Counter-Strike found a time bomb in the dust. But this time the terrorists improved the time bomb. The numbering sequence of time bombs is counted from 1 to N. If the current numbering sequence includes (substring!!) "49", the explosive power will increase by one point.
Asked how powerful it was when it exploded.
Input description
Input T represents group T (T <= 1e4)
Each group input one n (1 <= n<= 2^ 63-1)
Output description
The power of output explosion
Sample input
1
1
sample output
0 number in
interval How many 49, digital dp boards appeared in between
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#define ls (p<<1)
#define rs (p<<1|1)
#define ll long long
using namespace std;
const int maxn = 2e6+5;
const int INF = 0x3f3f3f3f;
ll dp[50][15];
ll a[30];
// void init(){
// dp[0][0]=1;
// dp[0][1]=dp[0][2]=0;
// for(int i=1;i<25;i++){
// dp[i][0]=10*dp[i-1][0]-dp[i-1][1];//长度为i不含有49的个数
// dp[i][1]=dp[i-1][0];//最高位为9但不含49的个数
// dp[i][2]=10*dp[i-1][2]+dp[i-1][1];//含有49的个数
// }
// }
ll dfs(ll pos,ll pre,bool limit){
if(pos==-1) return 1;
if(!limit&&dp[pos][pre]!=-1)
return dp[pos][pre];
ll up;
if(limit)
up=a[pos];
else up=9;
ll ans=0;
for(int i=0;i<=up;i++){
if(pre==4&&i==9)
continue;
else
ans+=dfs(pos-1,i,limit&&i==a[pos]);
}
if(!limit)
dp[pos][pre]=ans;
return ans;
}
void solve(){
int t;
cin>>t;
while(t--){
ll x,y;
cin>>x;
y=x;
memset(dp,-1,sizeof(dp));
ll len=0;
while(x){
a[len++]=x%10;
x/=10;
}
ll sum=dfs(len-1,0,1);
ll res=y-sum+1;
cout<<res<<endl;
}
}
int main()
{
ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
solve();
return 0;
}