HDOJ 3555 Bomb (数位DP)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3935    Accepted Submission(s): 1369

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

Recommend

zhouzeyong

 

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

unsigned long long int dp[30][3];

void init()

{

    memset(dp,0,sizeof(dp));

    dp[0][0]=1;

    for(int i=1;i<30;i++)

    {

        dp[0]=dp[i-1][0]*10-dp[i-1][1];                                 //不包含49的

        dp[1]=dp[i-1][0];                                                      //不包含49的但是开通为9的

        dp[2]=dp[i-1][2]*10+dp[i-1][1];                                 //包含49的

    }

}

int main()

{

    int T;

    unsigned long long int n;

    init();

    scanf("%d",&T);

while(T--)

{

    scanf("%I64d",&n);

    unsigned long long int ans=0;

    int dist[30];

    memset(dist,0,sizeof(dist));

    n++;

    int len=0;

    while(n)

    {

        dist[++len]=n%10;

        n/=10;

    }

    bool is49=false;

    int last=0;

    for(int i=len+1;i>=1;i--)

    {

        ans+=dist*dp[i-1][2];

        if(is49)

        {

            ans+=dp[i-1][0]*dist;

        }

        if(!is49&&dist>4)

        {

            ans+=dp[i-1][1];

        }

        if(last==4&&dist==9)

            is49=true;

        last=dist;

    }

    printf("%I64d ",ans);

}

    return 0;

}