[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

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n,len,l,r;
 LL f[][][];
 int d[];
 int zhan[],top;
 inline LL dfs(int now,int dat,int tot,int fp)
 {
     if (now==)return tot;
     if (!fp&&f[now][dat][tot]!=-)return f[now][dat][tot];
     LL ans=,mx=fp?d[now-]:;
     for (int i=;i<=mx;i++)
     {
         if (i==)ans+=dfs(now-,,tot,fp&&i==d[now-]);
         else ans+=dfs(now-,,tot+(dat==),fp&&i==d[now-]);
     }
     if (!fp)f[now][dat][tot]=ans;
     return ans;
 }
 inline LL calc(LL x)
 {
     if (x<=)return ;
     LL xxx=x;
     len=;
     while (xxx)
     {
         d[++len]=xxx%;
         xxx/=;
     }
     LL sum=;
     for (int i=;i<=d[len];i++)sum+=dfs(len,i,,i==d[len]);
     return sum;
 }
 main()
 {
     int T=read(),cnt=;
     while (T--)
     {
         r=read();
         if (r<l)swap(l,r);
         memset(f,-,sizeof(f));
         printf("Case %d: %lld\n",++cnt,calc(r));
     }
 }

LightOj 1032