HDU 5787 K-wolf Number

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题意

求区间[l,r]中有多少个数不存在连续k位含有相同的数位。

思路

比较裸的数位DP，但是因为刚开始没有想好思路就开始写后来调试挺长的时间。
因为K比较小，所以我们直接加4维状态分别表示前面取的四个数，每次加下一个的时候根据K判断一下就好了。有些不好处理的地方是010这种有前导零的，那么前导零我们统一用11这个数位取不到的数来表示就可以了。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e6 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;LL dig[25];LL l, r;int k;LL dp[25][12][12][12][12];LL dfs(int cur, int e, int a, int b, int c, int d, int flag){    //    if (cur <= 0)    {        //printf("%d %d %d %d %d %d\n", cur, e, a, b, c, d);        return flag;    }    if (!e && dp[cur][a][b][c][d] != -1)        return dp[cur][a][b][c][d];    int end = e ? dig[cur] : 9;    LL ans = 0;    for (int i = 0; i <= end; i++)    {        if (k == 2)        {            if (d == i) continue;            ans += dfs(cur - 1, i == end && e, b, c, d, i || flag ? i : 11, i || flag);        }        else if (k == 3)        {            if ((i == d || i == c)) continue;            ans += dfs(cur - 1, i == end && e, b, c, d, i || flag ? i : 11, i || flag);        }        else if (k == 4)        {            if ((i == d || i == c || i == b)) continue;            ans += dfs(cur - 1, i == end && e, b, c, d, i || flag ? i : 11, i || flag);        }        else if (k == 5)        {            if ((i == d || i == c || i == b || i == a)) continue;            //printf("nei   %d %d %d %d %d %d %d\n", i, end, (i == end) && e, a, b, c, d);            ans += dfs(cur - 1, i == end && e, b, c, d, i || flag ? i : 11, i || flag);        }    }    if (!e) dp[cur][a][b][c][d] = ans;    return ans;}LL Solve(LL x){    if (x == 0) return 0;    int len = 0;    while (x)    {        dig[++len] = x % 10;        x /= 10;    }    return dfs(len, 1, 11, 11, 11, 11, 0);}int main(){    //freopen("H:\\in.txt","r",stdin);    //freopen("H:\\out.txt","w",stdout);    while (scanf("%I64d%I64d%d", &l, &r, &k) != EOF)    {        memset(dp, -1, sizeof(dp));        printf("%I64d\n", Solve(r) - Solve(l - 1));    }    return 0;}

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