hacker cup 2015 资格赛

A：交换一个数的两位，问能得到最大和最小的数是多少。

水题：

// File Name: a.cpp
// Author: darkdream
// Created Time: 2015年01月10日 星期六 17时16分44秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;
LL a[20];
int T;
LL SP(LL i , LL j, LL tmp)
{
     int x =(tmp)/a[i] % 10 ; 
     int y =(tmp)/a[j] % 10 ;
     if(x == 0 && j == T)
         return tmp;
     return tmp - x * a[i] - y*a[j] + y *a[i] + x*a[j]; 
}
int count(LL tmp)
{
  int num = 0 ; 
   while(tmp)
   {
     num ++;
     tmp/= 10;
   }
   return num ;
}
int main(){
   int t ;
   scanf("%d",&t);
   a[1] = 1; 
   for(int i = 2;i <= 13 ;i++)
   {
     a[i] = a[i-1]*10;
   }
   for(int ca = 1 ; ca <= t ;ca ++)
   {
      LL tmp;
      scanf("%lld",&tmp);
      LL mx = tmp;
      LL mi = tmp;
      T = count(tmp);
      for(int i = 1;i <= T;i ++)
          for(int j = i + 1;j <= T ;j ++)
          {
            LL now = SP(i,j,tmp);
            if(now > mx)
                mx = now;
            if(now < mi)
                mi = now;
          }
      printf("Case #%d: %lld %lld
",ca,mi,mx);
   }

return 0;
}

B：给你n件物品，每个物品都有三种属性值，问从中选取任意件，能不能使得最后得到确定值

解题思路：二进制枚举

解题代码：

// File Name: b.cpp
// Author: darkdream
// Created Time: 2015年01月10日 星期六 17时55分37秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;
int Ga,Gb,Gc;
int a[30],b[30],c[30];
bool solve(int k)
{
   int t= 0 ; 
   int ta,tb,tc;
   ta = tb = tc = 0 ; 
   while(k)
   {
      if(k % 2)
      {
        ta += a[t];
        tb += b[t];
        tc += c[t];
      }
      k /= 2;
      t ++ ; 
   }
  //<F5> printf("%d %d %d
",ta,tb,tc);
   if(ta == Ga && tb == Gb&& tc == Gc)
   {
       return  1;
   }
   return 0 ; 
}
int main(){
  int T;
  scanf("%d",&T);
  for(int ca = 1;ca <= T; ca ++)
  {
      scanf("%d %d %d",&Ga,&Gb,&Gc);
      int n;
      scanf("%d",&n);
      for(int i= 0;i < n;i++)
      {
        scanf("%d %d %d",&a[i],&b[i],&c[i]);
      }
      int total = (1 << n)-1;
      int ok = 0 ; 
      for(int i =0 ;i <= total;i ++)
      {
         if(solve(i))
         {
           ok = 1; 
           break;
         }
      }
      if(ok)
        printf("Case #%d: yes
",ca);
      else printf("Case #%d: no
",ca);
  }
return 0;
}

