标签：dp

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5179

beautiful number

2
1 11
999999993 999999999

10
2

思路：数位dp模板题~~

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
using namespace std;

int dp[25][25][3];
int bit[25];

int dfs(int pos,int pre,int flag,int z)
{
  if(pos==0)return 1;
  if( !flag && dp[pos][pre][z]!=-1)return dp[pos][pre][z];
  int end=flag?bit[pos]:9;
  int i=0,ans=0;
  if(z==0)i=1;
  for(;i<=end;i++)
  {
       if(pre==0)
         ans+=dfs(pos-1,i,flag&&(i==end),z&&(i==0));
       else
       {
           if( pre>=i && pre%i==0)ans+=dfs(pos-1,i,flag&&(i==end),z&&(i==0));
       }
  }
  if(!flag)dp[pos][pre][z]=ans;
  return ans;
}

int cal(int x)
{
 int len=0;
 memset(bit,0,sizeof(bit));
 memset(dp,-1,sizeof(dp));
 while(x)
 {
   bit[++len]=x%10;
   x/=10;
 }
 return dfs(len,0,1,1);
}

int main()
{
  int l,r;
  int T;
  cin>>T;
  while(T--)
  {
      scanf("%d%d",&l,&r);
      int s1=cal(r);
      int s2=cal(l-1);
      printf("%d\n",s1-s2);
  }
  return 0;
}

hdu 5179 数位dp

标签：dp

原文地址：http://blog.csdn.net/liusuangeng/article/details/44014889