題目鏈接：http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=412

　　　　　　　　　　　　　　Same binary weight

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.

輸入

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

輸出

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

樣例輸入

1717
4
7
12
555555

樣例輸出

1718
8
11
17
555557

看到問題後沒有任何思路，在查看討論區以後發現是使用STL中的bitset來解決的

大體看完了bitset的用法

或許暴力的方法是可行的，每次+1逐一嘗試

 1 #define _CRT_SBCURE_NO_DEPRECATE
 2 #include <bitset>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iostream>
 6 
 7 using namespace std;
 
 8 
 9 int main()
10 {
11     int n;
12     while(scanf("%d",&n)!=EOF)
13     {
14         bitset<32> b(n);
15         int nlen = b.count();
16         int ilen;
17         for(int i = n+1;;i++)
18         {
19             bitset<32> bb(i);
20             ilen = bb.count();
21             if(nlen == ilen)
22             {
23                 cout << i << endl;
24                 break;
25             }
26         }    
27     }
28     
29     return 0;
30 }

但是提交後超時了，看來需要找更快的方法，看完別人的題解後來自己找一下規律解決問題

將十進制轉化為二進制

4 :00000100

8 :00001000

7 :00000111

11:00001011

12:00001100

17:00010001

1717:0011010110101

1718:0011010110110

1.從右到左發現的第一個01都變成了10

2.發現的第一個01的左邊是沒有變化的

3.01右邊的1都移動到了最右邊

（其實我也不知道怎麽發現的這麽神奇的規律 emmmm...）

一道比較神奇的題目

接下來的工作就簡單些了，根據發現的規律來實現代碼

#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;
}

【STL】NYOJ 412 Same binary weight (bitset)