

 9 #include <string>

10 #include <cstdio>

11 #include <vector>

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 LL N, M;

21 

22 /*

23  *　(a / b) % c 等价于 (a % (b * c)) / b;

24  * 　适用于ｂｃ不大而且ｂ在ｃ的剩余系中逆元不存在的情况

25  */

26 LL sum(LL n) {

27       LL m = M * 30;

28       n %= m;

29       return (n * (6 * n % m * n % m * n % m * n % m + 15 * n % m * n % m * n % m + 10 * n % m * n % m - 1 + m) % m) % m / 30;

30 }

31 

32 int main(void) {

33     ios::sync_with_stdio(false);

34     int T;

35     cin >> T;

36     while (T--) {

37         cin >> N >> M;

38         if (N == 1) {

39             cout << N % M << endl;

40             continue;

41         } else if (N == 2) {

42             cout << (N + 16) % M << endl;

43             continue;

44         } else if (N == 3) {

45             cout << (N + 16 + 81) % M << endl;

46             continue;

47         }

48 

49         LL res = 0, i = 1, fst = 0, last = N, j;

50         while (true) {

51             LL r = N / (i + 1), l = i;

52             j = N / i;

53             res = (res + (sum(last) - sum(r) + M) % M * (i % M) % M) % M;

54             if (r != fst || last != l)

55                 res = (res + (sum(l) - sum(fst) + M) % M * (j % M) % M) % M;

56             if (l >= r) break;

57             i++;

58             last = r;

59             fst = l;

60         }

61 

62         cout << res << endl;

63     }

64 

65     return 0;

66 }

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。

  1 /*************************************************************************

  2     > File Name: E.cpp

  3     > Author: Stomach_ache

  4     > Mail: [email protected]

  5     > Created Time: 2014年09月08日 星期一 17时18分23秒

  6     > Propose: 

  7  ************************************************************************/

  8 

  9 #include <cmath>

 10 #include <string>

 11 #include <cstdio>

 12 #include <fstream>

 13 #include <vector>

 14 #include <cstring>

 15 #include <iostream>

 16 #include <algorithm>

 17 using namespace std;

 18 /*Let's fight!!!*/

 19 

 20 #define rep(i, n) for (int i = (1); i <= (n); i++)

 21 #define per(i, n) for (int i = (n); i >= (1); i--)

 22 const int MAX_N = 200050;

 23 typedef long long LL;

 24 int N, M, w, A[MAX_N], c[MAX_N];

 25 struct node {

 26     int id, l, r, ltr, ltl, eql, eqr;

 27     node () {}

 28     node (int id, int l, int r):id(id), l(l), r(r),ltr(0), ltl(0) {}

 29     friend bool operator < (const node &x, const node &y) {

 30           return x.id < y.id;

 31     }

 32 }Q[MAX_N];

 33 

 34 int lowbit(int x) {

 35       return x & -x;

 36 }

 37 

 38 void add(int x) {

 39       while (x <= w) {

 40         c[x]++;

 41         x += lowbit(x);

 42     }

 43 }

 44 

 45 int sum(int x) {

 46       int res = 0;

 47       while (x > 0) {

 48         res += c[x];

 49         x -= lowbit(x);

 50     }

 51     return res;

 52 }

 53 

 54 int main(void) {

 55     ios::sync_with_stdio(false);

 56     while (cin >> N >> M) {

 57         vector<int> var;

 58         rep (i, N) {

 59             cin >> A[i];

 60             var.push_back(A[i]);

 61         }

 62         sort(var.begin(), var.end());

 63         var.resize(unique(var.begin(), var.end()) - var.begin());

 64         w = var.size();

 65 

 66         memset(c, 0, sizeof(c));

 67         LL res = 0;

 68         rep (i, N) {

 69             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1;

 70             res += i - 1 - sum(pos - 1) - (sum(pos) - sum(pos - 1));    

 71             add(pos);

 72         }

 73         

 74 //        cerr << res << endl;

 75 

 76         vector<int> lft[MAX_N], rgt[MAX_N];

 77         int x, y;    

 78         rep (i, M) {

 79             cin >> x >> y;

 80             if (x > y) swap(x, y);

 81             Q[i] = node(i, x, y);

 82             lft[x].push_back(i);

 83             rgt[y].push_back(i);

 84         }

 85 

 86         memset(c, 0, sizeof(c));

 87         rep (i, N) {

 88               int szl = lft[i].size(), szr = rgt[i].size();    

 89             for (int j = 0; j < szl; j++) {

 90                 int id = lft[i][j];

 91                 int pos = lower_bound(var.begin(), var.end(), A[Q[id].r]) - var.begin() + 1;

 92                 Q[id].eqr = sum(pos) - sum(pos - 1);

 93                 Q[id].ltr = sum(pos - 1);

 94             }

 95 

 96             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1;

 97             for (int j = 0; j < szr; j++) {

 98                 int id = rgt[i][j];

 99                 Q[id].ltr = sum(pos - 1) - Q[id].ltr;

100                 Q[id].eqr = sum(pos) - sum(pos - 1) - Q[id].eqr;

101             }

102             add(pos);

103         }

104 

105         memset(c, 0, sizeof(c));

106         per (i, N) {

107             int szl = lft[i].size(), szr = rgt[i].size();    

108             for (int j = 0; j < szr; j++) {

109                 int id = rgt[i][j];

110                 int pos = lower_bound(var.begin(), var.end(), A[Q[id].l]) - var.begin() + 1;

111                 Q[id].ltl = sum(pos - 1);

112                 Q[id].eql = sum(pos) - sum(pos - 1);

113             }

114             int pos = lower_bound(var.begin(), var.end(), A[i]) - var.begin() + 1;

115             for (int j = 0; j < szl; j++) {

116                 int id = lft[i][j];

117                 Q[id].ltl = sum(pos - 1) - Q[id].ltl;

118                 Q[id].eql = sum(pos) - sum(pos - 1) - Q[id].eql;

119             }

120             add(pos);

121         }

122     //    cerr << "say hello\n";

123 

124         sort (Q + 1, Q + M + 1);

125         rep (i, M) {

126             int l = Q[i].l, r = Q[i].r;

127             LL tmp = Q[i].ltr - Q[i].ltl - (r - l + 1 - Q[i].ltr - Q[i].eqr) + (r - l + 1 - Q[i].ltl - Q[i].eql); 

128             if (A[r] > A[l]) tmp--;

129             else if (A[r] < A[l]) tmp++;

130             cout << res + tmp << endl;

131         }

132     }

133 

134     return 0;

135 }

H

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