hdu 3652 B-number

B-number

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

Total Submission(s): 4070 Accepted Submission(s): 2321



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

[align=left]Output[/align]
Print each answer in a single line.

[align=left]Sample Input[/align]

13
100
200
1000

[align=left]Sample Output[/align]

1
1
2
2

[align=left]Author[/align]
wqb0039

[align=left]Source[/align]
2010 Asia Regional Chengdu Site —— Online Contest

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| Note

神奇的数位DP：

dp[i][j][0] 表示位数不超过i，余数为j的含13的个数

dp[i][j][1] 表示位数不超过i,余数为j的不含13且最高位为3的个数

dp[i][j][2] 表示位数不超过i,余数为j的不含13的个数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,m,num[10],md[12];
int dp[20][20][20];
int sqr(int x,int y)
{
int t=1;
for (int i=1;i<=y;i++)
t=t*10;
return t;
}
void calc()
{
md[0]=1;
for (int i=1;i<=9;i++)
md[i]=md[i-1]*10;
memset(dp,0,sizeof(dp));
dp[0][0][2]=1;
for (int i=0;i<=9;i++)
{
dp[1][i][0]=0,dp[1][i][2]=1,dp[1][i][1]=0;
}
dp[1][3][1]=1;
for (int i=2;i<=9;i++)
{
int t=md[i-1];
for (int j=0;j<=9;j++)
{
int k=(t*j)%13;
for (int l=0;l<=12;l++)
{

dp[i][(k+l)%13][0]+=dp[i-1][l][0]+(j==1?dp[i-1][l][1]:0);
dp[i][(k+l)%13][1]+=(j==3?dp[i-1][l][2]:0);
dp[i][(k+l)%13][2]+=dp[i-1][l][2]-(j==1?dp[i-1][l][1]:0);
}
}
}
}
int solve(int x)
{
int d[11],len=0;
for(int i=x;i;i/=10) d[len++]=i%10;
d[len]=0;
bool flag=false;
int ans=0,mod=0;
for(int i=len-1;i>=0;mod=(mod+d[i]*md[i])%13,i--)
{
for(int j=0;j<d[i];j++) ans+=dp[i][(13-(mod+j*md[i])%13)%13][0];
if(flag)//如果原数列中有13，那么之后的数就都是合法的
{
for(int j=0;j<d[i];j++) ans+=dp[i][(13-(mod+j*md[i])%13)%13][2];
continue;
}
if(d[i+1]==1 && d[i]>3) ans+=dp[i+1][(13-mod)%13][1];
if(d[i]>1) ans+=dp[i][(13-(mod+md[i])%13)%13][1];
if(d[i+1]==1 && d[i]==3) flag=true;
}
return ans;
}
int main()
{
calc();
while (scanf("%d",&n)!=EOF)
{
int t=solve(n+1);
printf("%d\n",t);
}
}