HDU 3555 Bomb（数位DP）

　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　Bomb

　　　　　　　　　　　　　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　Total Submission(s): 11804    Accepted Submission(s): 4212

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

【思路】

数位DP。

　　 预处理：

f[i][0] 表示i位数中不含49的数的数目

f[i][1] 表示i位数中不含49且最高位为9的数的数目

f[i][2] 表示i位数中含49的数的数目

根据n的每一位统计ans，具体见代码。

【代码】

 #include<cstdio>
 using namespace std;
 
 typedef long long LL;
 LL f[][];
 /*
      f[i][0] i位数 无49存在
     f[i][1] i位数 无49存在且末尾为9
     f[i][2] i位数 有49
 */
 int b[];
 
 void init() {
     f[][]=; f[][]=f[][]=;
     for(int i=;i<;i++) {
         f[i][]=*f[i-][]-f[i-][];
         f[i][]=f[i-][];
         f[i][]=*f[i-][]+f[i-][];
     }
 }
 
 int main() {
     int T; LL n;
     scanf("%d",&T);
     init();
     while(T--) {
         scanf("%I64d",&n);
         int len=;
         while(n)
             b[++len]=n% , n/=;
         b[len+]=;
         LL ans=; bool flag=;
         for(int i=len;i;i--) {
             ans += b[i]*f[i-][];                    // + ...49...
             if(flag) ans += f[i-][]*b[i];            // + 49...(无49...)
             else if(b[i]>) ans += f[i-][];        // + 4(9 无49)
             if(b[i+]== && b[i]==) flag=;
         }
         if(flag) ans++;                                //算上自身
         printf("%I64d\n",ans);
     }
     return ;
 }