loj 1032 数位dp

题目链接：http://lightoj.com/volume_showproblem.php?problem=1032

思路：数位dp, 采用记忆化搜索， dp[pos][pre][have] 表示 pos处，前一位为pre, 当前有have个满足条件的状态。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

long long dp[34][2][34];
int n, digit[34];

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

long long Solve(int n)
{
    int pos = 0;
    while (n) {
        digit[pos] = n % 2;
        n /= 2;
        pos++;
    }
    return dfs(pos - 1, 0, 0, 1);
}

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