hdu3652(数位dp)
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B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4184 Accepted Submission(s): 2397
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
题意:求从1到n含有13且能整除13的数有多少个。
思路:数位dp,找状态转移方程和边界条件就好。dp[i][j][k]含义看代码。
#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>using namespace std;typedef long long ll;int dp[22][13][3];/* dp[i][j][0]:前i位对13取余为j的不含13的个数。 dp[i][j][1]:前i位对13取余为j的不含13且i+1位是1的个数。 dp[i][j][2]:前i位对13取余为j的含13的个数。*/int digit[22];int dfs(int pos,int mod,int st,bool limit){ if(pos==0) return st==2&&mod==0; if(!limit&&dp[pos][mod][st]!=-1)return dp[pos][mod][st]; int ans=0; int end=limit?digit[pos]:9; for(int i=0; i<=end; i++) if((st==1&&i==3)||st==2) ans+=dfs(pos-1,(mod*10+i)%13,2,limit&&(i==end)); else if(i==1) ans+=dfs(pos-1,(mod*10+i)%13,1,limit&&(i==end)); else ans+=dfs(pos-1,(mod*10+i)%13,0,limit&&(i==end)); if(!limit) dp[pos][mod][st]=ans; return ans;}int get(int x){ int bj=0; while(x) digit[++bj]=x%10,x/=10; return dfs(bj,0,0,1);}int main(){ memset(dp,-1,sizeof(dp)); int n; while(~scanf("%d",&n)) printf("%d\n",get(n)); return 0;}
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