多校5 HDU5787 K-wolf Number 数位DP

 

  1 // 多校5 HDU5787 K-wolf Number 数位DP
  2 // dp[pos][a][b][c][d][f] 当前在pos，前四个数分别是a b c d
  3 // f 用作标记，当现在枚举的数小于之前的数时，就不用判断i与dig[pos]的大小
  4 // 整体来说就，按位往后移动，每次添加后形成的数都小于之前的数，并且相邻k位不一样，一直深搜到cnt位
  5 // http://blog.csdn.net/weizhuwyzc000/article/details/52097690
  6 
  7 
  8 
  9 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 10 #include <iostream>
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <sstream>
 14 #include <string>
 15 #include <algorithm>
 16 #include <list>
 17 #include <map>
 18 #include <vector>
 19 #include <queue>
 20 #include <stack>
 21 #include <cmath>
 22 #include <cstdlib>
 23 // #include <conio.h>
 24 using namespace std;
 25 #define clc(a,b) memset(a,b,sizeof(a))
 26 const double inf = 0x3f3f3f3f;
 27 #define lson l,mid,rt<<1
 28 // #define rson mid+1,r,rt<<1|1
 29 const int N = 2010;
 30 const int M = 1e6+10;
 31 const int MOD = 1e9+7;
 32 #define LL long long
 33 #define LB long double
 34 // #define mi() (l+r)>>1
 35 double const pi = acos(-1);
 36 const double eps = 1e-8;
 37 void fre(){freopen("in.txt","r",stdin);}
 38 void freout(){freopen("out.txt","w",stdout);}
 39 inline int read(){int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;}
 40 
 41 LL l,r,cnt;
 42 int k;
 43 int dig[19];
 44 LL dp[19][11][11][11][11][2];
 45 int vis[19][11][11][11][11][2];
 46 int cas;
 47 LL dfs(LL pos,int a,int b,int c,int d,int f){
 48     if(pos==cnt) return 1LL;
 49     if(vis[pos][a][b][c][d][f]==cas) return dp[pos][a][b][c][d][f];
 50     vis[pos][a][b][c][d][f]=cas;
 51     LL ans=0; 
 52     if(f){
 53         for(int i=0;i<10;i++){
 54             if(k==2) {if(d==i) continue;}
 55             else if(k==3) {if(d==i||c==i) continue;}
 56             else if(k==4) {if(d==i||c==i||b==i) continue;}
 57             else {if(d==i||c==i||b==i||a==i) continue;}
 58             if(i==0){
 59                 if(d==10) ans+=dfs(pos+1,a,b,c,d,f);
 60                 else ans+=dfs(pos+1,b,c,d,i,f);
 61             }
 62             else ans+=dfs(pos+1,b,c,d,i,f);
 63         }
 64     }
 65     else{
 66         for(int i=0;i<10;i++){
 67             if(k==2) {if(d==i) continue;}
 68             else if(k==3) {if(d==i||c==i) continue;}
 69             else if(k==4) {if(d==i||c==i||b==i) continue;}
 70             else {if(d==i||c==i||b==i||a==i) continue;}
 71             if(i<dig[pos]){
 72                 if(i==0){
 73                     if(d==10) ans+=dfs(pos+1,a,b,c,d,1);
 74                     else ans+=dfs(pos+1,b,c,d,i,1);
 75                 }
 76                 else ans+=dfs(pos+1,b,c,d,i,1);
 77             }
 78             else if(i==dig[pos]){
 79                 if(i==0){
 80                     if(d==10) ans+=dfs(pos+1,a,b,c,d,f);
 81                     else ans+=dfs(pos+1,b,c,d,i,f);
 82                 }
 83                 else
 84                     ans+=dfs(pos+1,b,c,d,i,f);
 85             }
 86         }
 87     }
 88     return dp[pos][a][b][c][d][f]=ans;
 89 }
 90 
 91 bool ck(){
 92     for(int i=0;i<cnt;i++){
 93         for(int j=i-1;j>=max(0,i-k+1);j--){
 94             if(dig[i]==dig[j]) return false;
 95         }
 96     }
 97     return true;
 98 }
 99 int tem[19];
100 int main(){
101     while(~scanf("%I64d%I64d%d",&l,&r,&k)){
102         cnt=0;
103         while(r) tem[cnt++]=r%10,r=r/10;
104         int k=0;
105         for(int i=cnt-1;i>=0;i--) dig[k++]=tem[i];
106         cas++;
107         LL ans1=dfs(0,10,10,10,10,0);
108         cnt=0,k=0;
109         while(l) tem[cnt++]=l%10,l=l/10;
110         
111         for(int i=cnt-1;i>=0;i--) dig[k++]=tem[i];
112         cas++;
113         LL ans2=dfs(0,10,10,10,10,0);
114         if(ck()) ans2--; 
115         printf("%I64d\n",ans1-ans2);
116     }
117     return 0;
118 }

 

posted @ 2016-08-03 19:29  yyblues  阅读(397)  评论(0)    收藏  举报