Description

Problem J
Bits
Input: Standard Input

Output: Standard Output

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 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 up toN.

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

 题意：统计0-n中每个数存在两个连续的1的总个数。

思路：和的思路是一样的，也是枚举两个1的位置。 注意当11出现的位置和原数一样的时候，还要考虑+1或者+2的情况，还有就是用两个较大的数来储存结果。

#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;
}

UVA - 11645 Bits