SRM summary 440

Question 1: http://www.topcoder.com/stat? C = problem_statement & PM = 10309.

Solution: According to DI = 1/2 * g * (Ti ^ 2), take the first one as a reference to obtain the proportional relationship between all the following and the first one.

Then calculate the first T1 according to sum (Ti) = T, and then find G according to D1 = 1/2 * g * (T1 ^ 2 ).

Later, I looked at other people's code and found that there is a simpler method. I first enlightened T.

# Include <iostream> <br/> # include <stdio. h> <br/> # include <vector> <br/> # include <cmath> <br/> using namespace STD; </P> <p> struct incrediblemachineeasy <br/> {<br/> double fun (INT Xi, int X1) <br/>{< br/> return SQRT (double) XI/X1); <br/>}< br/> double gravitationalacceleration (vector <int> height, int t) <br/>{< br/> double totel = 1; <br/> for (INT I = 1; I <peight. size (); I ++) <br/>{< br/> totel + = fun (height [I], height [0]); <br/>}< br/> // cout <totel <Endl; <br/> double T = (double) T/totel; <br/> return 2.0 * (double) height [0]/(T * t); <br/>}< br/> };

Question 2: http://www.topcoder.com/stat? C = problem_statement & PM = 10288.

Solution: search for records and make statistics.

# Include <iostream> <br/> # include <vector> <br/> # include <string> <br/> using namespace STD; </P> <p> struct mazewandreingeasy <br/> {<br/> int Sx, Sy; <br/> int; <br/> bool flag [55] [55]; <br/> int dir [4] [2]; // = }, {0}, {0,-1 }}; </P> <p> bool judge (int x, int y) <br/>{< br/> If (x> = 0 & x <SX & Y> = 0 & Y <Sy) <br/> return true; <br/> return false; <br/>}</P> <p> void DFS (vector <string> maze, int X, int y, int totel) <br/>{< br/> If (maze [x] [Y] = '*') <br/>{< br/> An = totel; <br/>}< br/> flag [x] [Y] = true; // EXE <br/> int COUNT = 0; <br/> bool f [10]; <br/> int XX, YY; <br/> memset (F, false, sizeof (f )); <br/> for (INT I = 0; I <4; I ++) <br/>{< br/> xx = x + dir [I] [0]; <br/> YY = Y + dir [I] [1]; <br/> If (Judge (XX, YY) = false) continue; <br/> If (maze [XX] [YY] = 'X') continue; <br/> If (flag [XX] [YY] = true) continue; <br/> count ++; <br/> F [I] = true; <br/>}< br/> int add = 0; <br/> If (count> 1) Add = 1; <br/> for (Int J = 0; j <4; j ++) <br/>{< br/> If (F [J] = false) continue; <br/> xx = x + dir [J] [0]; <br/> YY = Y + dir [J] [1]; <br/> DFS (maze, XX, YY, totel + Add ); <br/>}</P> <p> int decisions (vector <string> maze) <br/>{< br/> SX = maze. size (); <br/> Sy = maze [0]. size (); <br/> for (INT I = 0; I <SX; I ++) <br/> {<br/> for (Int J = 0; j <sy; j ++) <br/> {<br/> If (maze [I] [J] = 'M ') <br/>{< br/> memset (flag, false, sizeof (FLAG); </P> <p> dir [0] [0] =-1; dir [0] [1] = 0; <br/> dir [1] [0] = 1; dir [1] [1] = 0; <br/> dir [2] [0] = 0; dir [2] [1] =-1; <br/> dir [3] [0] = 0; dir [3] [1] = 1; </P> <p> int temp = 0; <br/> DFS (maze, I, j, temp ); <br/> return an; <br/>}< br/> };

Question 3: http://www.topcoder.com/stat? C = problem_statement & PM = 10289.

Solution: Use next_permutation () to generate a full arrangement and judge in sequence. A group of data in system test 238 has timed out. I also know that the big data will time out, but the DP Method hasn't come up yet, so I should stop this one first. Thank you!

# Include <iostream> <br/> # include <algorithm> <br/> # include <vector> <br/> using namespace STD; </P> <p> // denominator: size! </P> <p> struct wickedteacher <br/> {<br/>/* <br/> void print (iterator: vector <int> X) <br/>{< br/> cout <* x <""; <br/>}< br/> */<br/> long gcd (long, long B) <br/>{< br/> If (B = 0) return a; <br/> return gcd (B, A % B ); <br/>}< br/> string cheatprobability (vector <string> numbers, int K) <br/>{< br/> int size = numbers. size (); <br/> vector <int> test (size); <br/> for (INT I = 0; I <size; I ++) test [I] = I; <br/> Long Count = 0; </P> <p> int sum = 0; <br/> for (Int J = 0; j <size; j ++) <br/> for (INT r = 0; r <numbers [test [JJ]. size (); R ++) <br/>{< br/> sum = sum * 10 + numbers [test [JJ] [r]-'0 '; <br/> sum % = K; <br/>}</P> <p> If (sum = 0) Count ++; </P> <p> while (next_permutation (test. begin (), test. begin () + size) <br/>{< br/> // for_each (test. begin (), test. begin () + size, print); <br/>/* <br/> for (INT I = 0; I <test. size (); I ++) <br/>{< br/> cout <test [I]; <br/>}< br/> cout <Endl; <br/> */<br/> // test [0]... test [last] <br/> sum = 0; <br/> for (Int J = 0; j <size; j ++) <br/> {<br/> for (INT r = 0; r <numbers [test [J]. size (); R ++) <br/>{< br/> sum = sum * 10 + numbers [test [J] [r]-'0 '; <br/> sum % = K; <br/>}< br/> If (sum = 0) Count ++; <br/>}< br/> long totel = 1; <br/> for (int p = 2; P <= size; P ++) totel * = P; <br/> long temp = gcd (totel, count); <br/> totel/= temp; <br/> count/= temp; <br/> // count/totel <br/> string STR; <br/> If (totel = 0) <br/> Str. push_back ('0'); <br/> else <br/> while (totel) <br/>{< br/> Str. push_back (totel % 10 + '0'); <br/> totel/= 10; <br/>}< br/> Str. push_back ('/'); <br/> If (COUNT = 0) <br/> Str. push_back ('0'); <br/> else <br/> while (count) <br/> {<br/> Str. push_back (count % 10 + '0'); <br/> count/= 10; <br/>}< br/> reverse (Str. begin (), str. end (); <br/> return STR; <br/>}< br/> };