[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"

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
23     return x*f;
24 }
25 LL n,len;
26 LL f[63][10][2];
27 int d[63];
28 inline LL dfs(int now,int dat,int sat,int fp)
29 {
30     if (now==1)return sat;
31     if (!fp&&f[now][dat][sat]!=-1)return f[now][dat][sat];
32     LL ans=0,mx=(fp?d[now-1]:9);
33     for (int i=0;i<=mx;i++)
34     {
35         if (sat||!sat&&dat==4&&i==9)ans+=dfs(now-1,i,1,fp&&mx==i);
36         else ans+=dfs(now-1,i,0,fp&&mx==i);
37     }
38     if (!fp&&f[now][dat][sat]==-1)f[now][dat][sat]=ans;
39     return ans;
40 }
41 inline LL calc(LL x)
42 {
43     LL xxx=x;
44     len=0;
45     while (xxx)
46     {
47         d[++len]=xxx%10;
48         xxx/=10;
49     }
50     LL sum=0;
51     for (int i=0;i<=d[len];i++)
52     sum+=dfs(len,i,0,i==d[len]);
53     return sum;
54 }
55 int main()
56 {
57     memset(f,-1,sizeof(f));
58     int T=read();
59     while (T--)n=read(),printf("%lld\n",calc(n));
60 }

hdu 3555

posted on 2017-08-04 11:21 zhber 阅读(143) 评论(0) 编辑收藏举报