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。

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 
         
           2 #include 
          
            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\n",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+156     }57     return 0;58 }