nyoj412_bitset_

最新推荐文章于 2018-02-18 20:28:19 发布

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最新推荐文章于 2018-02-18 20:28:19 发布

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原文链接：http://www.cnblogs.com/TWS-YIFEI/p/5903750.html

本文介绍了一个算法问题：找到大于给定整数N的最小整数，该整数与N具有相同的二进制权重。文章提供了一段C++代码实现，通过位操作来高效地解决这个问题。

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

样例输出

#include <iostream>
#include <cstdio>
#include <bitset>

using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        bitset<32> b(n);
        int j;
        int cou=0;
        for(int i=0;i<32;i++){
            if(b[i]){
                cou++;
            }
            if(b[i]&&!b[i+1]){
                j=i;
                b[i]=false;
                b[i+1]=true;
                break;
            }
        }
        cou--;

            for(int k=0;k<cou;k++){
                b[k]=true;
            }
            for(int k=cou;k<j;k++){
                b[k]=false;
            }

        printf("%ld\n",b.to_ulong());
        //cout<<b.to_ulong()<<endl();

    }
    return 0;
}