【数位DP】——hdu3555——两种写法

                                                Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15609    Accepted Submission(s): 5671

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.

 

 题意：找出1~n内包含49的数字

 

注意：000算是三位数

 

代码一：（递推）

#include<iostream>
#include<string.h>
using namespace std;

long long a[20];
long long dp[20][3];

void Init()
{
    memset(dp,0,sizeof(dp));

    dp[0][0] = 1;

    for(int i =1 ; i<=19; i++)
    {
        dp[i][0] = dp[i-1][0]*10-dp[i-1][1];
        dp[i][1] = dp[i-1][0];
        dp[i][2] = dp[i-1][2]*10+dp[i-1][1];
    }
}

long long solve(long long n)
{
    long long ans=0;
    int flag;

    int len = 0;
    while(n)
    {
        a[++len] = n%10;
        n/=10;
    }

    a[len+1]=0;
    flag = 0;

    for(int i=len; i>=1; i--)
    {
        ans+=dp[i-1][2]*a[i];

        if(flag!=0)
            ans+=dp[i-1][0]*a[i];

        if(flag==0 && a[i]>4)
            ans+=dp[i-1][1];

        if(a[i+1]==4 && a[i]==9)
            flag = 1;
    }
    return ans;
}

int main()
{
    Init();

    int t;
    cin>>t;
    while(t--)
    {
        long long x;
        cin>>x;
        cout<<solve(x+1)<<endl;
    }
    return 0;
}

 

代码二：（递归）

#include<iostream>
#include<string.h>
using namespace std;

long long bit[20];
long long dp[20][3];

long long ms(int pos,int state,bool flag)
{
    if(pos==0)return state==2;

    if(flag!=0 && dp[pos][state]!=-1)return dp[pos][state];

    long long ans=0;

    int x=flag?9:bit[pos];

    for(int i=0;i<=x;i++)
    {
        if(state==2 || (state==1&&i==9))
            ans+=ms(pos-1,2,flag||i<x);
        else if(i==4)
            ans+=ms(pos-1,1,flag||i<x);
        else
            ans+=ms(pos-1,0,flag||i<x);
    }

    if(flag!=0)dp[pos][state]=ans;

    return ans;
}

long long cal(long long x)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return ms(len,0,0);
}

int main()
{
    int t;
    cin>>t;
    memset(dp,-1,sizeof(dp));
    for(int i=0;i<t;i++)
    {
        long long n;
        cin>>n;
        cout<<cal(n)<<endl;
    }
    return 0;
}