B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3368 Accepted Submission(s): 1887
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
解题：数位dp
dp[i][j][k]表示前i位 余数为j 状态为k
k取0表示不包含13
1表示i+1位取了1
2表示包含13
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 15; 4 int dp[maxn][maxn][3],b[maxn],n; 5 int dfs(int p,int st,int mod,bool flag) { 6 if(!p) return st == 2 && mod == 0; 7 if(flag && dp[p][mod][st] != -1) return dp[p][mod][st]; 8 int ret = 0,u = flag?9:b[p]; 9 for(int i = 0; i <= u; ++i) { 10 int tmp = (mod*10 + i)%13; 11 if(st == 2 || st == 1 && i == 3) ret += dfs(p-1,2,tmp,flag||(i < u)); 12 else if(i == 1) ret += dfs(p-1,1,tmp,flag||(i < u)); 13 else ret += dfs(p-1,0,tmp,flag||(i < u)); 14 } 15 if(flag) dp[p][mod][st] = ret; 16 return ret; 17 } 18 int solve(int x) { 19 int len = 0; 20 while(x) { 21 b[++len] = x%10; 22 x /= 10; 23 } 24 return dfs(len,0,0,0); 25 } 26 int main() { 27 memset(dp,-1,sizeof dp); 28 while(~scanf("%d",&n)) printf("%d ",solve(n)); 29 return 0; 30 }