Topcoder-srm-613-div2

250 points:
Simple question, the number of the three characters of the C a T is OK

/************************************************************************* > File name:250.cpp > Author: ALex > Mail: [email protected] > Created time:2015 May 08 Friday 20:29 05 seconds ************************************ ************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;classtarostring { Public:stringGetanswer (stringSTR) {intn = str.length ();intCnt1 =0, Cnt2 =0, Cnt3 =0;intPOS1 =0, Pos2 =0, POS3 =0; for(inti =0; I < n; ++i) {if(Str[i] = =' C ') {++cnt1;                POS1 = i; }Else if(Str[i] = =' A ') {++cnt2;                Pos2 = i; }Else if(Str[i] = =' T ') {++cnt3;                POS3 = i; }            }if(Cnt1! =1|| Cnt2! =1|| Cnt3! =1) {return "Impossible"; }if(Pos1 < Pos2 && Pos2 < POS3) {return "Possible"; }return "Impossible"; }};

500 points:
Violence enumerates the leftmost, then the smallest right end
All to the left and all to the right also to consider

/************************************************************************* > File name:500.cpp > Author: ALex > Mail: [email protected] > Created time:2015 May 08 Friday 20:37 19 seconds ************************************ ************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> Pint;classTarofriends { Public:intGetNumber ( vector <int>ArrintX) {intn = arr.size ();if(n = =1) {return 0; }intL, R;intAns = (1<< -); for(inti =0; I < n; ++i) { for(intSGN =0; SGN <2; ++SGN) {L = Arr[i];if(SGN)                    {L + = X; }Else{L-= X; } r =-(1<< -); for(intj =0; J < N; ++J) {if(j = = i) {Continue; }if(Arr[j]-x < l && Arr[j] + x < L) {r =-(1<< -); Break; }if(Arr[j]-X < L)                        {r = max (R, Arr[j] + X); }Else{r = max (R, Arr[j]-X); }                    }if(r = =-(1<< -)) {Continue;                } ans = min (ans, r-l); }            }returnAns }};

1000 points:
Dp[i][sta][k] Represents the enumeration to the I card, <=10 the number status of STA, a total of k different number of scenarios, transfer sub-I card to take or not take

/************************************************************************* > File name:1000.cpp > Author: ALex > Mail: [email protected] > Created time:2015 May 08 Friday 21:25 26 seconds *********************************** *************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL; LL dp[ -][1100][ the];classTarocards { Public: LL GetNumber ( vector <int>F vector <int>SintK) {intn = f.size (); for(inti =0; I <= N; ++i) { for(intj =0; J < (1<<Ten); ++J) { for(intK =0; K <= K; ++K) {Dp[i][j][k] =0; }}} dp[0][0][0] =1; for(inti =0; I < n; ++i) { for(intj =0; J < (1<<Ten); ++J) { for(intK =0; K <= K; ++K) {if(Dp[i][j][k]) {dp[i +1][j][k] + = dp[i][j][k];intNewsta = j;intKK = k;if(F[i] >Ten) {++kk; }Else if(! ((1<< (F[i]-1)) (& Newsta)) {++kk; Newsta |= (1<< ((f[i)-1))); }if(! ((1<< (S[i]-1)) (& Newsta)) {++kk; Newsta |= (1<< ((s[i)-1))); }if(KK > K) {Continue; } Dp[i +1][NEWSTA][KK] + = dp[i][j][k]; }}}} LL ans =0; for(inti =0; I < (1<<Ten); ++i) { for(intj =0; J <= K;                ++j) {ans + = dp[n][i][j]; }            }returnAns }};

Topcoder-srm-613-div2