B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4184 Accepted Submission(s): 2397
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
题意：求从1到n含有13且能整除13的数有多少个。
思路：数位dp，找状态转移方程和边界条件就好。dp[i][j][k]含义看代码。
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int dp[22][13][3];
/*
dp[i][j][0]:前i位对13取余为j的不含13的个数。
dp[i][j][1]:前i位对13取余为j的不含13且i+1位是1的个数。
dp[i][j][2]:前i位对13取余为j的含13的个数。
*/
int digit[22];
int dfs(int pos,int mod,int st,bool limit)
{
if(pos==0) return st==2&&mod==0;
if(!limit&&dp[pos][mod][st]!=-1)return dp[pos][mod][st];
int ans=0;
int end=limit?digit[pos]:9;
for(int i=0; i<=end; i++)
if((st==1&&i==3)||st==2)
ans+=dfs(pos-1,(mod*10+i)%13,2,limit&&(i==end));
else if(i==1)
ans+=dfs(pos-1,(mod*10+i)%13,1,limit&&(i==end));
else ans+=dfs(pos-1,(mod*10+i)%13,0,limit&&(i==end));
if(!limit)
dp[pos][mod][st]=ans;
return ans;
}
int get(int x)
{
int bj=0;
while(x)
digit[++bj]=x%10,x/=10;
return dfs(bj,0,0,1);
}
int main()
{
memset(dp,-1,sizeof(dp));
int n;
while(~scanf("%d",&n))
printf("%d\n",get(n));
return 0;
}