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:

[cpp] view plain copy print ?

1. #include<iostream> 
2. #include<stdio.h> 
3. #include<algorithm> 
4. #include<string.h> 
5. #include<math.h> 
6. #include<map> 
7. #include<queue> 
8. #include<stack> 
9. #define ll long long 
10. #define oo 1000000000 
11. #define pi acos(-1) 
12. using namespace std;       
13. ll ans,n,m,k,a[10]={0,1,2,0,3,4,5,6,0,7}; 
14. int main() 
15. {  
16.   //   freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);  
17.      while (~scanf("%I64d",&n)) 
18.      {     
19.            if (!n) break; 
20.            m=n;  ans=0;  k=1; 
21.            while (n) 
22.            { 
23.                  ans+=a[n%10]*k; 
24.                  n/=10; 
25.                  k*=8; 
26.            } 
27.            printf("%I64d: %I64d\n",m,ans);  
28.      } 
29.      return 0; 
30. } 

#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:

[cpp] view plain copy print ?

1. #include<iostream> 
2. #include<stdio.h> 
3. #include<algorithm> 
4. #include<string.h> 
5. #include<math.h> 
6. #include<map> 
7. #include<queue> 
8. #include<stack> 
9. #define ll long long 
10. #define oo 1000000000 
11. #define pi acos(-1) 
12. using namespace std;       
13. int t,n,s[20]={0,0,0,0,0,0,1,1,2,3,4,4,5};    
14. ll a,b,p1,p2; 
15. ll getsum(ll x) 
16. { 
17.      if (x<=12) return s[x]; 
18.      ll m,k; 
19.      m=x/2;  
20.      k=(ll)sqrt(x); 
21.      if (k%2==0) m-=2; 
22.         else m-=1; 
23.      return m; 
24. } 
25. int main() 
26. {   
27.     // freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);  
28.      scanf("%d",&t); 
29.      while (t--) 
30.      {  
31.             scanf("%I64d%I64d",&a,&b); 
32.             a--; 
33.             p1=getsum(a);   
34.             p2=getsum(b); 
35.             printf("%I64d\n",p2-p1); 
36.      } 
37.      return 0; 
38. } 

#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...

[cpp] view plain copy print ?

1. #include<iostream> 
2. #include<stdio.h> 
3. #include<algorithm> 
4. #include<string.h> 
5. #include<math.h> 
6. #include<map> 
7. #include<queue> 
8. #include<stack> 
9. #define ll long long 
10. #define oo 1000000000 
11. #define pi acos(-1) 
12. using namespace std;       
13. struct node 
14. { 
15.      int son[10],w; 
16. }p[600005]; 
17. 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}; 
18. char s[5005][10],str[10]; 
19. int main() 
20. {  
21.     // freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);  
22.      scanf("%d",&t); 
23.      while (t--) 
24.      {     
25.              int i,h,k,len; 
26.              scanf("%d%d",&n,&m); 
27.              for (i=1;i<=n;i++) scanf("%s",s[i]); 
28.              memset(p,0,sizeof(p)); 
29.              g=0; 
30.              while (m--) 
31.              { 
32.                     scanf("%s",str); 
33.                     len=strlen(str); 
34.                     h=0; 
35.                     for (i=0;i<len;i++) 
36.                     { 
37.                            if (!p[h].son[c[str[i]-'a']]) p[h].son[c[str[i]-'a']]=++g; 
38.                            h=p[h].son[c[str[i]-'a']]; 
39.                     } 
40.                     p[h].w++; 
41.              } 
42.              for (k=1;k<=n;k++) 
43.              { 
44.                     len=strlen(s[k]); 
45.                     ans=h=0; 
46.                     for (i=0;i<len;i++) 
47.                     { 
48.                            if (!p[h].son[s[k][i]-'0']) break; 
49.                            h=p[h].son[s[k][i]-'0']; 
50.                     } 
51.                     if (i==len) ans=p[h].w; 
52.                     printf("%d\n",ans); 
53.              } 
54.      } 
55.      return 0; 
56. }