[暑假集训--数位dp]LightOj1032 Fast Bit Calculations

A bit is a binary digit, taking a logical value of either 1 or 0 (also referred to as "true" or "false" respectively). And every decimal number has a binary representation which is actually a series of bits. If a bit of a number is 1 and its next bit is also 1 then we can say that the number has a 1 adjacent bit. And you have to find out how many times this scenario occurs for all numbers up to N.

Examples:

Number         Binary          Adjacent Bits

12                    1100                        1

15                    1111                        3

27                    11011                      2

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer N (0 ≤ N < 2³¹).

Output

For each test case, print the case number and the summation of all adjacent bits from 0 to N.

Sample Input

7

0

6

15

20

21

22

2147483647

Sample Output

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

Case 7: 16106127360

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
22     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
23     return x*f;
24 }
25 LL n,len,l,r;
26 LL f[50][2][50];
27 int d[110];
28 int zhan[50],top;
29 inline LL dfs(int now,int dat,int tot,int fp)
30 {
31     if (now==1)return tot;
32     if (!fp&&f[now][dat][tot]!=-1)return f[now][dat][tot];
33     LL ans=0,mx=fp?d[now-1]:1;
34     for (int i=0;i<=mx;i++)
35     {
36         if (i==0)ans+=dfs(now-1,0,tot,fp&&i==d[now-1]);
37         else ans+=dfs(now-1,1,tot+(dat==1),fp&&i==d[now-1]);
38     }
39     if (!fp)f[now][dat][tot]=ans;
40     return ans;
41 }
42 inline LL calc(LL x)
43 {
44     if (x<=0)return 0;
45     LL xxx=x;
46     len=0;
47     while (xxx)
48     {
49         d[++len]=xxx%2;
50         xxx/=2;
51     }
52     LL sum=0;
53     for (int i=0;i<=d[len];i++)sum+=dfs(len,i,0,i==d[len]);
54     return sum;
55 }
56 main()
57 {
58     int T=read(),cnt=0;
59     while (T--)
60     {
61         r=read();
62         if (r<l)swap(l,r);
63         memset(f,-1,sizeof(f));
64         printf("Case %d: %lld\n",++cnt,calc(r));
65     }
66 }

LightOj 1032