Hdu Bomb(数位DP)

Bomb

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

Total Submission(s): 16713 Accepted Submission(s): 6128

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

HintFrom 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

/*
数位DP.
f[i][j]表示以j开头的i位数的决策.
倒序存储
最后还要加一位统一开区间处理.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 25
#define LL long long
using namespace std;
LL f[MAXN][MAXN],ans,n;
int t,l,a[MAXN];
void pre()
{
    f[0][0]=1;
    for(int i=1;i<=20;i++)//枚举位数.
      for(int j=0;j<=9;j++)//枚举这一位数.
        for(int k=0;k<=9;k++)//枚举上一位数.
          {
            if(j==4&&k==9) continue;
            f[i][j]+=f[i-1][k];
          }
    return ;
}
void slove()
{
    n++;ans=n;
    l=0;memset(a,0,sizeof a);
    while(n) a[++l]=n%10,n/=10;
    a[l+1]=0;
    for(int i=l;i;i--)
    {
        for(int j=0;j<a[i];j++)
        {
            if(a[i+1]==4&&j==9) continue;
            ans-=f[i][j];
        }
        if(a[i]==9&&a[i+1]==4) break;
    }
    cout<<ans<<endl;
    return ;
}
int main()
{
    pre();scanf("%d",&t);
    while(t--) cin>>n,slove();
    return 0;
}