杭电 3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 6609    Accepted Submission(s): 2303

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.

第一个数位动规！

！！！

看代码理解的。差点儿相同懂那么点点点了。

AC代码例如以下：

传统模板！

！

！

///递推   46MS 328K

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;

int main()
{
    __int64 dp[20][3];//0表示该位数中不含49的情况数。1表示不含49但头为9的情况数。2表示含49的情况数
    int i,j;
    memset(dp,0,sizeof dp);
    dp[0][0]=1;
    for(i=1;i<=20;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];
        //cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;
    }
    int t;
    cin>>t;
    __int64 num[30];
    __int64 n;
    while(t--)
    {
        memset(num,0,sizeof num);
        cin>>n;
        n++;//这一步非常重要
        int tt=1;
        while(n)
        {
            num[tt++]=n%10;
            n/=10;
        }
        __int64 ans=0;
        int flag=0;int last=0;
        for(i=tt-1;i>=1;i--)
        {
            ans+=num[i]*dp[i-1][2];
            if(flag==1)
                ans+=dp[i-1][0]*num[i];
            if(flag==0&&num[i]>4)
                ans+=dp[i-1][1];
            if(num[i+1]==4&&num[i]==9)
                flag=1;
        }
        cout<<ans<<endl;
    }
    return 0;
}

///数位之记忆化搜索   78MS 276K

#include<cstdio>
#include<iostream>
#include<cstring>
#define mod 1000000009
#define ll __int64
#define M 100005
using namespace std;

ll dp[30][3];
ll num[30];

ll dfs(int pos,int statu,int limit)
{
    int i;
    if(!pos) return statu==2;
    if(!limit&&dp[pos][statu]!=0) return dp[pos][statu];
    int end=limit?

num[pos]:9;
    ll sum=0;
    for(i=0;i<=end;i++)
    {
        int flag=statu;
        if(flag==0&&i==4) flag=1;
        if(flag==1&&i==9) flag=2;
        if(flag==1&&i!=4&&i!=9) flag=0;
        sum+=dfs(pos-1,flag,limit&&i==end);
    }
    return limit?

sum:(dp[pos][statu]=sum);
}

ll _49(ll n)
{
    int pos=1;
    memset(dp,0,sizeof dp);
    while (n>0)
    {
        num[pos++]=n%10;
        n/=10;
    }
    //cout<<pos<<"~~~~~~~~~~~~~"<<endl;
    return dfs(pos-1,0,1);
}

int main()
{
    int i,j;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n;
        scanf("%I64d",&n);
        printf("%I64d\n",_49(n));
    }

    return 0;
}