beautiful number(HDU 5179 数位DP)

beautiful number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 883    Accepted Submission(s): 571

Problem Description

Let A=∑ni=1ai∗10n−i(1≤ai≤9)(n is the number of A's digits). We call A as “beautiful number” if and only if a[i]≥a[i+1] when 1≤i<n and a[i] mod a[j]=0 when 1≤i≤n,i<j≤n(Such as 931 is a "beautiful number" while 87 isn't).
Could you tell me the number of “beautiful number” in the interval [L,R](including L and R)?

 

Input

The fist line contains a single integer T(about 100), indicating the number of cases.
Each test case begins with two integers L,R(1≤L≤R≤109).

 

Output

For each case, output an integer means the number of “beautiful number”.

 

Sample Input

21 11999999993 999999999

 

Sample Output

102

 

//题意：

给一个数A，a[i]表示它 i 位上的数字，若a[i]>=a[i+1] (1<=i<n)且a[i] % a[j]==0(1<=i<j<=n)，我们称数A为beautiful number。现在给你L,R，求[L,R]上beautiful number的个数。

//思路：

简单的数位DP+记忆化搜索：

定义一个二位DP数组：dp[i][j] : i 表示数位，j 表示该位前一位数字的值。

设第 i 位的值为num，它前一位的值为front，那么若满足front%num==0说明到目前为止是可行的，否则的话有这个前缀的所有数都不满足要求。（如果front<num，那么front%num的值必不可能==0）；

此外，只有首位能取0，即若该数位前面的数位有非0值，那么此位及它后面的数位的值一定不能取0。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <vector>#include <algorithm>using namespace std;int cnt;int bit[20];int dp[20][20];int dfs(int pos, int front, int mod, int flag){if (pos <= 0)return (mod == 0);if (!flag&&dp[pos][front] != -1)return dp[pos][front];int num = flag ? bit[pos] : 9;int ans = 0;for (int i = 0; i <= num; i++){if (i == 0 && pos < cnt&&front != 0)continue;else if (pos == cnt || front == 0){//0不能取if (i == 0 && pos == 1){mod = 1;continue;}elsemod = 0;}else{if (front%i == 0)mod = 0;else{mod = 1;continue;}}ans += dfs(pos - 1, i, mod, flag&&i == num);}if (!flag)dp[pos][front] = ans;return ans;}int solve(int num){cnt = 0;memset(bit, 0, sizeof(bit));while (num){bit[++cnt] = num % 10;num = num / 10;}int ans;if (cnt == 0)ans = 0;elseans = dfs(cnt, 10, 0, 1);return ans;}int main(){int T;scanf("%d", &T);memset(dp, -1, sizeof(dp));while (T--){int l, r;scanf("%d%d", &l, &r);  int ans1 = solve(r);int ans2 = solve(l - 1);printf("%d\n", ans1 - ans2);}return 0;}