CodeForces 844A Diversity (超级超级水唯一需要想得就是开个标记数组吧)

最新推荐文章于 2020-10-03 21:53:16 发布

原创最新推荐文章于 2020-10-03 21:53:16 发布 · 618 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

codefroces 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个字符串处理问题，目标是最少修改次数使字符串包含至少k种不同的字母。文章提供了完整的算法思路及C++实现代码。

                     **A. Diversity**
          time limit per test :1 second
          memory limit per test: 256 megabytes
          input: standard input
          output: standard output

Calculate the minimum number of characters you need to change in the string s, so that it contains at least k different letters, or print that it is impossible.

String s consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.

Input
First line of input contains string s, consisting only of lowercase Latin letters (1 ≤ |s| ≤ 1000, |s| denotes the length of s).

Second line of input contains integer k (1 ≤ k ≤ 26).

Output
Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible.

**Examples**
**input**
yandex
6
**output**
0
**input**
yahoo
5
**output**
1
**input**
google
7
**output**
impossible

**
In the first test case string contains 6 different letters, so we don’t need to change anything.

In the second test case string contains 4 different letters: {‘a’, ‘h’, ‘o’, ‘y’}. To get 5 different letters it is necessary to change one occurrence of ‘o’ to some letter, which doesn’t occur in the string, for example, {‘b’}.

In the third test case, it is impossible to make 7 different letters because the length of the string is 6.

题意
就是给你个字符串然后给你个k代表有不重复的字符求你要达到这个k 最少需要
思路：
求出字符串长度如果比k小直接输出 impossible，如果比k大那就用标记数组来找不重复的最多有多少如果比k大就输出 0;如果比k小输出k-不重复的字符的个数

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

int flag[2000];

int main()
{
    char s[2000];
    int k;
    cin>>s;
    memset(flag,0,sizeof(flag));  //数组初始化
    cin>>k;
    int m=strlen(s);
    if(k>m)

        {
            cout<<"impossible"<<endl;
            return 0;
        }
    int ans=0;
    for(int i=0;i<m;i++)
    {
        if(flag[s[i]]==0)  //标记数组 
        {
            flag[s[i]]=1;
            ans++;
        }
    }
    if(ans>=k)
        cout<<0<<endl;
    else
    cout<<k-ans<<endl;
    return 0;
}