HDU 3652 B-number
B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.
Input
Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).
Output
Print each answer in a single line.
Sample Input
13
100
200
1000
Sample Output
1
1
2
2
總算用記憶化搜索搞定了一個難一點的數位dp了！。
AC代碼例如以下：
///記憶化搜素 500MS 272K
#include<iostream>
#include<cstdio>
#include<cstring>
#define mod 13
using namespace std;
int dp[20][15][4];
int num[20];
int dfs(int pos,int mo,int status,bool limit)
{
int i;
//cout<<pos<<"~~~~~~~"<<mo<<"~~~~~"<<status<<"~~~~~~~~"<<limit<<"~~~~~~~"<<dp[pos][status]<<endl;
if(!pos)
return status==2&&mo==0;
if(!limit&&dp[pos][mo][status]!=0) return dp[pos][mo][status];
int end = limit ? num[pos] : 9;
int sum=0;
for(i=0;i<=end;i++)
{
int a=mo;
int flag = status ;
if(flag==0&&i==1) flag=1;
if(flag==1&&i==3) flag=2;
if(flag==1&&i!=1&&i!=3) flag=0;
sum+=dfs(pos-1,(a*10+i)%mod,flag,limit&&i==end);
}
//cout<<"!!!!!!!!!!!"<<sum<<"!!!!!!!!"<<endl;
return limit ?
sum : dp[pos][mo][status] = sum;
}
int _13(int n)
{
int pos=1;
memset(dp,0,sizeof dp);
while (n>0)
{
num[pos++]=n%10;
n/=10;
}
//cout<<"~~~~~~~~~"<<pos-1<<"~~~~~~~~"<<endl;
return dfs(pos-1,0,0,true);
}
int main()
{
int i,j;
int n;
while(~scanf("%d",&n))
{
printf("%d\n",_13(n));
}
return 0;
}
HDU 3652 B-number