# 数位DP

cxlove基础数位DP第三题

= =预处理是个很有用的东西！然后就是分类讨论！

 1 /**************************************************************
2     Problem: 1026
3     User: Tunix
4     Language: C++
5     Result: Accepted
6     Time:0 ms
7     Memory:1272 kb
8 ****************************************************************/
9
10 //BZOJ 1026
11 #include<cmath>
12 #include<vector>
13 #include<cstdio>
14 #include<cstring>
15 #include<cstdlib>
16 #include<iostream>
17 #include<algorithm>
18 #define rep(i,n) for(int i=0;i<n;++i)
19 #define F(i,j,n) for(int i=j;i<=n;++i)
20 #define D(i,j,n) for(int i=j;i>=n;--i)
21 #define pb push_back
22 using namespace std;
23 int getint(){
24     int v=0,sign=1; char ch=getchar();
25     while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();}
26     while(isdigit(ch))  {v=v*10+ch-'0'; ch=getchar();}
27     return v*sign;
28 }
29 const int N=1e7+10,INF=~0u>>2;
30 const double eps=1e-8;
31 /*******************template********************/
32 int dp[15][15];
33 void init(){
34     F(i,0,9) dp[1][i]=1;
35     F(i,2,10)
36         F(j,0,9)
37             F(k,0,9)
38                 if (abs(j-k)>=2) dp[i][j]+=dp[i-1][k];
39 }
40 int solve(int n){
41     int len=0,bit[15];
42     for(;n;n/=10) bit[++len]=n%10;
43     bit[len+1]=0;
44     int ans=0;
45     //长度为1~len-1的
46     F(i,1,len-1)
47         F(j,1,9)
48             ans+=dp[i][j];
49     //长度为len，但最高位没达到上界
50     F(j,1,bit[len]-1) ans+=dp[len][j];
51     //长度为len，且最高位达到上界
52     D(i,len-1,1){
53         F(j,0,bit[i]-1)
54             if (abs(j-bit[i+1])>=2) ans+=dp[i][j];
55         if (abs(bit[i]-bit[i+1])<2) break;
56     }
57     return ans;
58 }
59 int main(){
60     init();
61     int l,r;
62     while(scanf("%d%d",&l,&r)!=EOF)
63         printf("%d\n",solve(r+1)-solve(l));
64     return 0;
65 }
66 
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posted @ 2015-03-01 17:55  Tunix  阅读(122)  评论(0编辑  收藏