[暑假集训--数位dp]UESTC250 windy数

windy定义了一种windy数。

不含前导零且相邻两个数字之差至少为 $2$

windy想知道，在 $A$

Input

包含两个整数， $A$

满足 $1 \leq A \leq B \leq 2000000000$

OutputSample Input

1 10

Sample Output

记一下上一位出现的数是啥

 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,l,r;
26 LL f[20][11][2];
27 int d[110];
28 inline LL dfs(int now,int dat,int lead,int fp)
29 {
30     if (now==1)return (!lead)||(lead&&dat==0);
31     if (!fp&&f[now][dat][lead]!=-1)return f[now][dat][lead];
32     LL ans=0,mx=fp?d[now-1]:9;
33     for (int i=0;i<=mx;i++)
34     {
35         if (!lead&&(i==dat-1||i==dat||i==dat+1))continue;
36         ans+=dfs(now-1,i,lead&&i==0&&now-1!=1,fp&&i==d[now-1]);
37     }
38     if (!fp)f[now][dat][lead]=ans;
39     return ans;
40 }
41 inline LL calc(LL x)
42 {
43     if (!x)len=1,d[len]=0;
44     else
45     {
46         LL xxx=x;
47         len=0;
48         while (xxx)
49         {
50             d[++len]=xxx%10;
51             xxx/=10;
52         }
53     }
54     LL sum=0;
55     sum+=dfs(len,0,1,0);
56     for (int i=1;i<=d[len];i++)
57         sum+=dfs(len,i,0,i==d[len]);
58     return sum;
59 }
60 int main()
61 {
62     while (~scanf("%lld%lld",&l,&r))
63     {
64         memset(f,-1,sizeof(f));
65         printf("%lld\n",calc(r)-calc(l-1));
66     }
67 }

UESTC 250

posted on 2017-08-04 12:54 zhber 阅读(190) 评论(2) 编辑收藏举报