Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3362 Accepted Submission(s): 1185

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

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.

Author

fatboy_cw@WHU

Topic link: http://acm.hdu.edu.cn/showproblem.php?pid=3555

meaning of the title

Find the number of numbers 49 in 1~N 1 <= N <= 2^63-1 to 

analyze
 the digital DP. This routine for finding the number of numbers that do not contain a certain number string is as follows.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long dp[25][3];
/*
 * dp[i][0], indicating that it does not contain 49
 * dp[i][1], means that it does not contain 49, and the highest bit is 9
 * dp[i][2], means containing 49
 */
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 ]; // Add 0~9 numbers in front, subtract 4 in front of 9 
        dp[ i][ 1 ]=dp[i- 1 ][ 0 ]; // Add 9 to the highest bit 
        dp[i][ 2 ]= 10 *dp[i- 1 ][ 2 ]+dp[i- 1 ][ 1 ]; // Add any number in front of 49, or add 4 in front of 9 
    }
}
int bit[ 25 ];
 long  long calc( long  long n)
{
    int len=0;
    while(n)
    {
        bit[++len]=n%10;
        n/=10;
    }
    bit[len+1]=0;
    bool flag=false;
    long long ans=0;
    for(int i=len;i>=1;i--)
    {
        ans+=dp[i-1][2]*bit[i];
        if(flag)ans+=dp[i-1][0]*bit[i];
        else
        {
            if(bit[i]>4)ans+=dp[i-1][1];
        }
        if(bit[i+1]==4&&bit[i]==9)flag=true;
    }
    if (flag)ans++; // Add n itself 
    return ans;
}
intmain ()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    long long n;
    scanf("%d",&T);
    init();
    while(T--)
    {
        scanf("%I64d",&n);
        printf("%I64d\n",calc(n));
    }
    return 0;
}

HDU 3555 Bomb (Digital DP)

Bomb

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