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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#356857#5422. Perfect PalindromeInexhaustible#AC ✓4ms3832kbC++141.3kb2024-03-18 14:04:502024-03-18 14:04:51

Judging History

你现在查看的是最新测评结果

  • [2024-03-18 14:04:51]
  • 评测
  • 测评结果:AC
  • 用时:4ms
  • 内存:3832kb
  • [2024-03-18 14:04:50]
  • 提交

answer

#include <set>
#include <map>
#include <list>
#include <queue>
#include <cmath>
#include <time.h>
#include <random>
#include <bitset>
#include <vector>
#include <cstdio>
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <memory.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define mk make_pair
#define fi first
#define se second

inline int read(){
	int t = 0,f = 1;
	register char c = getchar();
	while (c < 48 || c > 57) f = (c == '-') ? (-1) : (f),c = getchar();
	while (c >= 48 && c <= 57) t = (t << 1) + (t << 3) + (c ^ 48),c = getchar();
	return f * t;
}

inline int reads(char *s){
	int t = 0;
	register char c = getchar();
	while (c < 'a' || c > 'z') c = getchar();
	while (c >= 'a' && c <= 'z') s[++t] = c,c = getchar();
	return t;
}

const int N = 1e5 + 10,V = 26;
int T,n,buc[V];
char s[N];

void solve(){
	n = reads(s);
	for (int i = 0;i < V;i++) buc[i] = 0;
	for (int i = 1;i <= n;i++) ++buc[s[i] - 'a'];
	int mx = 0;
	for (int i = 0;i < V;i++) mx = max(mx,buc[i]);
	printf("%d\n",n - mx);
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in","r",stdin);
	// freopen("out.out","w",stdout);
#endif
	T = read();
	while (T--) solve();
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3736kb

input:

2
abcb
xxx

output:

2
0

result:

ok 2 number(s): "2 0"

Test #2:

score: 0
Accepted
time: 4ms
memory: 3832kb

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