总算用记忆化搜索搞定了一个难一点的数位dp了!!
AC代码如下:
///记忆化搜素 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;
}
原文地址:http://blog.csdn.net/hanhai768/article/details/38064239