The 37th ACM/ICPC Asia Regional Tianjin Site Online Contest - A.B.J

A
手动打表做出映射关系...然后八进制转十进制..

Program:

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000000000
#define pi acos(-1)
using namespace std;      
ll ans,n,m,k,a[10]={0,1,2,0,3,4,5,6,0,7};
int main()
{ 
  //   freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout); 
     while (~scanf("%I64d",&n))
     {    
           if (!n) break;
           m=n;  ans=0;  k=1;
           while (n)
           {
                 ans+=a[n%10]*k;
                 n/=10;
                 k*=8;
           }
           printf("%I64d: %I64d\n",m,ans); 
     }
     return 0;
}

B
这种题目的第一直觉就是找规律...打出前面一些sum...不难发现当k>12时..在0~k间的real number要么是k/2-1要么是k/2-2...那关键就在于寻找什么时候是-1,什么时候是-2...通过观察可以发现... 每逢平方数...-1,-2就会转变... 并且当 p^2 <= k < (p+1)^2 时..当p为奇数...0~k的real number为k/2-1...p为偶数...其为k/2-2....

能快速得到0~k间real number的个数..剩下的就简单了....

Program:

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000000000
#define pi acos(-1)
using namespace std;      
int t,n,s[20]={0,0,0,0,0,0,1,1,2,3,4,4,5};   
ll a,b,p1,p2;
ll getsum(ll x)
{
     if (x<=12) return s[x];
     ll m,k;
     m=x/2; 
     k=(ll)sqrt(x);
     if (k%2==0) m-=2;
        else m-=1;
     return m;
}
int main()
{  
    // freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout); 
     scanf("%d",&t);
     while (t--)
     { 
            scanf("%I64d%I64d",&a,&b);
            a--;
            p1=getsum(a);  
            p2=getsum(b);
            printf("%I64d\n",p2-p1);
     }
     return 0;
}

J
我是用trie树做的...直接用数组标记或者用map也能很和谐吧...

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000000000
#define pi acos(-1)
using namespace std;      
struct node
{
     int son[10],w;
}p[600005];
int ans,t,n,m,g,c[27]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};
char s[5005][10],str[10];
int main()
{ 
    // freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout); 
     scanf("%d",&t);
     while (t--)
     {    
             int i,h,k,len;
             scanf("%d%d",&n,&m);
             for (i=1;i<=n;i++) scanf("%s",s[i]);
             memset(p,0,sizeof(p));
             g=0;
             while (m--)
             {
                    scanf("%s",str);
                    len=strlen(str);
                    h=0;
                    for (i=0;i<len;i++)
                    {
                           if (!p[h].son[c[str[i]-'a']]) p[h].son[c[str[i]-'a']]=++g;
                           h=p[h].son[c[str[i]-'a']];
                    }
                    p[h].w++;
             }
             for (k=1;k<=n;k++)
             {
                    len=strlen(s[k]);
                    ans=h=0;
                    for (i=0;i<len;i++)
                    {
                           if (!p[h].son[s[k][i]-'0']) break;
                           h=p[h].son[s[k][i]-'0'];
                    }
                    if (i==len) ans=p[h].w;
                    printf("%d\n",ans);
             }
     }
     return 0;
}

1007的那个旅行商问题一直WA啊....找不到bug了....前几天刚做过的啊...用16*2^15就可以解决啊...好郁闷....求差错..求强力数据:

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000000000
#define pi acos(-1)
using namespace std;  
struct node
{
     ll t,c,d;
}p[17];
ll arc[105][105],n,m,money,goal,dp[40000][17];
void Floyd()
{
     ll k,i,j;
     for (k=1;k<=n;k++)
        for (i=1;i<=n;i++)
           for (j=1;j<=n;j++)
              if (arc[i][k]!=-1 && arc[k][j]!=-1)
                if (arc[i][j]==-1 || arc[i][j]>arc[i][k]+arc[k][j])
                    arc[i][j]=arc[i][k]+arc[k][j];
     return;
}
bool EXE_DP()
{
     ll i,j,ans,k,h,x,w,data;
     memset(dp,-1,sizeof(dp));
     p[0].t=1; p[0].c=p[0].d=0;
     w=1;
     dp[0][0]=money;
     for (k=1;k<=goal;k++)  
     {  
             h=k;  
             x=1;  
             i=0;   
             while (h)  
             {  
                   i++;  
                   if (h%2)  
                   {  
                          w=k-x;  
                          for (j=0;j<=n;j++)  
                            if (dp[w][j]!=-1 && arc[p[j].t][p[i].t]!=-1)
                              if (dp[w][j]-arc[p[j].t][p[i].t]>=p[i].d)
                              {
                                     data=dp[w][j]-arc[p[j].t][p[i].t]+p[i].c-p[i].d;
                                     if (dp[k][i]==-1 || dp[k][i]<data) dp[k][i]=data;  
                              }
                   }  
                   x*=2;  
                   h/=2;  
             }  
     }       
     for (i=0;i<=n;i++)
        if (arc[p[i].t][1]!=-1 && dp[goal][i]!=-1 && dp[goal][i]>=arc[p[i].t][1]) 
            return true;
     return false;
}
int main()
{  
     freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout); 
     ll t,x,y,h,w;
     scanf("%I64d",&t);
     while (t--)
     { 
           memset(arc,-1,sizeof(arc));
           scanf("%I64d%I64d%I64d",&n,&m,&money);
           while (m--)
           {
                  scanf("%I64d%I64d%I64d",&x,&y,&w);
                  if (arc[y][x]==-1 || w<arc[y][x]);
                      arc[x][y]=arc[y][x]=w;
           }
           for (x=1;x<=n;x++) arc[x][x]=0;
           Floyd();
           scanf("%I64d",&h);
           goal=1;
           for (x=1;x<=h;x++) 
           {
                  scanf("%I64d%I64d%I64d",&p[x].t,&p[x].c,&p[x].d); 
                  goal*=2;
           } 
           n=h;
           goal--; 
           if (EXE_DP()) printf("YES\n");
                    else printf("NO\n");
     }
     return 0;
}