"Digital DP" hdu3555 Bomb

Test instructions is looking for 0 to n how many of the numbers contain 49. Data range close to 10^20

DP State is 2-D dp[len][3]
Dp[len][0] represents the number of scenarios with a length of Len without 49
DP[LEN][1] represents the number of scenarios with a length of Len without 49 but with a number beginning with 9
DP[LEN][2] For a length of Len containing 49 of the scheme number

State transitions are as follows
Dp[i][0] = dp[i-1][0] * 10-dp[i-1][1]; Not include 49 if it does not contain 49 and can be filled in 0-9 but minus dp[i-1][1] because 4 will and 9 constitute 49
DP[I][1] = dp[i-1][0]; Not including starts with 9 This is just a 9 on the number that doesn't include 49.
DP[I][2] = dp[i-1][2] * ten + dp[i-1][1]; Include 49 already contains 49 of the number can be filled 0-9, or 9 the beginning of the fill 4

And then you start counting from high.

When counting to a bit, add dp[i-1][2] * Digit[i] is obviously right, because this one can fill 0-(digit[i]-1)
If this person is next to 49, then add dp[i-1][0] * Digit[i] is also clearly right.
If this one is not next to 49, but Digit[i] is bigger than 4, then when this one fills 4, you have to add dp[i-1][1]

--http://www.cnblogs.com/jackge/archive/2013/05/15/3080958.html

#include <cstdio>using namespace Std;typedef long long ll;ll f[21][3],n;void init () {f[0][0]=1;for (int i=1;i<= 20;++i)  {  f[i][0]=f[i-1][0]*10-f[i-1][1];  F[I][1]=F[I-1][0];  F[I][2]=F[I-1][2]*10+F[I-1][1];}  } int T;int Main () {init ();//freopen ("Hdu3555.in", "R", stdin); scanf ("%d", &t); int bit[23];for (; T;--T)  {  int m=0; ll ans=0; bool flag=0;  scanf ("%i64d", &n); ++n;  while (n)    {    bit[++m]=n%10;    n/=10;    }  bit[m+1]=bit[m+2]=0;  for (int i=m;i;--i)    {    if (bit[i+1]==9 && bit[i+2]==4)      flag=1;    ans+=f[i-1][2]* (LL) bit[i];    if (flag)      ans+= (LL) bit[i]*f[i-1][0];    else if (Bit[i] > 4)      ans+=f[i-1][1];    }  printf ("%i64d\n", ans);  } return 0;}

"Digital DP" hdu3555 Bomb