TC 668 #DIV2

250 points:

 You are going to send a message to your friend. The message is given as the stringmessage. To confuse potential eavesdroppers, you are going to scramble the message.

Scrambling of a message is performed using the vector <int> key. If a letter is at the (0-based) position i in the original message, it will appear at the positionkey[i] in the scrambled message. (The constraints given below guarantee that this process will produce a valid scrambled message.)

To make the encryption even more confusing, you are going to repeat the above processK times in a row. Givenmessage, key, and the intK, find and return the final encrypted message.分析：直接暴力

/*"abcde"{4, 3, 2, 1, 0}1Returns: "edcba"*/#include <iostream>#include <fstream>#include <cstring>#include <climits>#include <cmath>#include <queue>#include <ctime>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <algorithm>typedef long long LL;using namespace std;class VerySecureEncryption {public:   string encrypt( string message, vector <int> key, int K ) {       int sz=key.size();       string ans;       for(int i=0;i<K;i++)       {           for(int j=0;j<sz;j++)             ans[key[j]]=message[j];           for(int j=0;j<sz;j++)             message[j]=ans[j];       }       return message;   }};// Powered by FileEdit// Powered by moj 4.12 [modified TZTester]// Powered by CodeProcessor

600 points：

It's a bird! It's a plane! No, it's a square in a plane! Wait, is it really a square?

There are four distinct points in the plane. You are given their coordinates in the vector <int>sx andy: for each i between 0 and 3, inclusive, there is a point at (x[i],y[i]).

Return "It's a square" (quotes for clarity) if the four points are the vertices of a square. Otherwise, return "Not a square".分析：判断是不是正方形

上次BC看到一个较好的方法：连上对角两条边共六条边排序，前4条边相等，后两条边相等前是前4条边平方的2倍。

/*{0, 0, 2, 2}{0, 2, 0, 2}Returns: "It's a square"*/#include <iostream>#include <fstream>#include <cstring>#include <climits>#include <cmath>#include <queue>#include <ctime>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <algorithm>typedef long long LL;using namespace std;double A[10];double dis(int x1,int y1,int x2,int y2){    double x=x1-x2;    double y=y1-y2;    return x*x+y*y;}class IsItASquare {public:   string isSquare( vector <int> x, vector <int> y ) {       int tot=0;       for(int i=0;i<4;i++)         for(int j=i+1;j<4;j++)            A[tot++]=dis(x[i],y[i],x[j],y[j]);       sort(A,A+tot);       int flag;       if(A[0]==A[1]&&A[1]==A[2]&&A[2]==A[3]&&A[4]==A[5]&&A[5]==2*A[3])         flag=1;       else         flag=0;       return flag?"It's a square":"Not a square";   }};// Powered by FileEdit// Powered by moj 4.12 [modified TZTester]// Powered by CodeProcessor

1000 points：

One day, Bob the Coder was wondering whether abstract programming problems can have applications in practice. The next day, he was selected to be on a quiz show. He will win one million dollars if he answers the following question:

Given a vector <int> A with N elements and an int K, count the number of tuples (p, q, r) such that 0 <= p < q < r < N andA[p] *A[q] * A[r] is divisible byK.

Please compute and return the answer to Bob's question.分析：让你找三元组满足A[P]*A[Q]*A[R]整除K其实我们可以暴力前两个的乘积找到第三个

做下预处理

/*-A will contain between 3 and 2,000 elements, inclusive.K will be between 1 and 1,000,000, inclusive.Each element of A will be between 1 and 100,000,000, inclusive.{4, 5, 2, 25}100Returns: 2*/#include <iostream>#include <fstream>#include <cstring>#include <climits>#include <cmath>#include <queue>#include <ctime>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <algorithm>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )LL sum[1010][2010];int b[2010];LL a[2020];class AnArray {public:   int solveProblem( vector <int> A, int K ) {       CLEAR(sum,0);       int tot=0;       int n=A.size();       REP(i,n) a[i+1]=1LL*A[i];       for(int i=1;i<=K;i++)       {           if(K%i) continue;           b[tot]=i;           for(int j=1;j<=n;j++)           {               if(a[j]%i==0)                  sum[tot][j]=1;               sum[tot][j]+=sum[tot][j-1];           }           tot++;       }       int ans=0;       for(int i=1;i<=n;i++)       {           for(int j=i+1;j<=n;j++)           {               LL x=a[i]*a[j];               int t=K/__gcd(1LL*K,x);               int pos=lower_bound(b,b+tot,t)-b;               ans+=sum[pos][n]-sum[pos][j];           }       }       return ans;   }};// Powered by FileEdit// Powered by moj 4.12 [modified TZTester]// Powered by CodeProcessor