B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1231 Accepted Submission(s): 651
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
meaning of the title
Find numbers that contain 13 and are divisible by 13.
analyze
Divisible by 13, just record the value of %13. So how to find the one containing 13, we need to record the value of the current last digit, so that we can go from 1 -- "13.
dp[i][j][k][z]: i: the processed digit, j: the value after the number is modulo 13, k: whether it already contains the number ending in 13, z
#include <iostream> #include <string.h> #include <algorithm> #include <stdio.h> using namespace std; int dp[12][15][2][10]; int bit[12]; int dfs(int pos,int num,bool t,int e,bool flag) { if(pos==-1)return t&&(num==0); if(!flag && dp[pos][num][t][e]!=-1) return dp[pos][num][t][e]; int end=flag?bit[pos]:9; int ans=0; for(int i=0;i<=end;i++) ans+=dfs(pos-1,(num*10+i)%13,t||(e==1&&i==3),i,flag&&(i==end)); if(!flag)dp[pos][num][t][e]=ans; return ans; } int calc(int n) { int pos=0; while(n) { bit[pos++]=n%10; n/=10; } return dfs(pos-1,0,0,0,1); } intmain () { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)==1) { printf("%d\n",calc(n)); } return 0; }