C语言之Easy Comparison

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题目

Problem Description

Given a string S, your task is to simulate the following operations. At first, you should sort it by lexicographic order and generate an order string S’. Then, you should compare S’ with S character by character. Finally, you should output the number of different characters between S’ and S.
For example, if the original string S is “ACMICPC”, when we sort this string we could get a new string S’ “ACCCIMP”. Then we compare these two strings, we could find out that only two characters ‘A’ whose index is 0 and ‘C’ whose index is 1 are the same (index from 0), so there are 5 different characters.

Input

The first line of the input contains an integer T (T <= 10), indicating the number of cases. Each case begins with a line containing one integer n (1 <= n <= 100), the length of the string S. The next line contains the string, consisting of characters ‘A’ to ‘Z’.

Output

For each test case, print a line containing the test case number (beginning with 1) and the number of different characters between the two strings as said above.

Sample Input

4
2
AC
6
ACCEPT
7
ACMICPC
10
FZUACMICPC

Sample Output

Case 1: 0
Case 2: 0
Case 3: 5
Case 4: 10

答案

下面展示 实现代码。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#define Max 107
int T,N;
char arr1[Max];
char arr2[Max];
using namespace std; 
int main()
{
    
    	
	while(scanf("%d",&T) != EOF)
	{
    
    
		for(int i = 0; i < T; i++)
		{
    
    
			scanf("%d",&N);
			memset(arr1,0,sizeof(arr1));
			memset(arr2,0,sizeof(arr2));
			for(int j = 0; j < N; j++)
			{
    
    
				cin >> arr1[j];
			}
			strcpy(arr2,arr1);
			sort(arr2,arr2+N);
			int num = 0;//记录有几个字母不同 
			for(int j = 0; j < N; j++)
			{
    
    
				if(arr1[j] != arr2[j])
				{
    
    
					num++;
				}				
			}
			cout << "Case " << i+1 << ": " << num << endl;
		}
	}
	return 0;
 } 

本题感悟

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思路：

这题用了两个字符串进行一一比对，计算有几个不相同的。

以上。