HDU 5179 - beautiful number (DFS)

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题意

求美丽数的数量。

思路

我能说我看了十分钟第一个公式吗（ＴДＴ），还以为是数学题。吓得我都飞起来了。

后来又想想数应该不是很多，搜索算了。

先获得left和right的位数，然后往上填数。因为数字不大，我就直接用int变量代替数组了。

我的有点类似DFSID，按照数字的位数搜。直接写DFS也可以。

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <cmath>using namespace std;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 1e5 + 10;const int MOD = 9901;const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };int cases = 0;typedef pair<int, int> pii;int arr[MAXN];int DFS(int curDep, LL num, int bitLimit, int limit){    int ans = 0;    if (limit == 0) return 0;    int tmp = num%10;   //前一位    for (int i = 1; i <= 9; i++)    {        if (i <= tmp && tmp%i == 0 || curDep == 1)        {            num = num*10 + i;            if (curDep < bitLimit) return DFS(curDep+1, num, bitLimit, limit);            if (num > limit) continue;            if (curDep == bitLimit) ans++;  //达到指定的位数            num /= 10;  //回溯        }    }    return ans;}int get_bit(int num){    int tmp = num, cnt = 0;    while (tmp)    {        tmp /= 10;        cnt++;    }    return cnt;}int main(){    //ROP;    int T;    scanf("%d", &T);    while (T--)    {        int aans1 = 0, aans2 = 0;        int l, r;        scanf("%d%d", &l, &r);        int a = get_bit(l), b = get_bit(r);        for (int i = 1; i <= b; i++)            aans1 += DFS(1, 0, i, r);        for (int i = 1; i <= a; i++)            aans2 += DFS(1, 0, i, l-1);        printf("%d\n", aans1-aans2);    }    return 0;}

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