B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2464 Accepted Submission(s): 1346
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
Author
wqb0039
Source
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#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long LL;
/*
int dfs(int i, int s, bool e) {
if (i==-1) return s==target_s;
if (!e && ~f[i][s]) return f[i][s];
int res = 0;
int u = e?num[i]:9;
for (int d = first?1:0; d <= u; ++d)
res += dfs(i-1, new_s(s, d), e&&d==u);
return e?res:f[i][s]=res;
}
~~f为记忆化数组;
~~i为当前处理串的第i位(权重表示法,也即后面剩下i+1位待填数);
~~s为之前数字的状态(如果要求后面的数满足什么状态,也可以再记一个目标状态t之类,for的时候枚举下t);
~~e表示之前的数是否是上界的前缀(即后面的数能否任意填)。
~~for循环枚举数字时,要注意是否能枚举0,以及0对于状态的影响,有的题目前导0和中间的0是等价的,
但有的不是,对于后者可以在dfs时再加一个状态变量z,表示前面是否全部是前导0,也可以看是否是首位,
然后外面统计时候枚举一下位数。It depends.
*/
//dp[i][j][k],i表示位数,j表示余数,k=0表示“不含13”,k=1表示“不含13且末位为1”,k=2表示“含有13”
LL n,dp[40][13][3];
int num[50];
//分别表示当前考虑的位置,前一个数字,当前余数,是否有限制,是否已经出现13
LL dfs(int pos,int pre,int mod,bool limit,bool flag)
{
if(pos==0) return flag&&(mod==0);
if(!limit && flag && dp[pos][mod][0]!=-1) return dp[pos][mod][0];
if(!limit && !flag && pre!=1 && dp[pos][mod][2]!=-1) return dp[pos][mod][2];
if(!limit && !flag && pre==1 && dp[pos][mod][1]!=-1) return dp[pos][mod][1];
int end = limit?num[pos]:9;
LL ans = 0;
for(int i=0;i<=end;i++)
{
ans+=dfs(pos-1,i,(mod*10+i)%13,limit&&(i==end),flag||(pre==1&&i==3));
}
if(!limit)
{
if(flag) dp[pos][mod][0] = ans;
if(!flag && pre!=1) dp[pos][mod][2] = ans;
if(!flag && pre==1) dp[pos][mod][1] = ans;
}
return ans;
}
LL work()
{
int len = 0;
while(n)
{
num[++len] = n%10;
n/=10;
}
return dfs(len,0,0,true,false);
}
int main()
{
memset(dp,-1,sizeof(dp));
while(scanf("%I64d",&n)!=EOF)
{
printf("%I64d\n",work());
}
return 0;
}