C：给你一个网格图，其中每个网格中有墙，旋转的激光炮台，激光炮每秒旋转一次，问你最后从起点到终点最少能够几步到达

解题思路：图只有4种情况，所以只需要把四种情况全部打表出来，没走一步状态就跟着变化就行

解题代码：

// File Name: c.cpp
// Author: darkdream
// Created Time: 2015年01月10日 星期六 20时10分04秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;
int mp[5][105][105];
int bex , bey ; 
int enx,  eny ;
int n , m; 
int vis[105][105][105];
void fillmap(int k )
{
    for(int i = 1;i <= n;i ++)
    {
       for(int j = 1;j <= m ;j ++)
       {
          if(mp[k][i][j]  == 2) 
          {
            int x = i + 1; 
            while(mp[k][x][j] == 1 || mp[k][x][j] == -1)
            {
                   mp[k][x][j] = -1;
                x ++ ; 
            }
          } else if(mp[k][i][j] == 3)
          {
            int x = i - 1; 
            while(mp[k][x][j] == 1 || mp[k][x][j] == -1)
            {
                   mp[k][x][j] = -1;
                x -- ; 
            }
          } else if(mp[k][i][j] == 4)
          {
            int x = j + 1; 
            while(mp[k][i][x] == 1 || mp[k][i][x] == -1)
            {
                   mp[k][i][x] = -1;
                x ++; 
            }
          }else if(mp[k][i][j] == 5)
          {
             int x = j -1; 
             while(mp[k][i][x] == 1 || mp[k][i][x] == -1)
             {
                mp[k][i][x] = -1;
                x --;
             }
          }
       }
    }
}
void print(int k )
{
   for(int i = 1;i <= n;i ++)
   {
       for(int j = 1;j <= m ;j ++)
       {
          printf("%d ",mp[k][i][j]);
       }
       printf("
");
   }
   printf("
");
}
void changeto(int k )
{
    for(int i = 1;i <= n;i ++)
    {
       for(int j = 1;j <= m;j ++)
       {
           if(mp[k-1][i][j] == 0)    
               mp[k][i][j] == 0;
           else if(mp[k-1][i][j] == 1)
               mp[k][i][j] = 1;
           else if(mp[k-1][i][j] == 2)
               mp[k][i][j] = 5;
           else if(mp[k-1][i][j] == 3)
               mp[k][i][j] = 4;
           else if(mp[k-1][i][j] == 4)
               mp[k][i][j] = 2;
           else if(mp[k-1][i][j] == 5)
               mp[k][i][j] = 3;
           else if(mp[k-1][i][j] == -1)
               mp[k][i][j] = 1;
       }
    }
    fillmap(k);
    //print(k);
}
struct node{
  int x,y,step,flag;
  node(){
  
  }
  node(int _x ,int _y ,int _step ,int _flag)
  {
     x = _x; 
     y = _y;
     step = _step;
     flag = _flag; 
  }
}qu[101000];
int xadd[] = {1,-1,0,0};
int yadd[] = {0,0,-1,1};
int bfs()
{
    memset(vis,0,sizeof(vis));
    qu[1] = node(bex,bey,0,0);
    vis[0][bex][bey] = 1; 
    int l = 1; 
    int r = 1;
    while(l <= r)
    {
      // printf("%d %d
",qu[l].x,qu[l].y);
       if(qu[l].x == enx && qu[l].y == eny)
       {
             return  qu[l].step;
       }
       int k = (qu[l].flag + 1)%4; 
       for(int i = 0;i <= 3; i ++)
       {
           int tx = qu[l].x + xadd[i];
           int ty = qu[l].y + yadd[i];
           if(mp[k][tx][ty] == 1 &&  vis[k][tx][ty] != 1)
           {
              vis[k][tx][ty] = 1;
              r ++ ; 
              qu[r] = node(tx,ty,qu[l].step + 1, k);
           }
       }
       l ++ ; 
    }    
    return 0 ;
}
int main(){
    freopen("in.txt","r",stdin);
    freopen("out","w",stdout);
    int T;
    scanf("%d",&T);
    for(int ca = 1; ca <= T; ca ++)
    {
       scanf("%d %d",&n,&m);
       char str[1000];
       memset(mp,0,sizeof(mp));
       for(int i = 1;i <= n;i ++)
       {
          scanf("%s",&str[1]);
          for(int j = 1;j <= m;j ++)
          {
              if(str[j] == 'S') 
              {
                  bex = i ; 
                  bey = j ; 
                  mp[0][i][j] = 1;
              }else if(str[j] == 'G')
              {
                  mp[0][i][j] = 1;
                  enx = i ; 
                  eny = j;
              }else if(str[j] == '#')
              {
                  mp[0][i][j] = 0 ; 
              }else if(str[j] ==  '.')
                  mp[0][i][j] = 1 ;
              else if(str[j] == 'v')
                  mp[0][i][j] = 2; 
              else if(str[j] == '^')
                  mp[0][i][j] =3;
              else if(str[j] == '>')
                  mp[0][i][j] =4 ; 
              else if(str[j] == '<')
                  mp[0][i][j] =5;
          }
       }
          
          fillmap(0);
          changeto(1);
          changeto(2);
          changeto(3);
          //for(int i = 0 ;i <= 3;i ++ )
            //  print(i);
          int ok = bfs();
          printf("Case #%d: ",ca);
          if(ok == 0 )
              printf("impossible
");
          else printf("%d
",ok);
    }
return 0;
}