HDU 3555 数位dp

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15072 Accepted Submission(s): 5441

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

题意：求1~n闭区间内含有“49”的数的个数

题解：

dp[i][2] 长度为i 含有“49”的个数

dp[i][1] 长度为i 不含有“49”但是高位为“9”的个数

dp[i][0] 长度为i 不含有“49”的个数

数组 a[i] 从低位到高位存储 n 的每一位数字。

dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; //考虑第i位为“4” i-1位为“9”

dp[i][1]=dp[i-1][0];

dp[i][0]=dp[i-1][0]*10-dp[i-1][1];

对于n处理之前为什么要自增1

因为题目要求处理的是闭区间可能自增1当作开区间处理

http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<queue>
14 #define ll __int64
15 using namespace std;
16 int t;
17 ll n;
18 ll a[65];
19 ll dp[65][5];
20 void init()
21 {
22     dp[0][0]=1;
23     for(int i=1; i<=22; i++)
24     {
25         dp[i][0]=10*dp[i-1][0]-dp[i-1][1];
26         dp[i][1]=dp[i-1][0];
27         dp[i][2]=10*dp[i-1][2]+dp[i-1][1];
28     }
29 }
30 int main()
31 {
32     init();
33     while(scanf("%d",&t)!=EOF)
34     {
35         for(int i=1; i<=t; i++)
36         {
37             scanf("%I64d",&n);
38             memset(a,0,sizeof(a));
39             int len=1;
40             n++;
41             while(n)
42             {
43                 a[len]=n%10;
44                 n=n/10;
45                 len++;
46             }
47             int flag=0;
48             int last=0;
49             ll ans=0;
50             for(int j=len; j>=1; j--)
51             {
52                 ans+=dp[j-1][2]*a[j];
53                 if(flag)
54                     ans+=dp[j-1][0]*a[j];
55                 if(!flag&&a[j]>4)
56                     ans+=dp[j-1][1];
57                 if(last==4&&a[j]==9)
58                     flag=1;
59                 last=a[j];
60             }
61             printf("%I64d
",ans);
62         }
63     }
64     return 0;
65 }