LightOJ-1032-Fast Bit Calculations

A bit is a binary digit, taking a logical value of either1Or0(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 is1And its next bit is also1Then we can say that the number has1Adjacent bit. And you have to find out how many times this scenario occurs for all numbers upN.

Examples:

Number Binary Adjacent Bits

12 1100 1

15 1111 3

27 11011 2

Input

Input starts with an integerT (≤ 10000), Denoting the number of test cases.

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

Output

For each test case, print the case number and the summation of all adjacent bits from0ToN.

Sample Input	Output for Sample Input
7 0 6 15 20 21 22 2147483647	Case 1: 0 Case 2: 2 Case 3: 12 Case 4: 13 Case 5: 13 Case 6: 14 Case 7: 16106127360

Similar to 1140

#include 
 
  #include 
  
   #include 
   
    #include 
    
     using namespace std;typedef long long ll;vector
     
       digit;int n;ll pow2[32];ll dp[32][2];ll getSum(int pos){    ll res = 0;    for(int i = pos; i >= 0; i--){        res *= 2;        res += digit[i];    }    return res+1;}ll dfs(int pos,int pre,int zero,int done){    if(pos==-1) return 0;    if(!zero && !done && ~dp[pos][pre]) return dp[pos][pre];    ll res = 0;    int end = done?digit[pos]:1;    for(int i = 0; i <= end; i++){        if(pre==1&&i==1){            if(done&&i==end){                res += getSum(pos-1);            }else{                res += pow2[pos];            }        }        res += dfs(pos-1,i,zero&&i==0,done&&i==end);    }    if(!done && !zero) dp[pos][pre] = res;    return res;}ll solve(ll x){    digit.clear();    while(x){        digit.push_back(x%2);        x /= 2;    }    return dfs(digit.size()-1,0,1,1);}void init(){    pow2[0] = 1;    for(int i = 1; i <= 31; i++){        pow2[i] = pow2[i-1]*2;    }    memset(dp,-1,sizeof dp);}int main(){    int ncase,T=1;    cin >> ncase;    init();    while(ncase--){        cin >> n;        printf("Case %d: %lld\n",T++,solve(n));    }    return 0;}