RT，第一个是TLE代码，第二个是AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#define MAXN 131100
#define INF 10000000
using namespace std;

char str[MAXN];
int cnt[MAXN][30];
int n, T;

inline int dfs(int l, int r, int col) {
    if (l == r) {
    	if (cnt[r][col] - cnt[l - 1][col] == 0)
    		return 1;
    	return 0;
	}
	int mid = (l + r) >> 1, len = r - l + 1;
	int a = dfs(mid + 1, r, col + 1), b = dfs(l, mid, col + 1);
	return min(len / 2 - (cnt[mid][col] - cnt[l - 1][col]) + a, len / 2 - (cnt[r][col] - cnt[mid][col]) + b);
}

int main() {
    scanf("%d", &T);
    while (T--) {
        scanf("%d %s", &n, str + 1);
        memset(cnt, 0, sizeof(cnt));
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= 26; j++) {
                if (str[i] - 'a' + 1 == j)
                    cnt[i][j] = cnt[i - 1][j] + 1;
                else
                    cnt[i][j] = cnt[i - 1][j];
            }
        }
        printf("%d\n", dfs(1, n, 1));
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#define MAXN 131100
#define INF 10000000
using namespace std;

char str[MAXN];
int cnt[MAXN][30];
int n, T;

inline int dfs(int l, int r, int col) {
    if (l == r) {
    	if (str[r] - 'a' + 1 != col)
    		return 1;
    	return 0;
	}
	int mid = (l + r) >> 1, len = r - l + 1;
	int a = dfs(mid + 1, r, col + 1), b = dfs(l, mid, col + 1);
	int Ans1 = 0, Ans2 = 0;
	for (int i = l; i <= mid; i++) if (str[i] - 'a' + 1 == col) Ans1++;
	for (int i = mid + 1; i <= r; i++) if (str[i] - 'a'+ 1 == col) Ans2++;
	return min(len / 2 - Ans1 + a, len / 2 - Ans2 + b);
}

int main() {
    cin >> T;
    while (T--) {
        cin >> n >> str + 1;
        printf("%d\n", dfs(1, n, 1));
    }
    return 0;
}

是不是memset太慢了？如果是，memset的时间复杂度又该怎么计算？

有没有巨佬救救这位退役近一年的新高一菜鸡？