HDU3652 數字DP 記憶化搜尋

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

100200
1000

sample output

112

2  

題意：找出1~n範圍內含有13並且能被13整除的數字的個數

/*** 

hdu 3652  數字dp（含13且被能被13整除的數） 
題目大意：求出給定區間內的數字含有「13」並且能被13整除的個數 
解題思路：記憶化搜尋。 
dp[i][j][k][z]：i:處理的數字，j:該數對13取模以後的值，k:是否已經包含13,z結尾的數 
*/  
#include #include #include #include using namespace std;  
int dp[10][15][2][10],bit[10];  
int dfs(int pos,int mod,int t,int now,int flag)  
if(!flag)dp[pos][mod][t][now]=ans;  
return ans;  
}  int solve(int n)  
return dfs(len-1,0,0,0,1);  
}  int main()  
return 0;  
}

數字DP與記憶化搜尋 HDU3652

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