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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
ll dp[100][10];
int bit[105];
int t;
ll dfs(int pos,int sta,int limit)
{
if(pos==-1)
return sta==2;
if(!limit&&dp[pos][sta]!=-1)
return dp[pos][sta];
int up=limit?bit[pos]:9;
int nsta;
ll ans=0;
for(int i=0;i<=up;i++)
{
nsta=sta;
if(sta==2)
nsta=2;
else if(sta==1&&i==9)
nsta=2;
else if(sta==0&&i==4)
nsta=1;
else if(sta==1&&i!=4)
nsta=0;
ans+=dfs(pos-1,nsta,limit&&i==up);
}
if(!limit)
dp[pos][sta]=ans;
return ans;
}
ll solve(ll x)
{int pos=0;
while(x)
{
bit[pos++]=x%10;
x/=10;
}
return dfs(pos-1,0,1);
}
int main()
{
scanf("%d",&t);
while(t--)
{ll n;
memset(dp,-1,sizeof(dp));
scanf("%lld",&n);
printf("%lld\n",solve(n));
}
return 0;
}