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.

 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mxn = 25 ;
ll n,m,k,t,dp[mxn][3],lim[mxn];
ll dfs(int pos , int state , int limit)
{
    if(!pos) return 1;
    if(!limit && dp[pos][state]!=-1) return dp[pos][state];
    int up = limit ? lim[pos] : 9 ;
    ll ans = 0 ;
    for(int i=0;i<=up;i++)
    {
        if( ( state && i==9)) continue ;
        ans += dfs(pos-1, i==4 , limit && i==lim[pos] ) ;
    }
    if(!limit) dp[pos][state] = ans;
    return ans;
}
ll cal(ll x)
{
    int len = 0;
    memset(lim,0,sizeof(lim));
    memset(dp,-1,sizeof(dp));
    while(x) lim[++len] = x%10 , x/=10 ;
    return dfs(len,false,true);
}
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        cout<<n-cal(n)+1<<endl;
    }
    return 0;
}

 

Author

fatboy_cw@WHU