VerySecureEncryption 模拟

题意：

给你个串message，然后一个置换key，输出置换K次后的结果。

题解：

直接模拟就好。

代码：

#include<iostream>

#include<vector>

#include<string>

#include<algorithm>

using namespace std;

class VerySecureEncryption {

public:

    string encrypt(string message, vector<int> key, int K) {

        char res[][];

        for (int i = ; i < message.length(); i++)res[][i] = message[i];

        for (int i = ; i < K; i++)

            for (int j = ; j < message.length(); j++)

                res[i & ][key[j]] = res[(i + ) & ][j];

        string r;

        for (int i = ; i < message.length(); i++)

            r = r + res[(K - ) & ][i];

        return r;

    }

};

IsItASquare 计算几何

题意：

给你平面上四个点，问你是否能够构成正方形

题解：

取三个点，看是否能够构成等腰直角三角形，然后再check一下最后一个点。

代码：

#include<iostream>

#include<vector>

#include<string>

#include<algorithm>

using namespace std;

class Coordinate

{

public:

    double xCoordinate;

    double yCoordinate;

    Coordinate(double x = ,double y = )

    {

        this->xCoordinate = x;

        this->yCoordinate = y;

    }

    bool operator!=(Coordinate const &comp) const

    {

        return (this->xCoordinate != comp.xCoordinate ||

                this->yCoordinate != comp.yCoordinate);

    }

};

/**

 * @explanation 判断是否为等腰直角三角形.

 */

bool Judge(Coordinate const x,Coordinate const y,Coordinate const z)

{

    Coordinate *mVector = new Coordinate(x.xCoordinate - y.xCoordinate,

                                         x.yCoordinate - y.yCoordinate);

    Coordinate *nVector = new Coordinate(z.xCoordinate -

                                         (x.xCoordinate + y.xCoordinate)/,

                                         z.yCoordinate -

                                         (x.yCoordinate + y.yCoordinate)/);

    //判断是否为等腰三角形

    bool result = ((mVector->xCoordinate * nVector->xCoordinate +

                    mVector->yCoordinate * nVector->yCoordinate) == );

    //判断是否是直角三角形

    if(result)

        result = (mVector->xCoordinate * mVector->xCoordinate +

                  mVector->yCoordinate * mVector->yCoordinate)

                 == ((nVector->xCoordinate * nVector->xCoordinate +

                      nVector->yCoordinate * nVector->yCoordinate) * );

    delete mVector;

    delete nVector;

    return result;

}

bool IsSquare(Coordinate *array,int length)

{

    if(length != )

        return false;

    int a,b,c;

    if(Judge(array[],array[],array[]))

    {

        a = ;

        b = ;

        c = ;

    }

    else if(Judge(array[],array[],array[]))

    {

        a = ;

        b = ;

        c = ;

    }

    else if(Judge(array[],array[],array[]))

    {

        a = ;

        b = ;

        c = ;

    }

    else

        return false;

    return (array[] != array[c] && Judge(array[a],array[b],array[]));

}

class IsItASquare {

public:

    string isSquare(vector <int> x, vector <int> y) {

        Coordinate ar[];

        for(int i=;i<;i++)ar[i].xCoordinate=x[i],ar[i].yCoordinate=y[i];

        return IsSquare(ar,)?"It's a square":"Not a square";

    }

};

AnArra 乱搞

题意：

给你个序列A，统计A[i]*A[j]*A[k]%K==0的个数，i<j<k

题解：

采用存一半搜一半的思想，预处理出每个因子有多少倍数。然后枚举A[i],A[j]，令g=gcd(A[i]*A[j],K)，t=K/g，那么问题就是有多少数是t的倍数，这个刚刚已经预处理好了。

代码：

#include<iostream>

#include<map>

#include<algorithm>

using namespace std;

typedef long long ll;

int ma[];

ll gcd(ll a,ll b) {

    return b ==  ? a : gcd(b, a % b);

}

class AnArray {

public:

    int solveProblem(vector<int> A, int K) {

        for (auto a:A) {

            for (int i = ; i * i <= a; i++) {

                if (a % i == ) {

                    ma[i]++;

                    int t = a / i;

                    if (i * i != a && t <= )ma[t]++;

                }

            }

        }

        ll res = ;

        for (int i = ; i < A.size(); i++)

            for (int j = ; j < A.size(); j++) {

                if (i == j)continue;

                ll tmp = ;

                tmp = tmp * A[i] * A[j];

                int g = gcd(tmp, K);

                int t = K / g;

                res = res + ma[K / g];

                if (A[i] % t == )res--;

                if (A[j] % t == )res--;

            }

        return res / ;

    }

};

巴特西

Topcoder SRM 668 DIV 2

VerySecureEncryption 模拟

题意：

题解：

代码：

IsItASquare 计算几何

题意：

题解：

代码：

AnArra 乱搞

题意：

题解：

代码：

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