hdu3555——Bomb（数位dp）

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

求1~n中出现49的数的个数

根据题意，只要数字中出现了49，后面的数字可以是0~9的任何一个，所以关键是判断是否能出现49

设3个状态s，s==0时dp[i][0]表示长度为i但不含49的方案数，dp[i][1]表示长度为i但末尾为4的方案数，dp[i][2]表示出现49的方案数。

当s==0且当前选择的数字为4是，说明可能会进入49的状态，所以s改为1

当s==1且当前正好选择了9，说明出现了49，状态不会再改变

当s==1且当前既不为4也不为9，说明这个4肯定不会组成49，状态设为0

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
long long n,dp[25][25];
int dight[25];
long long dfs(int pos,int status,bool limit)
{
if(pos==-1)
return status==2;
if(!limit&&dp[pos][status]!=-1)
return dp[pos][status];
long long ans=0;
int s,end=limit?dight[pos]:9;
for(int i=0;i<=end;++i)
{
s=status;
if(status==1&&i==9)
s=2;
if(status==0&&i==4)
s=1;
if(status==1&&i!=4&&i!=9)
s=0;
ans+=dfs(pos-1,s,limit&&i==end);
}
if(!limit)
dp[pos][status]=ans;
return ans;
}
long long cal(long long n)
{
int cnt=-1;
while(n)
{
dight[++cnt]=n%10;
n/=10;
}
return dfs(cnt,0,1);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,-1,sizeof(dp));
scanf("%I64d",&n);
printf("%I64d\n",cal(n));
}
return 0;
}