Codeforces Round #212 (Div. 2) A. Two Semiknights Meet /B. Petya and Staircases

A题：A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.

题解：这就是英文理解题了。。。在8*8的格子里面有两个K，他们只能向上走2次并且左或者右两次，或者向下走两次并且左或者右两次，'#'表示不可走。问这两个K都要走，能不能相遇，能则输出‘YES’。

我们只用考虑这两个k的坐标关系就可以了。假设两个K的x坐标x1,x2相差为2或者1，那么怎么走x坐标都没法相等。推导相差4或者4的倍数就可以了。

A题：

#include<cstdlib>#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;int main(){    int T,x1,x2,y1,y2;    char c;    cin>>T;        while(T--)        {            bool flag = false;            for(int i=0;i<8;i++)                for(int j=0;j<8;j++)                {                    cin>>c;                    if(c=='K')                    {                        if(!flag)                        {                            x1=i;                            y1=j;                            flag= true;                        }                        else                        {                            x2=i,y2=j;                        }                    }                }            if(abs(x1-x2)%4==0&&abs(y1-y2)%4==0)                cout<<"YES"<<endl;            else                cout<<"NO"<<endl;        }}

B题：理解题意就可以做了。这个人跳台阶，第一个台阶和最后一个台阶都不能是脏的，并且连续3个台阶是脏的也GAME OVER。。

#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<algorithm>using namespace std;int a[3003];int n,m;int main(){    while(cin>>n>>m)    {        bool ans= false ;        for(int i=1;i<=m;i++)        {            cin>>a[i];        }        sort(a+1,a+1+m);        if(a[1]==1||a[m]==n)        {            cout<<"NO"<<endl;            continue;        }        int tmp=a[1];        int cnt=0;        for(int i=2;i<=m;i++)        {            if(a[i]-tmp==1)            {                cnt++;                tmp=a[i];            }            else if(a[i]-tmp>1)            {                cnt=0;                tmp=a[i];            }            if(cnt>=2)            {                ans=true;                break;            }        }        if(ans)            cout<<"NO"<<endl;        else            cout<<"YES"<<endl;    }}