动态规划晋级——HDU 3555 Bomb【数位DP详解】

 

hdu3555 Bomb

2012-07-12 19:37  372人阅读  评论(1)  收藏  举报

integer input output training numbers each

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1604    Accepted Submission(s): 592

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

 

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

 

Output

For each test case, output an integer indicating the final points of the power.

 

Sample Input

    
    
    
    

     
     
     
     3
1
50
500
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     0
1
15


     
     
     
     
 
      
 
       Hint 
      
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15. 
     
     
     
     
      
    
    
    
    

 

Author

fatboy_cw@WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

Recommend

zhouzeyong

顺便写道数位dp的题，直接dfs下去就可以了，没什么说的。

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;

 int dig[20];
 long long dp[20][2][12];

 long long F(int pos,bool have,int last,bool inf)
 {
     int i;
     if (pos==-1) return have;
     if (!inf && dp[pos][have][last]!=-1) return dp[pos][have][last];
     int end=inf?dig[pos]:9;
     long long ans=0;
     for (i=0;i<=end;i++)
     {
         if (last==4 && i==9) ans+=F(pos-1,true,i,inf && (i==end));
         else ans+=F(pos-1,have,i,inf && (i==end));
     }
     if (!inf)
     {
         dp[pos][have][last]=ans;
     }
     return ans;
 }

 long long Cal(long long t)
 {
     int i,j,pos=0;
     while(t)
     {
         dig[pos++]=t%10;
         t/=10;
     }
     return F(pos-1,0,0,1);

 }

 int main()
 {
     int i,j,T;
     __int64 n;
     scanf("%d",&T);
     while(T--)
     {
         memset(dp,-1,sizeof(dp));
         scanf("%I64d",&n);
         printf("%I64d\n",Cal(n));
     }
     return 0;
 }