Topcoder SRM 687 Div2

Topcoder SRM 687 Div2
By number: 2
250:

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace STD;classQuorum { Public:intCount vector <int>,int);};intQuorum::count ( vector <int>Arrintk) {sort (Arr.begin (), Arr.end ());intres =0; for(inti =0; I < min (k, (int) arr.size ()); i++) Res + = Arr[i];returnRes;} <%:testing-code%>//powered by [Kawigiedit] 2.0!

500:
Plus a vector to backtrack.

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace STD;classquacking { Public:intQuack (string);};stringstr ="Quack";intQuacking::quack (strings) {intn = s.size ();intres =0;intvis[ the]; for(inti =0; I < n; i++) {//printf ("I =%d\n", i);        if(Vis[i] = =1)Continue;Else if(S[i]! =' Q ')return-1;intnum =0; vector<int>Rev; for(intj = i; J < N; J + +) {if(Vis[j] = =1)Continue;if(S[j] = = Str[num]) {num++, vis[j] =1, Rev.push_back (j);//, printf ("%d", j);}if(num = =5) {rev.clear (); num =0; }        }//Puts ("");         for(intj =0; J < (int) Rev.size (); J + +) Vis[rev[j]] =0;    res++; } for(inti =0; I < n; i++)if(Vis[i] = =0)return-1;returnRes;} <%:testing-code%>//powered by [Kawigiedit] 2.0!

1000:
Shot to death is the probability of DP
After conversion, it is found that the original expression is F (p,k) indicates that the first k-1 days are not completed, the probability of K-day completion
Well, the probability of completing every day is 1/p.
DP[I][J] Indicates the arrival (I,J) state, the first task consumes less time than the first task's probability
Initialize dp[i][0] = 0, dp[0][i!=0] = 1

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace STD;classQueueing { Public:DoubleProbfirst (int,int,int,int);};Const intMAXN = ++5;DoubleDP[MAXN][MAXN];DoubleQueueing::p Robfirst (intLen1,intLen2,intP1,intP2) {DoubleQ1 =1.0/ (1.0* p1);DoubleQ2 =1.0/ (1.0* p2); for(inti =0; I <= len1; i++) { for(intj =0; J <= Len2; J + +) {if(J = =0) Dp[i][j] =0;Else if(i = =0) Dp[i][j] =1;Else{DoubleTemp1 = (Q1) * (Q2) * Dp[i-1][j-1];DoubleTemp2 = (1-Q1) * Q2 * DP[I][J-1];DoubleTemp3 = (Q1) * (1-Q2) * Dp[i-1][J]; DP[I][J] = (Temp1 + temp2 + Temp3)/(1.0- (1-Q1) * (1-Q2)); }        }    }returnDP[LEN1][LEN2];} <%:testing-code%>//powered by [Kawigiedit] 2.0!

Topcoder SRM 687 Div2