Fast Bit Calculations LightOJ

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

详解：http://www.cnblogs.com/jianglangcaijin/archive/2012/10/10/2719029.html

好厉害，能这样找递推公式!!!!!!

F[ i ]表示[ 0，(2^i)-1 ]这个区间上的数字的adjacent bits之和。

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

int n;
ll F[40];

void Inite(){
    F[1]=0;F[2]=1;
    for(int i=3;i<=31;i++) F[i]=F[i-1]*2+(1<<(i-2));
}

ll Solve(int x){
    ll ans=0;
    for(int i=31;i>=0;i--){
        if(x&(1<<i)){
            x=x-(1<<i);
            ans=ans+F[i]+max(0,x-(1<<(i-1))+1);
        }
    }
    return ans;
}

int main()
{   Inite();
    
    int kase;
    cin>>kase;
    for(int t=1;t<=kase;t++){
        cin>>n;
        printf("Case %d: %lld
",t,Solve(n));
    }
    return 0;
}

 DP模式：

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

int n,num[40];
ll dp[40][40][2];

ll DFS(int pos,int sum,int pre,bool F){
    if(pos==-1) return sum;
    if(!F&&dp[pos][sum][pre]!=-1) return dp[pos][sum][pre];
    
    
    int maxv=F?num[pos]:1;
    ll ans=0;
    for(int i=0;i<=maxv;i++){
        if(pre&&i) ans=ans+DFS(pos-1,sum+1,i,F&&i==maxv);
        else ans=ans+DFS(pos-1,sum,i,F&&i==maxv);
    }
    
    if(!F) dp[pos][sum][pre]=ans;
    return ans;
}

ll Solve(int temp){
    memset(dp,-1,sizeof(dp));
    int cnt=0;
    while(temp){
        num[cnt++]=temp%2;
        temp/=2;
    }
    return DFS(cnt-1,0,0,true);
}

int main()
{   int kase;
    cin>>kase;
    for(int t=1;t<=kase;t++){ cin>>n; printf("Case %d: %lld
",t,Solve(n)); }
    return 0;
}