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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 |
---|---|---|---|---|---|---|---|---|---|
#128653 | #5422. Perfect Palindrome | PetroTarnavskyi# | AC ✓ | 6ms | 3752kb | C++17 | 692b | 2023-07-21 14:34:54 | 2023-07-21 14:34:54 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))
typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
vector<int> cnt(26);
for (char c : s) {
cnt[c - 'a']++;
}
cout << SZ(s) - *max_element(ALL(cnt)) << "\n";
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3448kb
input:
2 abcb xxx
output:
2 0
result:
ok 2 number(s): "2 0"
Test #2:
score: 0
Accepted
time: 6ms
memory: 3752kb
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