HDU5179 beautiful number（数位DP）

beautiful number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 607 Accepted Submission(s): 379

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
2
1 11
999999993 999999999

Sample Output
10
2

Source
BestCoder Round #31

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数位DP水题，没什么太大意思。。

#include <cstring>#include <cstdio>#include <string.h>#include <iostream>using namespace std;int num[15];int dp[15][15];//mark==1没前导0 mark==0有前导0int dfs(int pos,int pre,int over,int mark){    if(pos<0)        return 1;    if(dp[pos][pre]!=-1&&!over)        return dp[pos][pre];    int last=over?num[pos]:9;    int ans=0;    for(int i=0;i<=last;i++)    {      if(mark==0||(pre>=i&&i!=0&&pre%i==0))      {          ans+=dfs(pos-1,i,over&&i==last,mark||i);      }    }    if(!over)        dp[pos][pre]=ans;    return ans;}int solve(int n){    int len=0;    while(n)    {        num[len++]=n%10;        n/=10;    }    return dfs(len-1,0,true,0);}int main(){    int t;    scanf("%d",&t);    memset(dp,-1,sizeof(dp));    while(t--)    {        int l,r;        scanf("%d%d",&l,&r);        printf("%d\n",solve(r)-solve(l-1));    }}