openjudge-noi-4.6-1455:An Easy Problem

最新推荐文章于 2022-03-24 16:33:37 发布

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最新推荐文章于 2022-03-24 16:33:37 发布

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分类专栏：懵逼了半天终于AC 暴力

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本文探讨了如何找出大于给定正整数I的最小整数J，使得J的二进制形式中1的数量与I相同。通过不断递增并检查每个数的二进制表示，直至找到符合条件的数。

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题意大概是这样的：给一个数I，转化为二进制，找到一个比它大的数，使其1的数量与I相等，输出最小的满足题意的十进制数.

总时间限制:

1000ms

内存限制:

65536kB

描述

As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of '1's in whose binary form is the same as that in the binary form of I.

For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 '1's. The minimum integer, which is greater than "1001110" and also contains 4 '1's, is "1010011", i.e. "83", so you should output "83".

输入

One integer per line, which is I (1 <= I <= 1000000).

A line containing a number "0" terminates input, and this line need not be processed.

输出

One integer per line, which is J.

样例输入

样例输出

来源

POJ Monthly,zby03

思考：反正就是一直把那个数mod2,算出1的个数.然后一个一个加上去，直到1的个数与原数相等.

代码，一次AC：

#include<cstdio>
int n,p;
int main()
{
	int i,q,m,r;
	while(1)
	{
		i=0;
		scanf("%d",&n);
		q=n;
		if(n==0) break;
		while(q)
		{
			if(q%2) i++;
			q/=2;
		}
		m=i;
		while(1)
		{
			i=0;
			n++;
			r=n;
			while(r)
			{
				if(r%2) i++;
				r/=2;
			}
			if(i==m)
			{
				printf("%d\n",n);
				break;
			}
		}
	}
}<cstdio>
int n,p;
int main()
{
	int i,q,m,r;
	while(1)
	{
		i=0;
		scanf("%d",&n);
		q=n;
		if(n==0) break;
		while(q)
		{
			if(q%2) i++;
			q/=2;
		}
		m=i;
		while(1)
		{
			i=0;
			n++;
			r=n;
			while(r)
			{
				if(r%2) i++;
				r/=2;
			}
			if(i==m)
			{
				printf("%d\n",n);
				break;
			}
		}
	}
}