B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2668 Accepted Submission(s): 1467
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
数位DP模板题、只不过好久好久没写数位DP了、都忘了
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int bit[15]; int dp[15][15][3]; int dfs(int pos,int s1,int s2,bool limit) //s1代表余数,s2代表状态 { if(pos==-1) { return (s1==0 && s2==2); } if(!limit && dp[pos][s1][s2]!=-1) return dp[pos][s1][s2]; int ans=0; int end=limit?bit[pos]:9; for(int i=0;i<=end;i++) { int nows2=0; if(s2==0) { if(i==1) nows2=1; } else if(s2==1) { if(i==1) nows2=1; else if(i==3) nows2=2; } else if(s2==2) nows2=2; ans+=dfs(pos-1,(s1*10+i)%13,nows2,limit && i==end); } if(!limit) dp[pos][s1][s2]=ans; return ans; } int cal(int n) { int len=0; while(n) { bit[len++]=n%10; n/=10; } return dfs(len-1,0,0,1); } int main() { int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)!=EOF) { printf("%d ",cal(n)); } return 0; }