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。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define FILL(a,b) memset(a,b,sizeof(a))
typedef long long ll;

ll dp[33][2][33];
int n,digit[33];


ll dfs(int pos,int pre,int sum,int doing)
{
    if(pos==-1)return sum;
    if(!doing&&dp[pos][pre][sum]!=-1)return dp[pos][pre][sum];
    ll ans=0;
    int end=doing?digit[pos]:1;
    for(int i=0;i<=end;i++){
        ans+=dfs(pos-1,i,sum+(pre==i&&i==1),doing&&i==end);
    }
    if(!doing)dp[pos][pre][sum]=ans;
    return ans;
}

ll Solve(int n)
{
    int pos=0;
    while(n){
        digit[pos++]=(n&1);
        n>>=1;
    }
    return dfs(pos-1,0,0,1);
}

int main()
{
    FILL(dp,-1);
    int _case,t=1;
    scanf("%d",&_case);
    while(_case--){
        scanf("%d",&n);
        printf("Case %d: %lld
",t++,Solve(n));
    }
    return 0;
}