HDOJ 3555 Bomb

数位DP的DFS写法。。。。

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4630    Accepted Submission(s): 1614

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 <cstdio>
#include <cstring>

using namespace std;

typedef long long int LL;

int bit[30];
LL dp[25][3];

LL dfs(int pos,int s,bool limit)
{
    if(pos==-1)
        return s==2;
    if(limit==false&&~dp[pos][s])
        return dp[pos][s];
    int end=limit?bit[pos]:9;
    LL ans=0;
    for(int i=0;i<=end;i++)
    {
        int news=s;
        if(s==0&&i==4) news=1;
        if(s==1&&i!=9) news=0;
        if(s==1&&i==9) news=2;
        if(s==1&&i==4) news=1;
        ans+=dfs(pos-1,news,limit&&i==end);
    }
    if(!limit)
        return dp[pos][s]=ans;
    else
        return ans;
}

int main()
{
    LL n;
    int T;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d",&n);
        int len=0;
        while(n)
        {
            bit[len++]=n%10;
            n/=10;
        }
        printf("%I64d\n",dfs(len-1,0,1));
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )