HDU 3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5247    Accepted Submission(s): 1815

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

 

 

题意：求1--N，包含49的个数，注意：数字4949只算一次。

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 typedef __int64 LL;

 5 LL dp[21][11];

 6 void prepare()

 7 {

 8     LL i,j,cur,s;

 9     memset(dp,0,sizeof(dp));

10     for(i=4;i<=9;i++)   dp[2][i]=1;

11     for(i=3;i<=20;i++)

12     {

13         for(j=0;j<=9;j++)

14         {

15             if(j==0)

16                 dp[i][j]+=dp[i-1][9];

17             else

18             {

19                 dp[i][j]=dp[i][j]+dp[i][j-1];

20                 if(j==4)

21                 {

22                     dp[i][j]=dp[i][j]+dp[i-1][8];

23                     for(cur=1,s=i-2;s>0;s--)

24                         cur=cur*10;

25                     dp[i][j]=dp[i][j]+cur;

26                 }

27                 else dp[i][j]=dp[i][j]+dp[i-1][9];

28             }

29         }

30     }

31 }

32 int main()

33 {

34     LL T;

35     LL n,i,j,k,tom,cur,glag;

36     char a[30];

37     prepare();

38     scanf("%I64d",&T);

39     while(T--)

40     {

41         scanf("%s",a+1);

42         k=strlen(a+1);

43         n=k;

44         for(i=1,tom=0,glag=0;i<=k;i++)

45         {

46             cur=(a[i]-'0')-1;

47             if(cur==-1);

48             else tom=tom+dp[n][cur];

49             if(glag==2) glag--;

50             if(glag==1) break;

51             if(a[i]=='4' && i+1<=k && a[i+1]=='9')

52             {

53                 glag=2;

54                 for(j=i+2,cur=0;j<=k;j++)

55                 {

56                     cur=cur*10+(a[j]-'0');

57                 }

58                 tom=tom+cur+1;

59             }

60             n--;

61         }

62         printf("%I64d\n",tom);

63     }

64     return 0;

65 }