NYOJ 529 月赛水题

最新推荐文章于 2021-05-03 18:29:53 发布

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最新推荐文章于 2021-05-03 18:29:53 发布

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原文链接：http://www.cnblogs.com/javaspring/archive/2012/04/24/2656357.html

本文探讨了一个关于二进制位操作的问题，目标是通过最小次数的翻转操作将一个整数转换为比其大一的整数。通过实例解析算法实现及优化策略。

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这道题可以说是月赛时最水的题了，，没什么意思。题目：

flip

时间限制：1000 ms | 内存限制：65535 KB

难度：2

描述

Give you a non-negative integer x and an operation. The only operation you can do is to reverse one bit in binary form of x

once(i.e 1->0, 0->1).

your goal is to turn x into x+1.

Calculate the minimum times of operations you need to do.

输入

The first line of the input is an integer T indicates the test cases.
Then follow T lines. Each line is a non-negative integer x as described above, note that 0<=x<10^9.

输出

Output the minimum times of operations you need to do to reach the goal.

样例输入

样例输出

2
1
3

ac代码：

 
#include <iostream>
#include <stdio.h>
#include <string.h>
int numa[50],numb[50];
void fun(int x,int a[50]){
	int k=0;
	while(x){
	  a[k++]=x%2;
	  x/=2;
	}
}
int main(){
	int numcase,n;
	scanf("%d",&numcase);
	while(numcase--){
	  scanf("%d",&n);
	  memset(numa,0,sizeof(numa));
	  memset(numb,0,sizeof(numb));
	  fun(n,numa);
	  fun(n+1,numb);
	  int s=0;
	  for(int i=0;i<50;++i)
		  if(numa[i]!=numb[i])
			  s++;
	  printf("%d\n",s);
	}
	return 0;
}

转载于:https://www.cnblogs.com/javaspring/archive/2012/04/24/2656357.html