HDU 3555 Bomb （数位DP）

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. 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

求含49的个数：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
//ll dp[22][0]; //含有49 
//ll dp[22][1]; //不含49且最高位为9 
//ll dp[22][2]; //不含49
ll dp[22][3];  
int bits[22];

void init()
{
	int i,j;
	dp[0][2]=1;
	for(i=1;i<=20;i++) {
		dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
		dp[i][1]=dp[i-1][2];
		dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
	}
}

ll work(ll n)
{
	int i,j,len=0;
	bool flag=false;
	ll ans=0;
	while(n) {
		bits[++len]=n%10;
		n=n/10;
	}
	bits[len+1]=0;
	for(i=len;i;i--) {
		ans+=bits[i]*dp[i-1][0];
		if(flag) ans+=bits[i]*dp[i-1][2];
		if(!flag&&bits[i]>4) ans+=dp[i-1][1];
		if(bits[i]==9&&bits[i+1]==4) flag=true;
	}
	return ans;
}

int main()
{
	int t,i,j;
	ll n;
	cin>>t;
	init();
	while(t--) {
		cin>>n;
		cout<<work(n+1)<<endl;
	}
	return 0;
}