HDU 5787 K-wolf Number 数位DP

K-wolf Number

Problem Description

 

Alice thinks an integer x is a K-wolf number, if every K adjacent digits in decimal representation of x is pairwised different.
Given (L,R,K), please count how many K-wolf numbers in range of [L,R].

 

Input

 

The input contains multiple test cases. There are about 10 test cases.

Each test case contains three integers L, R and K.

1≤L≤R≤1e18
2≤K≤5

 

Output

 

For each test case output a line contains an integer.

 

Sample Input

 

1 1 2
20 100 5

 

Sample Output

 

1
72

 

题意：

　　询问 [L,R] 间有多少个数 满足 连续k位数都不相同

题解：

　　数位DP

　　传递 前 k-1 分别是什么

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const int N = 1e5+, M = 6e4+, mod = 1e9+, inf = 1e9+;
typedef long long ll;
 
ll L,R;
ll k,d[N],K;
 
ll dp[][][][][];
bool vis[][][][][];
 
ll dfs(int dep,int f,int now[],int bo) {
    int x = now[], y = now[], z = now[], e = now[];
    if(dep < ) return ;
    if(f&&vis[dep][x][y][z][e]) return dp[dep][x][y][z][e];
    if(f) {
 
        vis[dep][x][y][z][e] = true;
        ll& ret = dp[dep][x][y][z][e];
 
        for(int i = ; i <= ; ++i) {
            int OK = ;
            for(int j = ; j < k-; ++j) if(now[j] == i) {OK = ; break;}
            if(OK == ) continue;
            if(!bo && !i)  ret += dfs(dep-,f,now,bo);
            else {
                int tmpnow[];
                for(int j = ; j <= ; ++j) tmpnow[j] = now[j+];
                tmpnow[] = ;
                tmpnow[k-] = i;
                ret += dfs(dep-,f,tmpnow,bo||i);
            }
        }
        return ret;
 
    }else {
        ll ret = ;
        for(int i = ; i <= d[dep]; ++i) {
 
            int OK = ;
            for(int j = ; j < k-; ++j) if(now[j] == i) {OK = ; break;}
            if(OK == ) continue;
            if(!bo && !i)  ret += dfs(dep-,i<d[dep],now,bo);
            else {
 
                int tmpnow[];
                for(int j = ; j <= ; ++j) tmpnow[j] = now[j+];
                tmpnow[] = ;
                tmpnow[k-] = i;
                ret += dfs(dep-,i<d[dep],tmpnow,bo||i);
            }
 
        }
        return ret;
    }
}
 
ll solve(ll x) {
    //if(x < 0) return 0;
    memset(vis,false,sizeof(vis));
    memset(dp,,sizeof(dp));
    int len = ;
    while(x) {
        d[len++] = x % ;
        x /= ;
    }
    int now[];
    for(int i = ; i <= ; ++i) now[i] = ;
    return dfs(len-,,now,);
}
 
int main()
{
 
    while(~scanf("%I64d%I64d%I64d",&L,&R,&k)) {
        //K = pre[k];
        printf("%I64d\n",solve(R) - solve(L-));
    }
 
    /* ll x;
     while(~scanf("%I64d%d",&x,&k)) {
        K = pre[k];
        printf("%I64d\n",solve(x));
    }
    */
 
}