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

 

Sample Output

 

Author

fatboy_cw@WHU

 

Source

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <string.h>

using namespace std;

typedef long long LL;

LL dp[][];

/**

 * dp[i][0],表示长度为i，不存在不吉利数字

 * dp[i][1],表示长度为i, 不存在不吉利数字，且最高位为9

 * dp[i][2],表示长度为i, 存在不吉利数字

 */

 void Init()

 {

     int i;

     dp[][] = ;

     for(i=; i<; i++)

     {

         dp[i][] = *dp[i-][] - dp[i-][];

         dp[i][] = dp[i-][];

         dp[i][] = *dp[i-][] + dp[i-][];

     }

 }

 LL Slove(LL n)

 {

     int i, a[]={}, len=, flag=;

     LL n1 = n, ans=;

     while(n1)

     {

         a[++len] = n1%;

         n1 /= ;

     }

     for(i=len; i>=; i--)

     {

         ans += dp[i-][]*a[i];

         if(flag)///高位已经出现49了，后面随意

            ans += dp[i-][]*a[i];

         else if(!flag && a[i]>)

            ans += dp[i-][];

         if(a[i+]== && a[i]==)

            flag = ;

     }

     return ans;

 }

int main()

{

    int T;

    Init();

    scanf("%d", &T);

    while(T--)

    {

        LL n;

        scanf("%I64d", &n);

        printf("%I64d\n", Slove(n+));

    }

    return ;

}