hdu 5787 K-wolf Number（数位dp）

K-wolf Number

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 807    Accepted Submission(s): 300

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 1 220 100 5

 

Sample Output

172

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

【题意】每次输入l,r,k 三个数，求(l,r)区间中有多少个数满足（相邻k位数字都不相同）。

【分析】很明显的数位dp，我们可以枚举前k-1位数字，作为我们保存的状态，然后套数位dp。（注意前导0的情况）

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<map>#include<set>#include<vector>#define F first#define S second#define mp make_pairusing namespace std;typedef __int64 LL;LL digit[100],dp[20][100010],k;//对于为什么存五位，有点bug，还没解决。

bool judge(LL d,LL s){    int x=k-1;    while(x--)    {        if(d==(s%10)) return false;        s/=10;    }    return true;}const int ten[] = {1, 10, 100, 1000, 10000, 100000};LL get(LL d,LL s,bool first){    if(first&&!d) return 0;    if(!first)    {//        LL sum=1;//        for(int i=0; i<k-2; i++) sum*=10;//        cout<<sum<<" "<<s<<endl;//        while(s>=sum) s-=sum;//        cout<<s<<endl;        return (s*10+d)%ten[k];    }    else    {        LL sum=0;        for(int i=0; i<k-1; i++)        {            sum=sum*10+d;        }        return sum;    }}LL dfs(LL cnt,LL s,bool e,bool first)//first 用来标记前导0{    if(!cnt) return 1LL;    if(!e && ~dp[cnt][s])        return dp[cnt][s];    LL ret=0;    LL to=e?digit[cnt]:9;    for(LL d=0; d<=to; d++)    {        if(!judge(d,s)&&!first)        {            continue;        }        LL as=get(d,s,first);        ret+=dfs(cnt-1,as,e&&d==to,first&&!d);    }    if(!e) dp[cnt][s]=ret;    return ret;}LL calc(LL x){    memset(dp, -1, sizeof(dp));    LL cnt = 0;    while(x) digit[++cnt] = x % 10, x /= 10;    return dfs(cnt, 0, 1, 1);}//bool judge1(int x, int k)//{//    int cnt = 0;//    for(; x; x /= 10) digit[++cnt] = x % 10;////    for(int i = 1; i <= cnt; ++i)//    {//        bool ok = true;//        for(int j = 1; j < k; ++j)//        {//            if(i - j >= 1 && digit[i] == digit[i - j])//            {//                ok = false;//                break;//            }//        }//        if(!ok) return false;//    }//    return true;//}//int sum[1000005];int main(){//    for(k = 1; k <= 5; ++k)//    {//        for(int i = 1; i <= 100000; ++i)//        {//            sum[i] = sum[i - 1] + judge1(i, k);//            if(sum[i]  + 1!= calc(i))//            {//                printf("%d: %d %d %d\n", i, sum[i] + 1, calc(i), k);//                printf("WA\n");//                break;//            }//        }//    }    LL l,r;    while(~scanf("%I64d%I64d%I64d",&l,&r,&k))    {        printf("%I64d\n",calc(r)-calc(l-1));    }    return 0;}