hdu3652 B-number(数位DP)

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6672    Accepted Submission(s): 3885

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

131002001000

 

Sample Output

1122

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

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还是数位DP，数位DP写着写着就熟能生巧了，只是多几维状态，这题也算基础的数位DP。

题意是让你求1到n中含有13且能被13整除的数的个数。

定下dp[len][flag1][pre][mod]表示还剩长度为len的数字没取，当前取的数是否可被13整除(flag1)，前面一个取的数为pre，当前数模13的余数mod

空间估算dp[10][2][10][15]（开大保险）

配上基本模板，就AC了此题。(*^▽^*)

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int digit[100],dp[100][2][15][10];int dfs(int len,bool flag1,int mod,bool flag2,int pre){if(!len) return (mod==0)&&(flag1);if(!flag2 && dp[len][flag1][mod][pre]!=-1) return dp[len][flag1][mod][pre];int top=9,sum=0;if(flag2) top=digit[len];for(int i=0;i<=top;i++){sum+=dfs(len-1,flag1||(i==3 && pre==1),(mod*10+i)%13,flag2&&(i==top),i);}if(!flag2) dp[len][flag1][mod][pre]=sum;return sum;}int solve(int x){int len=0;while(x){digit[++len]=x%10;x/=10;}return dfs(len,false,0,true,0);}int main(){int n;memset(dp,-1,sizeof(dp));while(~scanf("%d",&n)){printf("%d\n",solve(n));}return 0;}