Easy Comparison（ZCMU暑期提高六）

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基础算法-----字符和字符串专栏收录该内容

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本文介绍了一种简单的字符串处理算法：首先对输入字符串进行字典序排序生成新字符串，然后逐字符对比原始字符串与排序后的字符串，最后输出两者不同的字符数量。示例代码使用 C++ 实现，展示了完整的输入输出流程。

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

字符串的简单操作，看懂题肯定过的题看不懂题也能猜到题目意思（我就是）

A题速度太慢了，

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
int ca;
int main()
{
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		int n;
		string s;
		string t;
		s.clear();
		t.clear();
		cin >> n >> s;
		t = s;
		sort(s.begin(), s.end());
		int flag = 0;
		for (int i = 0;s[i];i++)  if (s[i] != t[i]) flag++;
		cout << "Case " << ++ca << ": " << flag << endl;
	}
	return 0;
}