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数位dp——hud3652
B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
模板与前一篇博客是一样的,只是dp数组的意义略微做了改动。
f[pos][p][is13]表示到第pos位,除以13的余数为p,是否出现过13的数字个数。
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 using namespace std; 6 int f[20][14][4],bit[20],n,l; 7 int get(int a,int b){ 8 if(b==2)return 2; 9 return a==1?1:a==3&&b==1?2:0;10 }11 int dfs(int pos,int lim,int p,int is13){12 if(pos<=0&&p==0&&is13==2)return 1;13 if(pos<=0)return 0;14 if(!lim&&f[pos][p][is13]!=-1)return f[pos][p][is13];15 int rng=(lim?bit[pos]:9),ret=0;16 for(int i=0;i<=rng;i++)17 ret+=dfs(pos-1,lim&&(i==rng),(p*10+i)%13,get(i,is13));18 if(!lim)f[pos][p][is13]=ret;19 return ret;20 }21 22 int main(){23 while(scanf("%d",&n)!=EOF){24 for(l=0;n;)bit[++l]=n%10,n/=10;25 memset(f,-1,sizeof(f));26 printf("%d\n",dfs(l,1,0,0));27 }28 }
数位dp——hud3652
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