QOJ.ac
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 |
---|---|---|---|---|---|---|---|---|---|
#74409 | #5422. Perfect Palindrome | magicduck# | AC ✓ | 9ms | 3456kb | C++14 | 971b | 2023-02-01 12:39:08 | 2023-02-01 12:39:09 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
template <typename T> inline void read(T &F) {
int R = 1; F = 0; char CH = getchar();
for(; !isdigit(CH); CH = getchar()) if(CH == '-') R = -1;
for(; isdigit(CH); CH = getchar()) F = F * 10 + CH - 48;
F *= R;
}
inline void file(string str) {
freopen((str + ".in").c_str(), "r", stdin);
freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 30, M = 1e6 + 10;
int cnt[N]; char s[M];
int main() {
//file("");
int T; read(T);
while(T--) {
memset(cnt, 0, sizeof(cnt));
scanf("%s", s + 1); int n = strlen(s + 1);
for(int i = 1; i <= n; i++) cnt[s[i] - 'a']++;
int res = 0;
for(int i = 0; i < 26; i++) res = max(res, cnt[i]);
cout << n - res << '\n';
}
#ifdef _MagicDuck
fprintf(stderr, "# Time: %.3lf s", (double)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 2ms
memory: 3456kb
input:
2 abcb xxx
output:
2 0
result:
ok 2 number(s): "2 0"
Test #2:
score: 0
Accepted
time: 9ms
memory: 3372kb
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