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

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.

 

 

//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

 

 #include <stdio.h>
 #include <string.h>

 __int64 dp[][];
 //dp[i][0]   含49，数的个数
 //dp[i][1] 不含49，但首位是9的数的个数
 //dp[i][2] 不含49，数的个数(包含了首位是9的数的个数)
 void Init()
 {
     dp[][]=;
     for (int i=;i<=;i++)
     {
         dp[i][]=dp[i-][]*+dp[i-][];
         dp[i][]=dp[i-][];
         dp[i][]=dp[i-][]*-dp[i-][];
     }
 }

 __int64 solve(__int64 n)
 {
     __int64 a[],len=;

     while (n)
     {
         a[++len]=n%;
         n/=;
     }
     a[len+]=;

     __int64 ans=;
     int flag=;

     for (int i=len;i>;i--)
     {
         ans+=dp[i-][]*a[i];
         if (flag)           //前面有49就所有情况都是49数了
             ans+=dp[i-][]*a[i];
         if (!flag&&a[i]>)  //加上4  9... 的情况
             ans+=dp[i-][];
         if (a[i]==&&a[i+]==)//判断是不是 49
             flag=;
     }
     return ans;
 }

 int main()
 {
     int t;
     __int64 n;
     Init();
     scanf("%d",&t);
     while (t--)
     {
         scanf("%I64d",&n);
         printf("%I64d\n",solve(n+));//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
     }
     return ;
 }