[暑假集训--数位dp]hdu3555 Bomb

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.

数位dp，找子序列‘49’

记一下有没有出现过'4',是否已经出现过"49"

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n,len;
 LL f[][][];
 int d[];
 inline LL dfs(int now,int dat,int sat,int fp)
 {
     if (now==)return sat;
     if (!fp&&f[now][dat][sat]!=-)return f[now][dat][sat];
     LL ans=,mx=(fp?d[now-]:);
     for (int i=;i<=mx;i++)
     {
         if (sat||!sat&&dat==&&i==)ans+=dfs(now-,i,,fp&&mx==i);
         else ans+=dfs(now-,i,,fp&&mx==i);
     }
     if (!fp&&f[now][dat][sat]==-)f[now][dat][sat]=ans;
     return ans;
 }
 inline LL calc(LL x)
 {
     LL xxx=x;
     len=;
     while (xxx)
     {
         d[++len]=xxx%;
         xxx/=;
     }
     LL sum=;
     for (int i=;i<=d[len];i++)
     sum+=dfs(len,i,,i==d[len]);
     return sum;
 }
 int main()
 {
     memset(f,-,sizeof(f));
     int T=read();
     while (T--)n=read(),printf("%lld\n",calc(n));
 }

hdu 3555