Topcoder SRM 668 DIV 2

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[2][12];
        for (int i = 0; i < message.length(); i++)res[1][i] = message[i];
        for (int i = 0; i < K; i++)
            for (int j = 0; j < message.length(); j++)
                res[i & 1][key[j]] = res[(i + 1) & 1][j];
        string r;
        for (int i = 0; i < message.length(); i++)
            r = r + res[(K - 1) & 1][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 = 0,double y = 0)
    {
        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)/2,
                                         z.yCoordinate -
                                         (x.yCoordinate + y.yCoordinate)/2);
    //判断是否为等腰三角形
    bool result = ((mVector->xCoordinate * nVector->xCoordinate +
                    mVector->yCoordinate * nVector->yCoordinate) == 0);

    //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) * 4);

    delete mVector;
    delete nVector;

    return result;
}

bool IsSquare(Coordinate *array,int length)
{
    if(length != 4)
        return false;
    int a,b,c;

    if(Judge(array[0],array[1],array[2]))
    {
        a = 0;
        b = 1;
        c = 2;
    }
    else if(Judge(array[0],array[2],array[1]))
    {
        a = 0;
        b = 2;
        c = 1;
    }
    else if(Judge(array[2],array[1],array[0]))
    {
        a = 1;
        b = 2;
        c = 0;
    }
    else
        return false;

    return (array[3] != array[c] && Judge(array[a],array[b],array[3]));
}

class IsItASquare {
public:
    string isSquare(vector <int> x, vector <int> y) {
        Coordinate ar[4];
        for(int i=0;i<4;i++)ar[i].xCoordinate=x[i],ar[i].yCoordinate=y[i];
        return IsSquare(ar,4)?"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[1000006];

ll gcd(ll a,ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

class AnArray {
public:
    int solveProblem(vector<int> A, int K) {
        for (auto a:A) {
            for (int i = 1; i * i <= a; i++) {
                if (a % i == 0) {
                    ma[i]++;
                    int t = a / i;
                    if (i * i != a && t <= 1000000)ma[t]++;
                }
            }
        }
        ll res = 0;
        for (int i = 0; i < A.size(); i++)
            for (int j = 0; j < A.size(); j++) {
                if (i == j)continue;
                ll tmp = 1;
                tmp = tmp * A[i] * A[j];
                int g = gcd(tmp, K);
                int t = K / g;
                res = res + ma[K / g];
                if (A[i] % t == 0)res--;
                if (A[j] % t == 0)res--;
            }
        return res / 6;
    }
};