LightOJ 1032 - Fast Bit Calculations 数位DP

http://www.lightoj.com/volume_showproblem.php?problem=1032

题意：问1~N二进制下连续两个1的个数

思路：数位DP，dp[i][j][k]代表第i位为j，前面已有k个1的个数。

/** @Date    : 2016-12-17-13.51
  * @Author  : Lweleth ([e-mail])
  * @Link    : https://github.com/
  * @Version :
  */
#include<bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset(x, -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int bit[50];
LL dp[50][2][50];
LL dfs(int pos, int pre, int cnt, int ise)
{
    if(pos <= 0)
        return cnt;
    if(!ise && dp[pos][pre][cnt]!=-1)
        return dp[pos][pre][cnt];
    LL ans = 0;
    int len = ise?bit[pos]:1;
    for(int i = 0; i <= len; i++)
    {
        if(pre && i)
            ans += dfs(pos - 1, i, cnt + 1, ise && i == len);
        else
            ans += dfs(pos - 1, i, cnt, ise && i== len);
    }
    if(!ise)
        dp[pos][pre][cnt] = ans;
    return ans;
}

LL sol(int n)
{
    int len = 0;
    MMG(dp);
    while(n)
    {
        bit[++len] = n % 2;
        n /= 2;
    }
    return dfs(len, 0, 0, 1);
}
int main()
{
    int T;
    int cnt = 0;
    cin >> T;
    while(T--)
    {
        int n;
        scanf("%d", &n);
        printf("Case %d: %lld\n", ++cnt, sol(n));
    }
    return 0;
}