HDU 3555 Bomb

http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7763    Accepted Submission(s): 2717

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.

思路：http://blog.csdn.net/winkloud/article/details/7866520

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
__int64 dp[25][3];
int bit[25];
void init()
{
    memset(dp,0,sizeof(dp));
    int i;
    dp[0][0]=1;
    for(i=1;i<=20;i++)
      {
          dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
          //dp[i][0] 表示i位数字中不含49的数字的个数  
          dp[i][1]=dp[i-1][0];
          //dp[i][1] 表示i位数字中以9开头的数字的个数  
          dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
          ////dp[i][2] 表示i位数字中含有49的数字的个数 
    }
}
__int64 solve(__int64 n)
{
  memset(bit,0,sizeof(bit));
    int len=1,i;
    __int64 temp=n;
    while(temp)
    {
        int r=temp%10;
        bit[len++]=r;
        temp=temp/10;
    }
    int flag=0;
    __int64 ans=0;
    bit[len]=0;
    for(i=len-1;i>=1;i--)
    {
         ans+=dp[i-1][2]*bit[i];
         if(flag)
          {
              ans+=dp[i-1][0]*bit[i];
          }
        //if(!flag&&bit[i+1]==4&&bit[i]>=9)
        //{
        //    ans+=dp[i][1];

       // }
        if(!flag&&bit[i]>4)
             {
                ans+=dp[i-1][1];
             }
        if(bit[i+1]==4&&bit[i]==9)
             flag=1;
      }
     return ans;

}
int main()
{
    int T;
    __int64 r;
    scanf("%d",&T);
   init();
   while(T--)
   {
       scanf("%I64d",&r);

       printf("%I64d\n",solve(r+1));
   }
   return 0;
}