hdu 3555 Bomb 数位dp

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

3150500

Sample Output

0115          

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：数位dp，dp[i][j][k],i表示第i位，j表示前一位的数，k表示前边是否出现过49

ac代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int num[20];long long dp[20][10][2];long long dfs(int len,int pre,int state,int limit){    if(len<0)    return state;    if(!limit&&dp[len][pre][state]!=-1)    return dp[len][pre][state];    int fp=limit?num[len]:9;    long long ret=0;    for(int i=0;i<=fp;i++)    ret+=dfs(len-1,i,state||(pre==4&&i==9),limit&&i==fp);    if(!limit)    dp[len][pre][state]=ret;    return ret;}long long get_num(long long x){    int len=0;    while(x)    {        num[len++]=x%10;        x=x/10;    }    return dfs(len-1,0,0,1);}int main(){    long long n,m;    memset(dp,-1,sizeof(dp));    int t;    cin>>t;    while(t--)    {       cin>>n;       long long ans=get_num(n);       cout<<ans<<endl;    }    return 0;}