LightOJ 1042 Secret Origins【位运算】

1042 - Secret Origins

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Time Limit: 0.5 second(s)Memory Limit: 32 MB

This is the tale of Zephyr, the greatest time traveler the world will never know. Even those who are aware of Zephyr's existence know very little about her. For example, no one has any clue as to which time period she is originally from.

But we do know the story of the first time she set out to chart her own path in the time stream. Zephyr had just finished building her time machine which she named - "Dokhina Batash". She was making the final adjustments for her first trip when she noticed that a vital program was not working correctly. The program was supposed to take a number N, and find what Zephyr called its Onoroy value.

The Onoroy value of an integer N is the number of ones in its binary representation. For example, the number 13 (1101₂) has an Onoroy value of 3. Needless to say, this was an easy problem for the great mind of Zephyr. She solved it quickly, and was on her way.

You are now given a similar task. Find the first number after N which has the same Onoroy value as N.

Input

Input starts with an integer T (≤ 65), denoting the number of test cases.

Each case begins with an integer N (1 ≤ N ≤ 10⁹).

Output

For each case of input you have to print the case number and the desired result.

Sample Input

Output for Sample Input

5

23

14232

391

7

8

Case 1: 27

Case 2: 14241

Case 3: 395

Case 4: 11

Case 5: 16

题意：

给出一个数字，让你求大于这个数，而且二进制中1的个数和这个数的相等的最小的数。

题解：

不知道怎么写题解了，确实没做出来，只想到了一部分怎么做，然后没能够实现，找了大神的做法，瞬间感觉自己什么都不会......

#include<cstdio>#include<cstring>#include<cmath>using namespace std;typedef long long ll;int main(){ll t;//freopen("shuju.txt","r",stdin);scanf("%lld",&t);for(ll k=1;k<=t;++k){ll n;scanf("%lld",&n);ll x=n&-n,y=n+x;n=((n&~y)/x>>1)|y;printf("Case %lld: %lld\n",k,n);}return 0;}

贪心的思路是可以解决的：

找到最后的一个1，然后向前找第一个0，找到后，向后找第一个1，两者交换位置，然后把找到0，然后把这个1 之后的所有的1 都放在最后边，得到的数就是需要的数....

（纸上谈兵...）