B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5894 Accepted Submission(s): 3389
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <map> 10 #define ll __int64 11 #define mod 1000000007 12 #define dazhi 2147483647 13 #define bug() printf("!!!!!!!") 14 #define M 100005 15 using namespace std; 16 ll bit[100]; 17 ll dp[100][5][13]; 18 ll n; 19 ll functi(int pos,ll pre,int flag ,int m) 20 { 21 if(pos==0) return (pre==2&&m==0); 22 if(flag&&dp[pos][pre][m]!=-1) return dp[pos][pre][m]; 23 ll x=flag ? 9 : bit[pos]; 24 ll ans=0; 25 for(ll i=0;i<=x;i++) 26 { 27 if(pre==2||(pre==1&&i==3)) 28 ans+=functi(pos-1,2,flag||i<x,(m*10+i)%13); 29 else if(i==1) 30 ans+=functi(pos-1,1,flag||i<x,(m*10+i)%13); 31 else 32 ans+=functi(pos-1,0,flag||i<x,(m*10+i)%13); 33 } 34 return dp[pos][pre][m]=ans; 35 } 36 ll fun(ll x) 37 { 38 memset(dp,-1,sizeof(dp)); 39 memset(bit,0,sizeof(bit)); 40 int len=0; 41 while(x) 42 { 43 bit[++len]=x%10; 44 x/=10; 45 } 46 return functi(len,0,0,0); 47 } 48 int main() 49 { 50 while(scanf("%I64d",&n)!=EOF) 51 { 52 printf("%I64d ",fun(n)); 53 } 54 return 0; 55 }