loj 1032(数位dp)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=25909

思路：好久没接触数位dp了=.=!，搞了这么久！一类记忆化搜索。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define FILL(a,b) memset(a,b,sizeof(a))
 7 typedef long long ll;
 8 
 9 ll dp[33][2][33];
10 int n,digit[33];
11 
12 
13 ll dfs(int pos,int pre,int sum,int doing)
14 {
15     if(pos==-1)return sum;
16     if(!doing&&dp[pos][pre][sum]!=-1)return dp[pos][pre][sum];
17     ll ans=0;
18     int end=doing?digit[pos]:1;
19     for(int i=0;i<=end;i++){
20         ans+=dfs(pos-1,i,sum+(pre==i&&i==1),doing&&i==end);
21     }
22     if(!doing)dp[pos][pre][sum]=ans;
23     return ans;
24 }
25 
26 ll Solve(int n)
27 {
28     int pos=0;
29     while(n){
30         digit[pos++]=(n&1);
31         n>>=1;
32     }
33     return dfs(pos-1,0,0,1);
34 }
35 
36 int main()
37 {
38     FILL(dp,-1);
39     int _case,t=1;
40     scanf("%d",&_case);
41     while(_case--){
42         scanf("%d",&n);
43         printf("Case %d: %lld\n",t++,Solve(n));
44     }
45     return 0;
46 }

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