美团20240420春招笔试【硬件开发方向】
编程题:
修改字母只能每次加1,都是小写字母,输入操作k次,问能否恰好k次将字符s修改为和字符串t相同
输入:
3
2 1
aa
ba
2 3
aa
ba
3 101
bbb
aaa
3 // 总共有几次修改字母
2 1 // 字符串长度为2,操作次数为1
aa // 字符串s
ba // 字符串t
2 3 // 字符串长度为2,操作次数为3
aa // 字符串s
ba // 字符串t
3 101 // 字符串长度为3,操作次数为101
bbb // 字符串s
aaa // 字符串t
输出:
// 三次修改结果
Yes
No
Yes
示例代码(不一定全对)
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class Solution {
public:
char correct(char a) {
if (a < 'z') {
return a + 1;
}
else {
return 'a';
}
}
vector<string> correct_string(vector<string> s_vec, vector<string> t_vec, vector<int> n_vec, vector<int> k_vec, int q) {
vector<string> result(q, "No");
vector<vector<int>> cnt_vec(q);
for (int i = 0; i < q; i++) {
int cnt = k_vec[i];
for (int j = 0; j < n_vec[i]; j++) {
if (s_vec[i][j] == t_vec[i][j]) {
cnt_vec[i].push_back(0);
continue;
}
else {
char temp = s_vec[i][j];
int cnt_1 = 0;
while (temp != t_vec[i][j]) {
temp = correct(temp);
cnt_1++;
}
cnt_vec[i].push_back(cnt_1);
}
}
int sum = 0;
for (int s = 0; s < n_vec[i]; s++) {
sum += cnt_vec[i][s];
}
if ((sum - cnt) % 26 == 0) {
result[i] = "Yes";
}
}
return result;
}
};
int main() {
int q;
cin >> q;
int temp = q;
vector<string> s_vec;
vector<string> t_vec;
vector<int> n_vec;
vector<int> k_vec;
while (temp--) {
int n, k;
cin >> n >> k;
n_vec.push_back(n);
k_vec.push_back(k);
string s;
cin >> s;
s_vec.push_back(s);
string t;
cin >> t;
t_vec.push_back(t);
}
Solution sol;
vector<string> res = sol.correct_string(s_vec, t_vec, n_vec, k_vec, q);
for (int i = 0; i < q; i++) cout << res[i] << endl;
}
// 64 位输出请用 printf("%lld")