Beautiful Palindrome Number

Time Limit: 3000/1500 MS ( Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 657 Accepted Submission(s): 369

Problem Description

A positive integer x can represent as 
  
   (a1a2…akak…a2a1)10
   or 
  
   (a1a2…ak?1akak?1…a2a1)10
   of a 10-based notational system, we always call x is a Palindrome Number. If it satisfies 
  
   0
   
    , we call x is a Beautiful Palindrome Number.
    
Now, we want to know how many Beautiful Palindrome Numbers are between 1 and 
   
  
  
   10N
  .

Input

The first line in the input file is an integer 
  
   T(1≤T≤7)
  , indicating the number of test cases.
Then T lines follow, each line represent an integer 
  
   N(0≤N≤6)
  .

Output

For each test case, output the number of Beautiful Palindrome Number.

Sample Input

2
1
6

Sample Output

9
258

Statistic | Submit | Clarifications | Back
代码：

#include
  
   
#include
   
     #include
    
      #include
     
       #include
       #include
       
         #include
        
          #include
         
           #include
          
            #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f using namespace std; const int maxn=1000000+10; char str[maxn]; bool judge(int k) { sprintf(str,"%d",k); int len=strlen(str); for(int i=0;i
           
            
 第二题我竟然直接向想暴力，简直too young too simple。。这题思路是先保存所有操作，然后最后询问的时候我们就逆推这些操作，最后得到要询问的数的下标，那么就引刃而解了，还有一个难点是怎么找出操作后的下标关系，估计这个应该是最难的。其实可以看出，比如1 2 3 4 5 6 7 8 ，那么经过fun1的变换后得到的序列是是1 3 5 7 2 4 6 8，那么可以看出后一半的下标在变换前是（当前位置-一半位置）*2，那么前一半的位置的原来坐标是（当前位置-1）*2+1,，那么逆序操作就好做多了，就是n-id-1，那么就简单了，相当于是离线的操作吧，又长见识了。。。
             
            还有不知道为嘛用int一直wa，longlong就过，不是取余了吗？？ 
            题目： 
             
             Operation the Sequence 
            
 Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
            
 Total Submission(s): 603 Accepted Submission(s): 102
            
 
            
 
            
 Problem Description 
            You have an array consisting of n integers: 
             
              a1=1,a2=2,a3=3,…,an=n
             . Then give you m operators, you should process all the operators in order. Each operator is one of four types:
Type1: O 1 call fun1();
Type2: O 2 call fun2();
Type3: O 3 call fun3();
Type4: Q i query current value of a[i], this operator will have at most 50.
Global Variables: a[1…n],b[1…n];
fun1() {
    index=1;
  for(i=1; i<=n; i +=2) 
    b[index++]=a[i];
  for(i=2; i<=n; i +=2)
    b[index++]=a[i];
  for(i=1; i<=n; ++i)
    a[i]=b[i];
}
fun2() {
  L = 1;R = n;
  while(L
             
                  Swap(a[L], a[R]); 
              
    ++L;--R;
              
  }
              
}
              
fun3() {
              
  for(i=1; i<=n; ++i) 
              
    a[i]=a[i]*a[i];
              
}
              Input 
            The first line in the input file is an integer 
             
              T(1≤T≤20)
             , indicating the number of test cases.
The first line of each test case contains two integer 
             
              n(0
              
               , 
              
             
             
              m(0
              
               .
               
Then m lines follow, each line represent an operator above.
              
             
            
              Output 
             For each test case, output the query values, the values may be so large, you just output the values mod 1000000007(1e9+7). Sample Input 
             1
3 5
O 1
O 2
Q 1
O 3
Q 1 Sample Output 
             2
4 
             
               Statistic | Submit | Clarifications | Back
              
             
 代码： 
              
             
             #include
              
               
#include
               
                 #include
                
                  #include
                 
                   #include
                   #include
                   
                     #include
                    
                      #include
                     
                       #include
                      
                        #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f #define mod 1000000007 using namespace std; const int maxn=100000+10; int t,n,m,cnt,b,op[maxn]; ll a[maxn]; ll solve(int cnt,int val) { int half=(n+1)/2; ll mul=0; for(int i=cnt;i>=1;i--) { if(op[i]==1) { if(val>half) val=(val-half)*2; else val=(val-1)*2+1; } else if(op[i]==2) val=n-val+1; else mul++; } ll ans=a[val]; for(int i=1;i<=mul;i++) ans=ans*ans%mod; return ans; } int main() { char s[2]; scanf("%d",&t); while(t--) { cnt=0; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) a[i]=i; while(m--) { scanf("%s%d",s,&b); if(s[0]=='O') op[++cnt]=b; else { ll ans=solve(cnt,b); printf("%I64d\n",ans); } } } return 0; }


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