O-Bomb(数位dp)

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15062 Accepted Submission(s): 5437

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

500

Sample Output

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.

//题意 t 组数据，在 1- n 中出现了49的数据个数，n 很大，2^63-1 , 遍历肯定不行。

//听说这是数位dp的一道水题，这也是水题吗，我服了

自己想了一阵，只想到了一点点。看了别人的也看了很久，才明白，这篇博客非常详细，我就不赘述了，认真看一定能看懂。

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

我的与他的dp数组有细微差别，因为我自己理解了写了一遍，稍微注意一下。

代码 15ms

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 __int64 dp[25][3];
 5 //dp[i][0]   含49，数的个数
 6 //dp[i][1] 不含49，但首位是9的数的个数
 7 //dp[i][2] 不含49，数的个数(包含了首位是9的数的个数)
 8 void Init()
 9 {
10     dp[0][2]=1;
11     for (int i=1;i<=24;i++)
12     {
13         dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
14         dp[i][1]=dp[i-1][2];
15         dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
16     }
17 }
18 
19 __int64 solve(__int64 n)
20 {
21     __int64 a[25],len=0;
22 
23     while (n)
24     {
25         a[++len]=n%10;
26         n/=10;
27     }
28     a[len+1]=0;
29 
30     __int64 ans=0;
31     int flag=0;
32 
33     for (int i=len;i>0;i--)
34     {
35         ans+=dp[i-1][0]*a[i];
36         if (flag)           //前面有49就所有情况都是49数了
37             ans+=dp[i-1][2]*a[i];
38         if (!flag&&a[i]>4)  //加上4  9... 的情况
39             ans+=dp[i-1][1];
40         if (a[i]==9&&a[i+1]==4)//判断是不是 49 
41             flag=1;
42     }
43     return ans;
44 }
45 
46 int main()
47 {
48     int t;
49     __int64 n;
50     Init();
51     scanf("%d",&t);
52     while (t--)
53     {
54         scanf("%I64d",&n);
55         printf("%I64d
",solve(n+1));//因为函数功能是 (0,n) 区间的49数，题目要求的是[1,n]，所以+1
56     }
57     return 0;
58 }

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