hdu3652 B-number 数位dp

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4376 Accepted Submission(s): 2520

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

和以前的数位dp的模板是一样的，只是怎么表示能被13整除？我们可以用余数来表示状态。

#include<bits/stdc++.h>
using namespace std;

int dp[15][3][13];
int digit[20];
int f(int n) {
int len=0;
while(n) {
digit[len++]=n%10;
n/=10;
}
digit[len]=0;
return len-1;
}
/**
25 13 ==12
120+4%13==117 7
254 13 ==7
*/
int dfs(int pos,int sta,int mod,int lim) {
if(pos<0)return sta==2&&mod==0;
if(!lim&&dp[pos][sta][mod]^-1)return dp[pos][sta][mod];
int len=lim?digit[pos]:9;
int ans=0;
for(int i=0; i<=len; ++i) {
int ts=sta;
if(ts==0&&i==1)ts=1;
else if(ts==1&&i==3)ts=2;
else if(ts==1&&(i!=1&&i!=3))ts=0;
ans+=dfs(pos-1,ts,(mod*10+i)%13,lim&&(i==len));
}
return lim?ans:(dp[pos][sta][mod]=ans);
}
int main() {
int n;
memset(dp,-1,sizeof(dp));
while(~scanf("%d",&n)) {
printf("%d\n",dfs(f(n),0,0,1));
}
return 0;
}