HDU 3555 Bomb 数位Dp

D - Bomb

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 3555

Appoint description:  System Crawler  (2016-04-27)

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.

判断是否子串是否含有49

按位dp即可 需要判断前面位数的3种状态

ACcode

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
ll dp[25][3];///第二个参数表示 2-》》含有49、1--》》 不含有49但前位是4   0--》》其他
int date[25];
ll dfs(int pos,int flag,bool limit){
    if(!pos)return flag==2;
    if(!limit&&dp[pos][flag]!=-1)
        return dp[pos][flag];
    ll ans=0;
    int ed=limit?date[pos]:9;
    for(int i=0;i<=ed;++i){
        int tmp=flag;
        if(flag==1&&i==9)
            tmp=2;
        else if(flag==0&&i==4)
            tmp=1;
        else if(flag==1&&i!=4&&i!=9)
            tmp=0;
        ans+=dfs(pos-1,tmp,limit&&i==ed);
    }
    return limit?ans:dp[pos][flag]=ans;
}
void doit(){
    ll n;
    scanf("%I64d",&n);
    int len=0;
    while(n){
        date[++len]=n%10;
        n/=10;
    }
    printf("%I64d\n",dfs(len,0,1));
}
int main(){
    int loop;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&loop);
    while(loop--)doit();
    return 0;
}