链接: http://acm.hdu.edu.cn/showproblem.php?pid=5179
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 860 Accepted Submission(s): 559
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
题意：求[a,b]内的数的前一位都是后一位数的倍数的数的个数
分析：用模板
不同的是 用dp[pos][pre] （pre代表的是前一位数） 而不是以前用的状态0,1,2了
并且还要判断是否有 前导0
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <sstream>
#include <string>
#include <set>
#include <queue>
using namespace std;
#define mem(a,n) memset(a,n,sizeof(a))
#define pb(x) push_back(x)
typedef long long LL;
typedef unsigned long long ull;
const int mod=1e9+7;
const double eps=1e-6;
const LL INF=0x3f3f3f3f;
const int N=200+5;
int dp[20][20],digit[20];
int dfs(int pos,int pre,bool lead,bool bounded)
///lead==1代表有前导0 lead==0则没有
{
if(pos==0) return 1;
int& ret=dp[pos][pre];
if(!bounded&&ret!=-1)
return ret;
int ans=0;
int end=bounded?digit[pos]:9;
for(int i=0; i<=end; i++)
{
if(lead||i&&pre>=i&&pre%i==0)
ans+=dfs(pos-1,i,lead&&!i,bounded&&i==end);
}
if(!bounded&&!lead) ret=ans;///不是上界且没有前导0的
return ans;
}
int cal(int x)
{
int len=0;
while(x)
{
digit[++len]=x%10;
x/=10;
}
return dfs(len,0,true,true);
}
int main()
{
int a,b,T;
mem(dp,-1);
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&a,&b);
printf("%d\n",cal(b)-cal(a-1));
}
return 0;
}
