网站建设开发设计营销公司山东推广普通话的意义是什么
原题链接
纯纯水一下；
昨天晚上的比赛，由于半夜打的，精神状态不好，wa了俩发直接睡觉去了，现在白天写写发现，不难，水中水
模拟题吧，题目怎么说就这么作
Kristina has a string ss of length nn, consisting only of lowercase and uppercase Latin letters. For each pair of lowercase letter and its matching uppercase letter, Kristina can get 11 burl. However, pairs of characters cannot overlap, so each character can only be in one pair.
For example, if she has the string ss = "aAaaBACacbE", she can get a burl for the following character pairs:
- s1s1 = "a" and s2s2 = "A"
- s4s4 = "a" and s6s6 = "A"
- s5s5 = "B" and s10s10 = "b"
- s7s7= "C" and s9s9 = "c"
Kristina wants to get more burles for her string, so she is going to perform no more than kk operations on it. In one operation, she can:
- either select the lowercase character sisi (1≤i≤n1≤i≤n) and make it uppercase.
- or select uppercase character sisi (1≤i≤n1≤i≤n) and make it lowercase.
For example, when kk = 2 and ss = "aAaaBACacbE" it can perform one operation: choose s3s3 = "a" and make it uppercase. Then she will get another pair of s3s3 = "A" and s8s8 = "a"
Find maximum number of burles Kristina can get for her string.
Input
The first line of input data contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.
The description of the test cases follows.
The first line of each test case contains two integers nn (1≤n≤2⋅1051≤n≤2⋅105) and kk (0≤k≤n0≤k≤n) — the number of characters in the string and the maximum number of operations that can be performed on it.
The second line of each test case contains a string ss of length nn, consisting only of lowercase and uppercase Latin letters.
It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.
Output
For each test case, print exactly one integer on a separate line: the maximum number of burles that Kristina can get for her string ss.
Example
input
Copy
5
11 2
aAaaBACacbE
2 2
ab
4 1
aaBB
6 0
abBAcC
5 3
cbccb
output
Copy
5 0 1 3 2
Note
The first test case is explained in the problem statement.
In the second test case, it is not possible to get any pair by performing any number of operations.
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<stack>
#include<string>
#include<algorithm>
#include<unordered_map>
#include<map>
#include<cstring>
#include<queue>
#include<set>
#include<stdlib.h>
#define dbug cout<<"hear!"<<endl;
#define rep(a,b) for(int i=a;i<=b;i++)
#define rrep(a,b) for(int j=a;j<=b;j++)
#define per(a,b) for(int i=a;i>=b;i--)
#define pper(a,b) for(int j=a;j>=b;j--)
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 2e5 + 100;
const int INF = 0x3f3f3f3f;
ll gcdd(ll a, ll b)
{if (b) while ((a %= b) && (b %= a));return a + b;
}
const int mod = 998244353;
ll t, n,m,a,b, c, x, k, cnt,ans, ant, sum, q, p, idx;
ll arr[N], brr[N], crr[N];int main()
{cin >> t;while (t--){cin >> n >> m;string s;cin >> s;map<int, int>mp;rep(0, n - 1){mp[s[i]]++;}ans = 0;for (int i = 'A';i <= 'Z';i++){while (mp[i] && mp[i + 32]){ans++;mp[i]--;mp[i + 32]--;}}for (int i = 'A';i <= 'Z';i++){while (m > 0 && mp[i] - 2 >= 0){ans++;mp[i] -= 2;m--;}}for (int i = 'a';i <= 'z';i++){while (m > 0 && mp[i] - 2 >= 0){ans++;mp[i] -= 2;m--;}}cout << ans<<endl;}
}