hdu3555Bomb(数位dp)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 18659    Accepted Submission(s): 6891

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

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU 

 

思路同hdu2089 http://www.cnblogs.com/L-Memory/p/7189656.html

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
long long f[20][20],dig[20];
long long ans,len,n,m,t;

void init()
{
    ans=0;len=0;
    memset(f,0,sizeof f);
    f[0][0]=1;
    for(long long i=1;i<=20;i++)
      for(long long j=0;j<10;j++)
        for(long long k=0;k<10;k++)
          if(!(j==4 && k==9)) f[i][j]+=f[i-1][k];
}

long long solve(long long x)
{
    while(x)
    {
        dig[++len]=x%10;
        x/=10;
    }dig[len+1]=0;
    for(long long i=len;i;i--)
    {
        for(long long j=0;j<dig[i];j++)
          if(!(dig[i+1]==4 && j==9)) ans+=f[i][j];
        if(dig[i]==9 && dig[i+1]==4) break;
    }return ans;
}

int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        init();
        cout<<n+1-solve(n+1)<<endl;
    }
    return 0;
}