HDOJ 3555 Bomb

数位DP的DFS写法。。。。

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4630    Accepted Submission(s): 1614

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

 

 

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; typedef  long  long  int LL; int bit[ 30]; LL dp[ 25][ 3]; LL dfs( int pos, int s, bool limit) {      if(pos==- 1)          return s== 2;      if(limit== false&&~dp[pos][s])          return dp[pos][s];      int end=limit?bit[pos]: 9;     LL ans= 0;      for( int i= 0;i<=end;i++)     {          int news=s;          if(s== 0&&i== 4) news= 1;          if(s== 1&&i!= 9) news= 0;          if(s== 1&&i== 9) news= 2;          if(s== 1&&i== 4) news= 1;         ans+=dfs(pos- 1,news,limit&&i==end);     }      if(!limit)          return dp[pos][s]=ans;      else          return ans; } int main() {     LL n;      int T;     memset(dp,- 1, sizeof(dp));     scanf( "%d",&T);      while(T--)     {         scanf( "%I64d",&n);          int len= 0;          while(n)         {             bit[len++]=n% 10;             n/= 10;         }         printf( "%I64d\n",dfs(len- 1, 0, 1));     }      return  0; }

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