HDU(3555)，数位DP

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

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15372    Accepted Submission(s): 5563

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

 

 

数位DP，之前做了一个相反的题目，结果WA了，后来发现数据不对，是long long 型数据。

 

#include <stdio.h>
#include <string.h>

int bit[];
long long dp[][];

long long dfs(int len,bool is4,bool ismax)
{
    if(len==) return ;     ///搜索成功
    if(!ismax&&dp[len][is4]>=) return dp[len][is4];

    long long cnt = ;
    int maxnum = ismax? bit[len]:;
    for(int i=; i<=maxnum; i++)
    {
        if((is4&&i==)) continue;
        cnt +=dfs(len-,i==,ismax&&i==maxnum);
    }
    return ismax?cnt:dp[len][is4]=cnt;
}

long long f(long long n)
{
    int len = ;
    while(n)
    {
        bit[++len] = n%;
        n/=;
    }
    return dfs(len,false,true);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long n;
        scanf("%lld",&n);
        memset(dp,-,sizeof(dp));
        printf("%lld\n",n-f(n)+);
    }
    return ;
}