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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#616644#5422. Perfect PalindromeGodwangAC ✓19ms3744kbC++234.8kb2024-10-06 09:30:082024-10-06 09:30:09

Judging History

你现在查看的是最新测评结果

  • [2024-10-06 09:30:09]
  • 评测
  • 测评结果:AC
  • 用时:19ms
  • 内存:3744kb
  • [2024-10-06 09:30:08]
  • 提交

answer

#include <iostream>
using namespace std;
#include <set>
#include <algorithm>
#include <cmath>
#include <map>
#include <cstdio>
#include <string>
#include <cstring>
#include <string.h>
#include <stdlib.h>
#include <iomanip>
#include <fstream>
#include <stdio.h>
#include <stack>
#include <queue>
#include <ctype.h>
#include <vector>
#include <random>
#include<list> 
#define ll long long
#define ull unsigned long long
#define pb push_back
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define lowbit(x) ((x)&(-x))
ll extend_gcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
ll fastpow(ll a, ll n, ll mod)
{
    ll ans = 1;
    a %= mod;
    while (n)
    {
        if (n & 1)
            ans = (ans * a) % mod; //% mod
        a = (a * a) % mod;         //% mod
        n >>= 1;
    }
    return ans;
}

inline void write(__int128 x)
{
    if (x > 9)
    {
        write(x / 10);
    }
    putchar(x % 10 + '0');
}
__int128 sqrt(__int128 m)
{
    __int128 leftt = 0, rightt = ((__int128)1) << 51, ret = -1, mid;
    while (leftt < rightt)
    {
        mid = (leftt + rightt) / 2;
        if (mid * mid > m)
        {
            rightt = mid;
        }    
        else
        {
            leftt = mid + 1;
            ret = mid;
        }
    }
    return ret;
}

const double eps = 1e-6;
int sgn(double x)
{
    if(fabs(x)<eps)
    {
        return 0;
    }
    else return x<0?-1:1;
}

struct Point
{
    double x,y;
    Point()
    {

    }
    Point(double x,double y):x(x),y(y)
    {

    }
    Point operator + (Point B)
    {
        return Point(x+B.x,y+B.y);
    }
    Point operator - (Point B)
    {
        return Point(x-B.x,y-B.y);
    }
    bool operator == (Point B)
    {
        return sgn(x-B.x)==0&&sgn(y-B.y)==0;
    }
    bool operator < (Point B)
    {
        return sgn(x-B.x)<0||(sgn(x-B.x)==0&&sgn(y-B.y)<0);
    }
};
typedef Point Vector;
double Cross(Vector A,Vector B)//叉积
{
    return A.x*B.y-A.y*B.x;
}
double Distance(Point A,Point B)
{
    return hypot(A.x-B.x,A.y-B.y);
}
int Convex_hull(Point *p,int n,Point *ch)
{
    n=unique(p,p+n)-p;
    sort(p,p+n);
    int v=0;

    for(int i=0;i<n;i++)
    {
        while (v>1&&sgn(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1]))<=0)
        {
            v--;
        }
        ch[v++]=p[i];
    }

    int j=v;

    for(int i=n-2;i>=0;i--)
    {
        while (v>j&&sgn(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1]))<=0)
        {
            v--;
        }
        ch[v++]=p[i];
    }
    if(n>1)
    {
        v--;
    }
    return v;
}

int kmp(string s, string p)
{
    int ans = 0, lastt = -1;
    int lenp = p.size();
    vector<int > Next(lenp+3,0);
    rep(i, 1, lenp - 1)
    {
        int j = Next[i];
        while (j && p[j] != p[i])
        {
            j = Next[j];
        }
        if (p[j] == p[i])
        {
            Next[i + 1] = j + 1;
        }
        else
        {
            Next[i + 1] = 0;
        }
    }
    int lens = s.size();
    int j = 0;
    rep(i, 0, lens - 1)
    {
        while (j && s[i] != p[j])
        {
            j = Next[j];
        }
        if (s[i] == p[j])
        {
            j++;
        }
        if (j == lenp)
        {
            ans++;
        }
    }
    return ans;
}

int dir[4][2] =
    {
        {-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 左右上下
// int dir[8][2]={
//         {-1, 0}, {0, 1}, {1, 0}, {0, -1},{-1,-1},{-1,1},{1,-1},{1,1}
// };
        
#define endl '\n'//交互题请删除本行
const ll inf = 1000000000000000000ll;
const ll mod1 = 998244353ll, P1 = 131, mod2 = 1e9 + 7ll, P2 = 13331;
ll inverse(ll x)
{
    return fastpow(x,mod1-2,mod1);
}

const int N = 4e6 + 10, M = 1e6 + 10;

///////////////////////////////////

int tt;

int n;


///////////////////////////////////



///////////////////////////////////

void init()
{

}

///////////////////////////////////

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//交互题请删除本行
    //freopen("ain.txt", "r", stdin); freopen("aout.txt", "w", stdout);
    
    cin>>tt;
    while (tt--)
    {
        map<char,int > ma;
        string s;
        cin>>s;
        for(auto i:s)
        {
            ma[i]++;
        }
        int maxx=0;
        for(auto i:