LightOJ 1042 - Secret Origins(数论＆进制转换）

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 number N, 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 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 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

 题意

给一个数，把它转换成二进制。找到一个比它大且二进制１的个数与之一样的最小数。

思路

把这个数拆成二进制，找到第一个０１，再把０１颠倒成１０，之后所有１全丢在最后几位。然而我代码有个迷之错误是错的。

如果是没有０１（如８，７）就在最高位前补个０．

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
int s[100];
long long quick(int x){
	if(x==0)return 1;
    long long  s=1,a=2;
    while(x!=1){
        if(x%2==1)s*=a;
        a=a*a;
        x/=2;
    }
    return a*s;
    }
int main(){
	int i,j,k,n,T,flag;
	long long m;
	scanf("%d",&T);k=T;
	while(T--){
		scanf("%d",&n);m=n;
		int l=-1,num=0;		
		long long sum=0;
		while(1){
			if(n%2==0){
			s[++l]=0;
			if(s[l-1]==1)
			break;
				}
			else{
			s[++l]=1;
			sum+=quick(l);
			num++;
			}
			n/=2;
		}
		m=m+quick(l)-quick(l-1);
		long long cha=(sum-quick(l-1))-quick(num-1)+1;
		m-=cha;
		printf("Case :%d %lld\n",k-T,m);
		}
return 0;
}

made by　罗旅洲