hdu 3555 bomb 數位dp

題目大意：給定n個數x，求出1..x含有49的數的個數。

題目大意：數位dp，思路和hdu2089差不多。求出1..x不含49的數的個數，相減即可。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
#include<cassert>
#include<climits>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define INF (2139062143)
#define phiF (1000000006)
#define MAXN (1000000+10)
typedef long long LL;

LL dp[50][20],digit[100];

void calc(){
	dp[0][0]=1;
	For (i,28)
	 Rep (j,10) 
	  Rep (k,10){
			if (!(j==4&&k==9))
			   dp[i][j]+=dp[i-1][k];
	  }
}

LL work(LL x){
	int len=0;
	while (x){
		digit[++len]=x%10;
		x/=10;
	}
	digit[len+1]=0;
	LL ans=0;
	ForD(i,len){
		Rep(j,digit[i])
		if (!(j==9&&digit[i+1]==4))
		 ans+=dp[i][j];
		 
		if (digit[i]==9&&digit[i+1]==4) break; 
	}
	return ans;
}


int main(){
	calc();
	int n;
	scanf("%d",&n);
	while (n--){
		LL x;
		scanf("%lld",&x);
		printf("%lld\n",x-work(x+1)+1);
	}
	return 0;
}