hdu3555——Bomb（数位dp）

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

这题跟不要62有点区别，最后的状态要判断下，相比之下复杂了点。并且不能直接算出没有49的数，当然也可能是我没想到正确的做法

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000000005
#define Mod 10001
using namespace std;
int dight[30];
long long dp[30][3];
long long dfs(int pos,int s,bool limit)
{
if(pos<0)
return s==2; //求符合条件的数与求不符合条件的区别
if(!limit&&dp[pos][s]!=-1)
return dp[pos][s];
int end;
long long ret=0;
if(limit)
end=dight[pos];
else
end=9;
for(int d=0;d<=end;++d)
{
int have=s;
if(s==1&&d==9)
have=2; //有了49，后面就随便填了
if(s==0&&d==4)
have=1; //有了4，要注意后面是否会有9
if(s==1&&d!=4&&d!=9)
have=0; //不用注意了
ret+=dfs(pos-1,have,limit&&d==end);
}
if(!limit)
dp[pos][s]=ret;
return ret;

}
long long solve(long long a)
{
memset(dight,0,sizeof(dight));
int cnt=0;
while(a!=0)
{
dight[cnt++]=a%10;
a/=10;
}
return dfs(cnt-1,0,1);
}
int main()
{
memset(dp,-1,sizeof(dp));
int t;
scanf("%d",&t);
long long n;
while(t--)
{
scanf("%I64d",&n);
printf("%I64d\n",solve(n));
}
return 0;
}