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 theworld will never know. Even those who are aware of Zephyr's existence know verylittle about her. For example, no one has any clue as to which time period sheis originally from.
But we do know the story of the first time she set out tochart her own path in the time stream. Zephyr had just finished building hertime machine which she named - "Dokhina Batash". She was making thefinal adjustments for her first trip when she noticed that a vital program wasnot working correctly. The program was supposed to take a numberN, andfind what Zephyr called its Onoroy value.
The Onoroy value of an integer N is the number of onesin its binary representation. For example, the number 13 (11012) hasan Onoroy value of 3. Needless to say, this was an easy problem for the greatmind of Zephyr. She solved it quickly, and was on her way.
You are now given a similar task. Find the first numberafter N which has the same Onoroy value asN.
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 andthe 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 |
題目大意: 給你一個n，求出大于n但是二進制的1的個數跟n相同的數。
例如: 010110->011001
首先資料很大，模拟的話肯定逾時；
看到二進制就要有位運算的聯想，，，如果沒有聯想到位運算的話，，，就很難有思路了
然後通過貪心找規律，，，最後通過代碼實作也需要思維和技巧，是一道品質很高的位運算題，，，
通過觀察我們發現隻要出現第一01時把01改成10 并把次01後面的1都壓到最後面就可以了
例如01(01)(111100)->01(10)(001111), 通過這樣就可以得出答案。
.#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main()
{
int t, ncase=1;
scanf("%d", &t);
while(t--)
{
int n;
scanf("%d", &n);
int cnt=0;
for(int i=1;i<=30;i++)
{
if(n&1)
{
cnt++;
if(((n>>1)&1)==0)
{
n=(n>>1)+1;
n=n<<i;
for(int j=0;j<cnt-1;j++)
{
n=n|(1<<j);
}
break;
}
}
n>>=1;
}
printf("Case %d: %d\n",ncase++,n);
}
return 0;
}