hdu3555 Bomb (数位dp入门题)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 19698    Accepted Submission(s): 7311

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.

代码（dfs解法）：

#include<iostream>
#include<cstring>
#define LL long long
using namespace std;

LL bit[25], dp[25][2];

LL dfs(int pos, int is4, int lim)//pos当前位, is4上一位是否为4, lim上一位是否取到最大值
{
    if(pos<0) return 1;
    if(!lim && dp[pos][is4]!=-1) return dp[pos][is4];
    int las=lim?bit[pos]:9;//若上以一位没有取最大值位则可以0~9循环
    LL res=0;
    for(int i=0; i<=las; ++i)
        if(!(is4 && i==9))//搜索不含49的，结果取反
            res+=dfs(pos-1, i==4, lim&&i==las);
    if(!lim) dp[pos][is4]=res;
    return res;
}

int main()
{
    int T;
    memset(dp, -1, sizeof(dp));
    cin>>T;
    while(T--)
    {
        LL n;
        cin>>n;
        LL len=0, m=n;
        while(n)
        {
            bit[len++]=n%10;
            n/=10;
        }
        LL ans=m-dfs(len-1, 0, 1)+1;//结果取反, 且搜索过程中会包含0, 故+1
        cout<<ans<<endl;
    }
    return 0;
}

代码（传统dp）：

#include<iostream>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;

LL num[25], dp[25][3];

int main()
{
    int T;
    memset(dp, 0, sizeof(dp));
    dp[0][0]=1;
    for(int i=1; i<21; ++i)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//dp[i][0]表示i位数不包含49且最高位不是9的数目
        dp[i][1]=dp[i-1][0];//dp[i][1]表示i位数不包含49且最高位是9的数目
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//dp[i][2]表示i位数包含49的数目
    }
    cin>>T;
    while(T--)
    {
        LL n;
        cin>>n;
        int len=0;
        memset(num, 0, sizeof(num));
        while(n)
        {
            num[++len]=n%10;
            n/=10;
        }
        int las=0;
        bool flag=false;
        LL ans=0;
        for(int i=len; i>=1; --i)
        {
            ans+=(dp[i-1][2]*num[i]);//若n=789, 则i=3时此处计算700以内的结果
            if(flag) ans+=dp[i-1][0]*num[i];//若之前已包含49
            if(!flag && num[i]>4) ans+=dp[i-1][1];//若之前未包含49
            if(las==4 && num[i]==9) flag=true;
            las=num[i];
        }
        if(flag) ans++;
        cout<<ans<<endl;
    }
    return 0;
}