hdu3652——B-number(数位dp变形)

2016-12-02 12:52:59来源:网络收集作者:管理员人点击

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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


加了一个还要被13整除的条件,那就在状态数组多加一项表示被13整除的情况,当这项等于0说明能被13整除


#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
#define MAXN 10000000005
#define Mod 10001
using namespace std;
int dight[30],dp[30][3][13];
int dfs(int pos,int s,bool limit,int mod)
{
if(pos==0)
return s==2&&mod==0;
if(!limit&&dp[pos][s][mod]!=-1)
return dp[pos][s][mod];
int end,ret=0;
if(limit)
end=dight[pos];
else
end=9;
for(int d=0;d<=end;++d)
{
int have=s,nmod=(mod*10+d)%13;
if(s==1&&d==3)
have=2;
if(s==0&&d==1)
have=1;
if(s==1&&d!=1&&d!=3)
have=0;
ret+=dfs(pos-1,have,limit&&d==end,nmod);
}
if(!limit)
dp[pos][s][mod]=ret;
return ret;
}
int solve(int a)
{
memset(dight,0,sizeof(dight));
int cnt=1;
while(a!=0)
{
dight[cnt++]=a%10;
a/=10;
}
return dfs(cnt-1,0,1,0);
}
int main()
{
memset(dp,-1,sizeof(dp));
int n;
while(~scanf("%d",&n))
{
printf("%d/n",solve(n));
}
return 0;
}
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