Topcoder SRM 687 div2

通过数：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;

class Quorum {
public:
    int count(vector <int>, int);
};

int Quorum::count(vector <int> arr, int k) {
    sort(arr.begin(), arr.end());
    int res = 0;
    for(int i = 0 ; i < min(k, (int)arr.size()) ; i++)
        res += arr[i];
    return res;
}

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500：

加一个vector用来回溯就好

#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;

class Quacking {
public:
    int quack(string);
};
string str = "quack";
int Quacking::quack(string s) {
    int n = s.size();
    int res = 0;
    int vis[5000];
    for(int i = 0 ; i < n ; i++) {
//        printf("i = %d\n", i);
        if(vis[i] == 1) continue;
        else if(s[i] != ‘q‘) return -1;
        int num = 0;
        vector<int>rev;
        for(int j = 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(int j = 0 ; j < (int)rev.size() ; j++) vis[rev[j]] = 0;
        res++;
    }
    for(int i = 0 ; i < n ; i++) if(vis[i] == 0) return -1;
    return res;
}

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1000：

打死想不到是概率DP

转换后发现原式表示的是f(p,k)表示前k-1天均未完成，第k天完成的概率

那么，每天完成的概率就是1/p

Dp[i][j]表示到达(i,j)状态时，第一个任务消耗时间严格小于第一个任务的概率

初始化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;

class Queueing {
public:
    double probFirst(int, int, int, int);
};
const int MAXN = 1000 + 5;
double dp[MAXN][MAXN];
double Queueing::probFirst(int len1, int len2, int p1, int p2) {
    double q1 = 1.0 / (1.0 * p1);
    double q2 = 1.0 / (1.0 * p2);
    for(int i = 0 ; i <= len1 ; i++) {
        for(int j = 0 ; j <= len2 ; j++) {
            if(j == 0) dp[i][j] = 0;
            else if(i == 0) dp[i][j] = 1;
            else {
                double temp1 = (q1) * (q2) * dp[i - 1][j - 1];
                double temp2 = (1 - q1) * q2 * dp[i][j - 1];
                double temp3 = (q1) * (1 - q2) * dp[i - 1][j];
                dp[i][j] = (temp1 + temp2 + temp3) / (1.0 - (1 - q1) * (1 - q2));
            }
        }
    }
    return dp[len1][len2];
}

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