【STL】NYOJ 412 Same binary weight (bitset)

题目链接：http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=412

  　　　　　　　　　　　　　　Same binary weight

时间限制：300 ms  |  内存限制：65535 KB 

难度：3

 

描述 

The binary weight of a positive  integer is the number of 1's in its binary representation.for example,the decmial

number 1 has a binary weight of 1,and the decimal number 1717 (which is 11010110101 in binary) has a binary weight

of 7.Give a positive integer N,return the smallest integer greater than N that has the same binary weight as N.N will

be between 1 and 1000000000,inclusive,the result is guaranteed to fit in a signed 32-bit interget.

 

输入

The input has multicases and each case contains a integer N.

输出

For each case,output the smallest integer greater than N that has the same binary weight as N.

样例输入

1717
4
7
12
555555

样例输出

1718
8
11
17
555557

看到问题后没有任何思路，在查看讨论区以后发现是使用STL中的bitset来解决的

大体看完了bitset的用法

或许暴力的方法是可行的，每次+1逐一尝试

#define _CRT_SBCURE_NO_DEPRECATE
#include <bitset>
#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        bitset<32> b(n);
        int nlen = b.count();
        int ilen;
        for(int i = n+1;;i++)
        {
            bitset<32> bb(i);
            ilen = bb.count();
            if(nlen == ilen)
            {
                cout << i << endl;
                break;
            }
        }    
    }
    
    return 0;
}

但是提交后超时了，看来需要找更快的方法，看完别人的题解后来自己找一下规律解决问题

将十进制转化为二进制

4 :00000100

8 :00001000

7 :00000111

11:00001011

12:00001100

17:00010001

1717:0011010110101

1718:0011010110110

1.从右到左发现的第一个01都变成了10

2.发现的第一个01的左边是没有变化的

3.01右边的1都移动到了最右边

（其实我也不知道怎么发现的这么神奇的规律   emmmm...）

一道比较神奇的题目

接下来的工作就简单些了，根据发现的规律来实现代码 

#define _CRT_SBCURE_NO_DEPRECATE
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>

using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        bitset<32> b(n);
        int i,coun = 0;
        for(i = 0;i < 32;i++)
        {
            if(b[i] == 1)
                coun++;
            if(b[i] == 1 && b[i+1] == 0)
            {
                b[i] = 0;
                b[i+1] = 1;
                break;
            }
        }
        int j = i-1;
        for(int i = 0;i <= j;i++)
        {
            if(i < coun-1)
                b[i] = 1;
            else
                b[i] = 0;
        }
        cout << b.to_ulong() << endl;
    }

    
    return 0;
}