SRM 458 Div 2

# Include <vector> <br/> # include <list> <br/> # include <map> <br/> # include <set> <br/> # include <deque> <br/> # include <stack> <br/> # include <bitset> <br/> # include <algorithm> <br/> # include <functional> <br/> # include <numeric> <br/> # include <utility> <br/> # include <sstream> <br/> # include <iostream> <br/> # include <iomanip> <br/> # include <cstdio> <br/> # include <cmath> <br/> # include <cstdlib> <br/> # includ E <ctime> <br/> # include <string> <br/> # include <cstring> <br/> using namespace STD; </P> <p> class <br/> desertification <br/> {<br/> Public: <br/> int desertarea (vector <string> Island, int T) <br/>{< br/> vector <string> P; <br/> If (T> = 20) <br/> T = 21; <br/> while (t --) <br/>{< br/> P = Island; <br/> for (INT I = 0; I <p. size (); I ++) <br/>{< br/> for (Int J = 0; j <p [0]. size (); j ++) <br/>{< br/> If (Island [I] [J] = 'D') <br/>{< br /> If (I> 0) <br/> P [I-1] [J] = 'D'; <br/> If (j> 0) <br/> P [I] [J-1] = 'D'; <br/> if (I + 1 <p. size () <br/> P [I + 1] [J] = 'D'; <br/> If (J + 1 <p [0]. size () <br/> P [I] [J + 1] = 'D '; </P> <p >}< br/> island = P; <br/>}< br/> int COUNT = 0; <br/> for (INT I = 0; I <p. size (); I ++) <br/>{< br/> for (Int J = 0; j <p [0]. size (); j ++) <br/>{< br/> If (P [I] [J] = 'D') <br/> count ++; <br/>}< br/> return count; <br/>}< br/> }; <br/>/* <br/> another idea is: Search directly, but nested layer-4 loops are required! <Br/> */

 

The first problem is that it is obviously a little slow to write, and then it is like this ..

At that time, I was wondering whether to adopt the second method!

This problem eventually fails because [0] is missing... Sorry .. This is the first time in history that rank1 has failed...

 

# Include <vector> <br/> # include <list> <br/> # include <map> <br/> # include <set> <br/> # include <deque> <br/> # include <stack> <br/> # include <bitset> <br/> # include <algorithm> <br/> # include <functional> <br/> # include <numeric> <br/> # include <utility> <br/> # include <sstream> <br/> # include <iostream> <br/> # include <iomanip> <br/> # include <cstdio> <br/> # include <cmath> <br/> # include <cstdlib> <br/> # includ E <ctime> </P> <p> using namespace STD; </P> <p> class bouncingbils <br/>{< br/> public: <br/> double expectedbounces (vector <int>, INT); <br/>}; </P> <p> double bouncingbils: expectedbounces (vector <int> X, int t) <br/>{< br/> int n = x. size (), ANS = 0, T, I, J, dir [99]; <br/> sort (X. begin (), X. end (); <br/> for (t = 0; t <(1 <n); t ++) <br/> {<br/> for (I = 0; I <n; I ++) <br/>{< br/> If (T & (1 <I) <br/> dir [I] = 1; <br /> Else dir [I] =-1; <br/>}< br/> for (I = 0; I <n; I ++) <br/> for (j = I + 1; j <n; j ++) <br/> If (dir [I] = 1 & dir [J] =-1 & T * 2> = x [J]-X [I]) ans ++; <br/>}< br/> return 1.0 * ANS/(1 <n ); <br/>}</P> <p>/* <br/> the above Code is indeed simplified! Clever Use of bit operations (T & (1 <I) <br/> the essence of this problem is that it does not need to be sorted !! <Br/> only one combination of directions can ensure collision! The expectation is obviously 0.25. <br/> the core code is only: <br/> int n = x. length, I, j; <br/> double ans = 0.0; <br/> for (INT I = 0; I <n; I ++) <br/> {<br/> for (Int J = 0; j <n; j ++) <br/> {<br/> If (X [I] <X [J] & X [J]-X [I] <= 2 * t) <br/> ans + = 0.25; <br/>}< br/> return ans; <br/> */

The second question was answered quickly and then made .. Because I have been counting twice, the last result is 2 <size (). As a result, the person in the room is misled to go to Cha and then fails .. Is it too bad for me ..

 

Import Java. util. *; <br/> public class producttriplet <br/> {<br/> Public long func (INT Minx, int Maxx, int miny, int Maxy, int minz, int maxz) <br/>{< br/> If (minx! = 1) <br/> return func (1, Maxx, miny, Maxy, minz, maxz)-func (1, Minx-1, miny, Maxy, minz, maxz ); <br/> If (miny! = 1) <br/> return func (Minx, Maxx, 1, Maxy, minz, maxz)-func (Minx, Maxx, 1, miny-1, minz, maxz ); <br/> If (minz! = 1) <br/> return func (Minx, Maxx, miny, Maxy, 1, maxz)-func (Minx, Maxx, miny, Maxy, 1, minz-1 ); <br/> long x = Maxx, y = Maxy, Z = maxz, ANS = 0, x, y; <br/> for (x = 1; x * x <= Z & x <= x; X ++) <br/> {<br/> Y = math. min (z/x, y); <br/> If (Y> = x) <br/> ans + = (Y-x + 1 ); <br/>}< br/> for (y = 1; y * Y <= Z & Y <= y; y ++) <br/>{< br/> X = math. min (z/y, x); <br/> If (x> Y) <br/> ans + = (x-y ); <br/>}< br/> return ans; <br/>}< br/> Public long counttriplets (INT Minx, int Maxx, Int miny, int Maxy, int minz, int maxz) <br/>{< br/> long ans = func (Minx, Maxx, miny, Maxy, minz, maxz ); <br/> return ans; <br/>}< br/>/* </P> <p> A very delicate answer, the first idea is to convert a count with intervals into a single interval. <Br/> now, we have gone through an iterative process <br/> the second part is to find the answer (1, Maxx, 1, Maxy, 1, maxz! <Br/> the answer must be solved within O (n. This is divided into a group of x> Y, and a group of x <= y! </P> <p> for (x = 1; x * x <= Z & x <= x; X ++) <br/>{< br/> Y = math. min (z/x, y); <br/> If (Y> = x) <br/> ans + = (Y-x + 1 ); <br/>}< br/> */

At that time, I did not think of a method to reduce the complexity, So I simply did not write it at will. This problem is still very clever! Learning !!!

 

This is the first chance to take the money .. The result is a very low-level error! Sorry, this is already happening several times...

Code more and exercise yourself!

 

PS: This is the first time I used notepad to write code. I had to stick to it for so long using Zhang yuze's computer .. Good results .. Haha!