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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 |
---|---|---|---|---|---|---|---|---|---|
#149170 | #5422. Perfect Palindrome | JadeReaper | AC ✓ | 2ms | 4048kb | C++23 | 894b | 2023-08-24 06:05:45 | 2023-08-24 06:05:48 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
using ll = long long;
using dbl = long double;
// #define int ll
using pii = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vii = vector<pii>;
using vvii = vector<vii>;
using vll = vector<ll>;
#define ff first
#define ss second
#define pb push_back
#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define tc \
int t; \
cin >> t; \
while (t--)
#define fightFight cin.tie(0)->sync_with_stdio(0)
void solve() {
string s;
cin >> s;
const int N = 26;
vi freq(N, 0);
for (auto &i : s) freq[i - 'a']++;
int best = sz(s);
for (auto &i : freq) best = min(best, sz(s) - i);
cout << best << "\n";
}
signed main() {
fightFight;
tc solve();
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3768kb
input:
2 abcb xxx
output:
2 0
result:
ok 2 number(s): "2 0"
Test #2:
score: 0
Accepted
time: 2ms
memory: 4048kb
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