hdu_3562_B-number(记忆化搜索|数位DP)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3652

题意:给你一个n,为比n小的能整除13并数字中有13的数有多少个

题解:记忆化搜索:记dp[i][j][k][l]为当前为第i位i+1位的数为j,余数为k,是否含有13的数的个数,然后搜索下去就得出答案了

 1 #include<cstdio>
 2 #include<cstring>
 3 #define F(i,a,b) for(int i=a;i<=b;i++)
 4 
 5 int dp[12][12][13][2],n,dig[12],len;
 6 
 7 int dfs(int pos,int pre,int mod,bool have,bool inf){
 8     if(!pos)return (have&&!mod);
 9     if(!inf&&dp[pos][pre][mod][have]!=-1)return dp[pos][pre][mod][have];
10     int end=inf?dig[pos]:9,ans=0;
11     F(i,0,end)if(pre==1&&i==3)ans+=dfs(pos-1,i,(mod*10+i)%13,1,inf&&(i==dig[pos]));
12         else ans+=dfs(pos-1,i,(mod*10+i)%13,have,inf&&(i==dig[pos]));
13     if(!inf)dp[pos][pre][mod][have]=ans;
14     return ans;
15 }
16 
17 int main(){
18     memset(dp,-1,sizeof(dp));
19     while(~scanf("%d",&n)){
20         for(len=0;n;)dig[++len]=n%10,n/=10;
21         printf("%d
",dfs(len,0,0,0,1));
22     }
23     return 0;
24 }
View Code


原文地址:https://www.cnblogs.com/bin-gege/p/5696132.html