Ultraviolet A-11645 bits

Description

Problem J
Bits
Input:Standard Input

Output:Standard output

 

ABitIs 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 it's 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 upN.

 

Examples:

Number binary adjacent bits

12 1100 1

15 1111 3

27 11011 2

 

Input

For each test case, you are given an integer number (0 <= n <= (2 ^ 63)-2), as described in the statement. the last test case is followed by a negative integer in a line by itself, denoting the end of input file.

 

Output

For every test case, print a line of the form "case X: Y", where X is the serial of output (starting from 1) and y is the cumulative summation of all adjacent bits from 0 to n.

Sample input output for sample input

0 6 15 20 21 22 -1

Case 1: 0 Case 2: 2 Case 3: 12 Case 4: 13 Case 5: 13 Case 6: 14

Calculate the total number of two consecutive 1 numbers for each number between 0 and N.

The idea is the same as the idea. It is also the place where two 1 values are enumerated. Note that when the location of 11 is the same as that of the original number, you must consider the situation of + 1 or + 2, and use two large numbers to store the results.

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;long long aa,bb;void cal(long long n) {bb += n;if (bb >= (1000000000000ll)) {aa += bb / (1000000000000ll);bb %= (1000000000000ll);}}int main() {int cas = 1;long long n;long long a,b,c,m;while (cin >> n && n >= 0) {aa = bb = 0;m = 1, a = n;for (int i = 0; i < 62; i++) {cal((n>>2)*m);if ((n & 3) == 3)cal((a&((1ll<<i)-1))+1);m <<= 1;n >>= 1;}printf("Case %d: ", cas++);if(aa) {cout << aa;printf("%012lld\n",bb);}else cout<<bb<<endl;}return 0;}

Ultraviolet A-11645 bits