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?

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.

For each test case, output an integer indicating the final points of the power.

3
1
50
500

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.
 

fatboy_cw@WHU

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU



所谓数位dp就是dp+组合数学

首先是预处理出dp:
dp[i][0]:i位数，不包含49的有几个
dp[i][1]:i位数，不包含49但是以9开头的有几个
dp[i][2]:i位数，包含49的有几个
（i位数包含0开头的情况）

组合数学部分：
算1到x有几个数自己含有49时，
为了方便描述，用一个串s储存x，其中0为起始位置，串长为len，
s[i]到s[len-1]组成一个数字y，算出数字恰有len-i位，且不大于y时，且自己含有49的有多少个
枚举所有i，累加即可

注意：
当49已经在之前出现时，要特殊处理
若x本身含有49，记得算上x本身
y可以是0开头，因为若每次都这样，是不会算多的，相当于每次都把下一次的i算了一些情况

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
ll dp[20][3];
ll work(string s)
{
	int len=s.length();
	ll ans=0;
	bool flag=0;
	for(int i=0;i<len;i++)
	{
		if(flag)
			ans+=(dp[len-i-1][0]+dp[len-i-1][2])*(s[i]-'0');
		else
		{
			ans+=dp[len-i-1][2]*(s[i]-'0');
			if(s[i]>'4')
				ans+=dp[len-i-1][1];
		}
		if(!flag&&i>0&&s[i-1]=='4'&&s[i]=='9')
			flag=1;
	}
	if(flag)
		ans++;
	return ans;
}
int main()
{
	dp[0][0]=1;
	for(int i=1;i<19;i++)
	{
		dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
		dp[i][1]=dp[i-1][0];
		dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
	}
	int T;
	cin>>T;
	while(T--)
	{
		string s;
		cin>>s;
		cout<<work(s)<<endl;
	}
}