凉心A题(最小的二进制数)

最新推荐文章于 2024-11-20 11:50:22 发布

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本文探讨了如何通过交换相邻字符或替换特定模式来最小化一个由'0'和'1'组成的正确二进制字符串。文章提供了一个算法示例，展示了如何将所有'1'尽可能地向前移，并统计'0'和'1'的数量，最终输出最小可能的二进制字符串。

String can be called correct if it consists of characters “0” and “1” and there are no redundant leading zeroes. Here are some examples:

“0”, “10”, “1001”. You are given a correct string s. You can perform
two different operations on this string:
swap any pair of adjacent characters (for example, “101” “110”); replace “11” with “1” (for example, “110” “10”).

Let val(s) be such a number that s is its binary representation.
Correct string a is less than some other correct string b iff
val(a) < val(b). Your task is to find the minimum correct string that
you can obtain from the given one using the operations described
above. You can use these operations any number of times in any order
(or even use no operations at all).
Input
The first line contains integer number n (1 ≤ n ≤ 100) — the length of string s. The second line contains the string s consisting
of characters “0” and “1”. It is guaranteed that the string s is
correct.
    Output
    Print one string — the minimum correct string that you can obtain from the given one.

    Examples
    


  Input
 4 1001



  Output
 100



  Input
 1 1



  Output
 1

  Note
  Int he first example you can obtain the answer by the following sequence of operations: "1001"  "1010"  "1100"  "100". In the second example you can't obtain smaller answer no matter what operations you use

思路：将1提前0，然后统计个数.

> #include <cstdio> 
#include <iostream>
using namespace std;
const int N=100; 
int main(){
 int n;
 cin >> n;
 char a[N];
  scanf("%s",a); 
   int zero=0,one=0;
 for(int i=0; a[i]; i++){
  if(a[i] == '0')
   zero++;
  else
   one++;
 }
 if(one > 0){ 
  putchar('1');
  for(int i=0; i<zero; i++)
   putchar('0');} 
 else
  putchar('0');
 return 0;
}

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