Bomb HDU

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? 

InputThe 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. 
OutputFor 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.

这道题放在写过找0和找1的后面。。。我在写这题之前刚写过62的那道题，结果就感觉比较简单了。。。果然水题比较好写，至少比上一道好写。。。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <string.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 20
#define PI acos(-1)
#define mod 2520
#define LL long long
/********************************************************/
LL dp[80][20][5],d[80];

LL dfs(int now, int up, int p, int fp)
{
    if(now==1) return p;
    if(!fp&&dp[now][up][p]!=-1) return dp[now][up][p];
    LL ans=0;
    int len=fp?d[now-1]:9;

    for(int i=0;i<=len;i++)
    {
        if((up==4&&i==9)||p==1)
            ans+=dfs(now-1, i, 1, fp&&i==len);
        else
            ans+=dfs(now-1, i, 0, fp&&i==len);
    }

    if(!fp) dp[now][up][p]=ans;
    return ans;
}

LL solve(LL X)
{
    int len=0;
    memset(dp,-1,sizeof(dp));

    while(X)
    {
        d[++len]=X%10;
        X/=10;
    }

    LL sum=0;
    for(int i=0;i<=d[len];i++)
        sum+=dfs(len, i, 0, i==d[len]);
    return sum;
}

int main()
{
    LL n,m;
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld",&n),
        printf("%lld
", solve(n));
    }
    return 0;
}