[poj 2453] An Easy Problem

最新推荐文章于 2022-08-28 15:09:52 发布

转载最新推荐文章于 2022-08-28 15:09:52 发布 · 102 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/whc200305/p/7325262.html

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。通过巧妙的位运算技巧，实现了O(1)时间复杂度的1数量计算，并通过迭代查找满足条件的最小整数。

An Easy Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8371		Accepted: 5009

Description

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".

Input

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.

Output

One integer per line, which is J.

Sample Input

Sample Output

Source

POJ Monthly,zby03

这题就是说给你一个1e6以内的数I,设x为I 在二进制下1的个数.要找到大于n的最小的ans,满足ans在二进制下有x个1.

这一题我似乎是直接水过的...

如果看了Matrix67的帖子,就知道判断一个整数在二进制下有几个1只要O1的时间.

那么,我们不断的n++,判断一下就好了.最坏复杂度O(TI).

 1 #include<cstdio>
 2 using namespace std;
 3 int n,m,cnt;
 4 int calc(int x){
 5     x=(x&1431655765)+((x>>1)&1431655765); 
 6     x=(x&858993459)+((x>>2)&858993459);
 7     x=(x&252645135)+((x>>4)&252645135); 
 8     x=(x&16711935)+((x>>8)&16711935); 
 9     x=(x&65535)+((x>>16)&65535);
10     return x;
11 }
12 int main(){
13     while (scanf("%d",&n)&&n){
14         cnt=calc(n),n++;
15         while (calc(n)!=cnt) n++;
16         printf("%d\n",n);
17     }
18     return 0;
19 }

View Code

转载于:https://www.cnblogs.com/whc200305/p/7325262.html