nyoj 412 Same binary weight （）

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

/**
    分析： 该题是，求一个大于题目所给的数   &&  其二进制中1的个数与所给的数的二进制中1的个数相同 
    算法1：
        ①、算出的二进制数从左到右遍历
        ②、if  第i位数为1     &&   第i + 1 位数位0： 
                swap (A[i], A[i + 1])
                pos = i
                break
        ③、将第0位到第pos位的所有1移动到最右边 
**/ 

C/C++代码实现（算法1）：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <stack>
#include <queue>
#include <map>

using namespace std;

int n;

int main () {
    while (~scanf ("%d", &n)) {
        int A [10000] = {0}, ans = 0, m = n, pos, k = 0, cnt = 0;
        
        while (m) {
            A [k ++] = m%2;
            m /= 2;
        }        
        for (int i = 0; i < k; ++ i) {
            if (A [i] && !A[i + 1]) {
                swap (A[i], A[i + 1]);
                pos = i;
                break;
            }
            if (A[i]) ++ cnt;
        }
        for (int i = 0; i < pos; ++ i) {
            if (cnt) {
                A [i] = 1;
                cnt --;
            } else  {
                A [i] = 0;
            }
        }
        for (int i = 0; i <= k; ++ i) {
            ans += pow (2, i) * A[i]; 
        }
        printf ("%d
", ans); 
    }
}

算法2：

/**
　　区别于算法一的是计算二进制的算法   ==>  bitset <32> A (n)    (说明：将n转换为二进制储存在A中)
    二进制转换为十进制   ==>   A.to_ulong() 
**/

C/C++实现算法2：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <stack>
#include <queue>
#include <map>
#include <bitset>

using namespace std;

int n;

int main () {
    while (~scanf ("%d", &n)) {
        bitset <32>  A (n); // 将n转化为2进制储存在A中 
        int pos, cnt = 0;
        for (int i = 0; i <= 32; ++ i) {
            if (A [i] && !A[i + 1]) {
                A [i] = 0;
                A [i + 1] = 1;
                pos = i;
                break;
            }
            if (A[i]) ++ cnt;
        }
        for (int i = 0; i < pos; ++ i) {
            if (cnt) {
                A [i] = 1;
                cnt --;
            } else  {
                A [i] = 0;
            }
        }
    
        printf ("%d
", A.to_ulong()); // 将得到的二进制转化为无符号十进制 
    }
}