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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#696310 | #5422. Perfect Palindrome | hhhhyf | AC ✓ | 3ms | 3784kb | C++20 | 997b | 2024-10-31 22:05:13 | 2024-10-31 22:05:13 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define all(a) a.begin(), a.end()
void print() { cout << '\n'; }
template <typename T, typename...Args>
void print(T t, Args...args) { cout << t << ' '; print(args...); }
using ll = long long;
const int N = 200010;
int dir[4][2] = {
{1, 0}, {0, 1}, {-1, 0}, {0, -1}
};
template <typename T> bool chmax(T &x, T y) { if (y > x) { x = y; return true; } return false; }
template <typename T> bool chmin(T &x, T y) { if (y < x) { x = y; return true; } return false; }
template <typename T = int>
vector<T> readVector(int n) {
vector<T> a(n);
for(T &x: a) cin >> x;
return a;
}
void solve () {
string s;
cin >> s;
int n = s.size();
vector<int> cnt(26);
for (char c: s) {
cnt[c - 'a'] ++;
}
cout << n - *max_element(all(cnt)) << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while(t --) {
solve();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3608kb
input:
2 abcb xxx
output:
2 0
result:
ok 2 number(s): "2 0"
Test #2:
score: 0
Accepted
time: 3ms
memory: 3784kb
input:
11107 lfpbavjsm pdtlkfwn fmb hptdswsoul bhyjhp pscfliuqn nej nxolzbd z clzb zqomviosz u ek vco oymonrq rjd ktsqti mdcvserv x birnpfu gsgk ftchwlm bzqgar ovj nsgiegk dbolme nvr rpsc fprodu eqtidwto j qty o jknssmabwl qjfv wrd aa ejsf i npmmhkef dzvyon p zww dp ru qmwm sc wnnjyoepxo hc opvfepiko inuxx...
output:
8 7 2 8 4 8 2 6 0 3 7 0 1 2 5 2 4 6 0 6 2 6 5 2 5 5 2 3 5 6 0 2 0 8 3 2 0 3 0 6 5 0 1 1 1 2 1 8 1 7 5 3 4 4 1 8 5 5 8 8 6 3 0 2 3 2 1 5 0 0 9 3 3 4 8 4 0 4 2 6 6 0 8 7 4 3 9 3 4 2 5 8 8 8 6 1 4 4 2 7 2 8 6 4 4 8 7 8 4 9 3 8 0 7 7 2 6 0 0 5 4 0 7 5 4 2 1 6 7 5 2 4 4 7 3 3 2 5 4 8 5 0 3 5 1 2 3 0 4 7 ...
result:
ok 11107 numbers