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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#740132 | #5422. Perfect Palindrome | Zawos# | AC ✓ | 4ms | 3804kb | C++14 | 782b | 2024-11-13 01:42:31 | 2024-11-13 01:42:32 |
Judging History
answer
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
using namespace std;
using namespace __gnu_pbds;
using ll=long long;
using ld=long double;
using vi=vector<int>;
template<class T> using oset =tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update> ;
//上
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while(t--){
string s;
cin >> s;
vector<int> occ(26);
int n = s.size();
for(int i = 0; i < n; i++){
occ[s[i]-'a']++;
}
int mx =*max_element(occ.begin(),occ.end());
cout <<(int)s.size()-mx<<'\n';
}
}
Details
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Test #1:
score: 100
Accepted
time: 1ms
memory: 3600kb
input:
2 abcb xxx
output:
2 0
result:
ok 2 number(s): "2 0"
Test #2:
score: 0
Accepted
time: 4ms
memory: 3804kb
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