TC SRM 400 DIV2

250PT:从(0,0)出发，出租车的速度是一定的，步行的速度也是固定的，所以最快的肯定是全部步行，或者步行至一个出租车处，然后坐车。
不会坐两辆出租车的。枚举所有出租车即可。
[cpp]?
class GrabbingTaxi{?
public:?
??? int minTime(vector tXs, vector tYs, int gX, int gY, int walkTime, int taxiTime){?
??????? int ans=(abs(gX)+abs(gY))*walkTime;?
??????? for(int i=0;i ??????????? ans=min(ans,(abs(tXs[i])+abs(tYs[i]))*walkTime+(abs(gX-tXs[i])+abs(gY-tYs[i]))*taxiTime);?
??????? return ans;?
??? }?
};?

500Pt：形如p^q的如果p为素数，q大于1，称为XX数。问n是否为XX数，找出p,q。
由于 n有一定的范围，最多不超过2^60,也就是q是肯定小于60的，便可以枚举q判断是否满足条件
[cpp]?
class StrongPrimePower{?
public:?
??? bool isprime(LL num){?
??????? if(num==2)?
??????????? return true;?
??????? if(!(num&1))?
??????????? return false;?
??????? for(int i=3;i<=(int)sqrt(num*1.0)+1;i++)?
??????????? if(num%i==0)?
??????????????? return false;?
??????? return true;?
??? }?
??? LL Pow(LL? a,int b){?
??????? return b==0?1:a*Pow(a,b-1);?
??? }?
??? vector baseAndExponent(string n){?
??????? LL num;?
??????? sscanf(n.c_str(),"%lld",&num);?
??????? vectorans;?
??????? for(int i=2;i<=60;i++){?
??????????? LL p=(int)(eps+pow(num,1.0/i));?
??????????? if(Pow(p,i)==num&&isprime(p)){?
??????????????? ans.push_back((int)p);?
??????????????? ans.push_back(i);?
??????????????? return ans;?
??????????? }?
??????? }?
??????? return ans;?
??? }?
};?

1000pt：跪烂的题，开关问题，最多是8*8，其实可以枚举第一行(2^8)，然后再判断是否满足即可。我偏来个高斯消元，结果发现有多解，然后就枚举自由变元，结果第一组数据就越界，猜测是栈溢出？？？？以下的代码是只能判断无解和唯一解的。
[cpp]?
int way[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,-1},{-1,1}};?
class LightedPanels{?
public:?
??? int a[70][70],n,row,col;?
??? void debug(){?
??????? for(int i=0;i ??????????? for(int j=0;j<=n;j++)?
??????????????? printf("%d ",a[i][j]);?
??????????? printf("\n");?
??????? }?
??? }?
??? int gauss(){?
??????? int i,j;?
??????? for(i=0,j=0;i ??????????? int k=i;?
??????????? for(;k ??????????????? if(a[k][j])?
??????????????????? break;?
??????????? if(a[k][j]){?
??????????????? for(int r=0;r<=n;r++)?
??????????????????? swap(a[i][r],a[k][r]);?
??????????????? for(int r=0;r ??????????????????? if(r!=i&&a[r][j])?
??????????????????????? for(k=0;k<=n;k++)?
??????????????????????????? a[r][k]^=a[i][k];?
??????????????? i++;?
??????????? }?
??????????? debug();?
??????????? cout< ??????? }?
??????? for(j=i;j ??????????? if(a[j][n])?
??????????????? return -1;?
??????? int ans=0;?
??????? for(int i=0;i ??????????? if(a[i][n])?
??????????????? ans++;?
??????? return ans;?
??? }?? www.2cto.com
??? int minTouch(vector board){?
??????? row=board.size();col=board[0].size();?
??????? n=row*col;?
??????? memset(a,0,sizeof(a));?
??????? for(int i=0;i ??????????? for(int j=0;j ??????????????? a[i*col+j][i*col+j]=1;?
??????????????? if(board[i][j]=='.')?
??????????????????? a[i*col+j][n]=1;?
??????????????? for(int k=0;k<8;k++){?
??????????????????? int x=i+way[k][0],y=j+way[k][1];?
??????????????????? if(x>=0&&y>=0&&x ??????????????????????? a[i*col+j][x*col+y]=1;?
??????????????? }?
??????????? }?
??????? return gauss();?
??? }?
};?