Hdu 3555-Bomb 数位dp

题目: http://acm.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): 12696    Accepted Submission(s): 4533

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 

 

 

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题解：

数位dp

和“不要62”那道题很类似。套模版即可。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL digit[26],f[26][12][2];
LL read()
{
    LL s=0,fh=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();}
    return s*fh;
}
LL Dfs(LL len,LL last,LL flag,LL Judge)
{
    if(len==0)return Judge;
    if(flag==0&&f[len][last][Judge]!=-1)return f[len][last][Judge];
    LL end=flag?digit[len]:9,i,ret=0;
    for(i=0;i<=end;i++)ret+=Dfs(len-1,i,flag&&(i==end),Judge||(last==4&&i==9));
    if(flag==0)f[len][last][Judge]=ret;
    return ret;
}
LL Calc(LL N)
{
    LL len=0;
    memset(digit,0,sizeof(digit));
    while(N>0){digit[++len]=N%10;N/=10;}
    return Dfs(len,0,1,0);
}
int main()
{
    LL T,N;
    T=read();
    while(T--)
    {
        N=read();
        memset(f,-1,sizeof(f));
        printf("%lld\n",Calc(N));
    }
    fclose(stdin);
    fclose(stdout);
    return 0;
}