UVA11645 Bits【位运算+大数】

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 adjacentbit. 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

For each test case, you are given an integer number (0 ≤ N ≤ ((263)−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 (startingfrom 1) and Y is the cumulative summation of all adjacent bits from 0 to N.

Sample Input

0

6

15

20

21

22

-1

Sample Output

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

问题链接：UVA11645 Bits。

问题简述：输入正整数n，求0-n中，有多少个连续的11。

问题分析：数学计算方法有点没搞懂，先占个位置。

程序说明：算出的结果超出了long long的范围，用两个long long存储输出。

题记：（略）

AC的C++语言程序如下：

/* UVA11645 Bits */

#include <iostream>
#include <stdio.h>

using namespace std;

const long long MOD = 1e13;

long long a, b;

inline void add(long long x)
{
    b += x;
    a += b / MOD;
    b %= MOD;
}

inline void solve (long long n) {
    long long digit = 1, v = n;

    a = b = 0;

    while(n) {
        add((n >> 2) * digit);
        if((n & 3) == 3)
            add((v & (digit - 1)) + 1);
        digit <<= 1;
        n >>= 1;
    }

    if (a)
        printf("%lld%013lld
", a, b);
    else
        printf("%lld
", b);
}

int main()
{
    long long n;
    int caseno = 0;

    while (scanf("%lld", &n) !=EOF && n >= 0) {
        printf("Case %d: ", ++caseno);
        solve(n);
    }

    return 0;
}