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.

思路：

数位DP，模板题。看了一天的数位dp，大概有点思路了，还有很多细节需要琢磨。推荐一个博客 数位DP入门
数位dp的主要思想就是，按位来枚举数字。比如 235，从高位到地位枚举的时候 可以分为如下步骤：枚举百分位的时候取值范围是0，1，2 ，档百分位是0的时候，十分位取值可以为 0~9 为1 的时候同理，但是当百分位为2的时候，这时候就有限制了此时十分位的枚举范围只能是0，1，2，3了， 如此把所有位数枚举完。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int digit[65];

long long dp[65][2];

long long dfs(int pos, bool have, bool limit)
{
    if (pos < 1) 
        return 1;
    if (!limit && dp[pos][have] != -1) return dp[pos][have];
    int num = limit?digit[pos]:9;
    long long tmp = 0;
    for (int i = 0; i <= num; ++i) {
        if (have && i == 9) continue;
        tmp += dfs(pos-1,i==4, limit && (i == num));
    }
    if (!limit) dp[pos][have] = tmp;

    return tmp;
}

long long solve(long long x)
{

    int len = 0;
    while(x) {
        digit[++len] = x%10;
        x /= 10;
    }


    memset(dp,-1,sizeof(dp));
    return dfs(len,false,true);
}

int main()
{   
    int t;
    cin >> t;
    long long x;
    while(t--) {
        cin >> x;
        cout << x - solve(x) + 1 << endl;
    }

    return 0;
}