topcoder-srm-613-div2

250分：
简单题，判断C A T这三个字符的个数就行了

/*************************************************************************
    > File Name: 250.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月08日 星期五 20时29分05秒
 ************************************************************************/

#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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

class TaroString {
    public:
        string getAnswer(string str) {
            int n = str.length();
            int cnt1 = 0, cnt2 = 0, cnt3 = 0;
            int pos1 = 0, pos2 = 0, pos3 = 0;
            for (int i = 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分：
暴力枚举最左端，然后求最小的右端
全部向左和全部向右的也要考虑

/*************************************************************************
    > File Name: 500.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月08日 星期五 20时37分19秒
 ************************************************************************/

#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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> Pint;

class TaroFriends {
    public:
        int getNumber(vector <int> arr, int X) {
            int n = arr.size();
            if (n == 1) {
                return 0;
            }
            int l, r;
            int ans = (1 << 30);
            for (int i = 0; i < n; ++i) {
                for (int sgn = 0; sgn < 2; ++sgn) {
                    l = arr[i];
                    if (sgn) {
                        l += X;
                    }
                    else {
                        l -= X;
                    }
                    r = -(1 << 30);
                    for (int j = 0; j < n; ++j) {
                        if (j == i) {
                            continue;
                        }
                        if (arr[j] - X < l && arr[j] + X < l) {
                            r = -(1 << 30);
                            break;
                        }
                        if (arr[j] - X < l) {
                            r = max(r, arr[j] + X);
                        }
                        else {
                            r = max(r, arr[j] - X);
                        }
                    }
                    if (r == -(1 << 30)) {
                        continue;
                    }
                    ans = min(ans, r - l);
                }
            }
            return ans;
        }
};

1000分：
dp[i][sta][k]表示枚举到第i张卡，<=10的数字状态为sta，共有k个不同的数字的方案数，转移分第i张卡取或者不取

/*************************************************************************
    > File Name: 1000.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月08日 星期五 21时25分26秒
 ************************************************************************/

#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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

LL dp[55][1100][110];

class TaroCards {
    public:
        LL getNumber(vector <int> F, vector <int> S, int K) {
            int n = F.size();
            for (int i = 0; i <= n; ++i) {
                for (int j = 0; j < (1 << 10); ++j) {
                    for (int k = 0; k <= K; ++k) {
                        dp[i][j][k] = 0;
                    }
                }
            }
            dp[0][0][0] = 1;
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < (1 << 10); ++j) {
                    for (int k = 0; k <= K; ++k) {
                        if (dp[i][j][k]) {
                            dp[i + 1][j][k] += dp[i][j][k];
                            int newsta = j;
                            int kk = k;
                            if (F[i] > 10) {
                                ++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 (int i = 0; i < (1 << 10); ++i) {
                for (int j = 0; j <= K; ++j) {
                    ans += dp[n][i][j];
                }
            }
            return ans;
        }
};