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

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】

 1 #include<cstdio>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 LL f[25][3];
 6 /*
 7      f[i][0] i位数 无49存在
 8     f[i][1] i位数 无49存在且末尾为9
 9     f[i][2] i位数 有49
10 */
11 int b[25];
12 
13 void init() {
14     f[0][0]=1; f[0][1]=f[0][2]=0;
15     for(int i=1;i<25;i++) {
16         f[i][0]=10*f[i-1][0]-f[i-1][1];
17         f[i][1]=f[i-1][0];
18         f[i][2]=10*f[i-1][2]+f[i-1][1];
19     }
20 }
21 
22 int main() {
23     int T; LL n;
24     scanf("%d",&T);
25     init();
26     while(T--) {
27         scanf("%I64d",&n);
28         int len=0;
29         while(n)
30             b[++len]=n%10 , n/=10;
31         b[len+1]=0;
32         LL ans=0; bool flag=0;
33         for(int i=len;i;i--) {
34             ans += b[i]*f[i-1][2];                    // + ...49... 
35             if(flag) ans += f[i-1][0]*b[i];            // + 49...(无49...)
36             else if(b[i]>4) ans += f[i-1][1];        // + 4(9 无49)
37             if(b[i+1]==4 && b[i]==9) flag=1;
38         }
39         if(flag) ans++;                                //算上自身
40         printf("%I64d
",ans);
41     }
42     return 0;
43 }