2014
11-30

# B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1744    Accepted Submission(s): 966

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

#include<iostream>
#include<cmath>
//#include<vector>
#include<cstdio>
#include<cstring>
//#include<algorithm>
using namespace std;
//#define M 1000000007
int dp[20][20][20][2]= {0};
void getdp()  //打表
{
int i,j,k,l,s=1;
dp[0][0][0][0]=1;
for(i=1; i<=10; i++)
{
for(j=0; j<=9; j++)
{
for(k=0; k<=9; k++)
{
for(l=0; l<13; l++)
{
if(k==3&&j==1)
dp[i][j][(l+(j*s)%13)%13][1]+=dp[i-1][k][l][0];
else dp[i][j][(l+(j*s)%13)%13][0]+=dp[i-1][k][l][0];
dp[i][j][(l+(j*s)%13)%13][1]+=dp[i-1][k][l][1];
}
}
}
s*=10;
}
}
int main()
{
int m,i,j,l;
getdp();
while(scanf("%d",&m)>0)
{
int len=1,len1[20]= {0},x=1,s=m,cg=0;
if(m%13==0) cg=1;
while(m)
{
len1[len++]=m%10;
m/=10;
x*=10;
}
int ans=0,bg=0;
for(i=len-1; i>=1; i--)
{
for(j=0; j<len1[i]; j++)
{
ans+=dp[i][j][(13-((s/x)*x)%13)%13][1];
if(bg||(len1[i+1]==1&&j==3)) ans+=dp[i][j][(13-((s/x)*x)%13)%13][0];  //判断这位是不是13
}
if(len1[i]==3&&len1[i+1]==1) bg=1;  //判断是否在前面已出现13
x/=10;
}
if(cg&&bg) ans++;
printf("%d\n",ans);
}
return 0;
}


1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。