HDU 3555 Bomb（数位DP）

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

 

 

Source

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

 

 

Recommend

zhouzeyong

 

 

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

 

简单的数位DP入门题。

注释见代码：

/*

 * HDU 3555

 * 求1~N中含有数字49的个数     1 <= N <= 2^63-1

 */

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

long long dp[25][3];

/*

 * dp[i][0],表示不含有49

 * dp[i][1],表示不含有49，且最高位为9

 * dp[i][2],表示含有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];//在前面加0~9的数字，减掉在9前面加4

        dp[i][1]=dp[i-1][0];//最高位加9

        dp[i][2]=10*dp[i-1][2]+dp[i-1][1];//在本来含有49的前面加任意数，或者在9前面加4

    }

}

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++;//加上n本身

    return ans;

}

int main()

{

    //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;

}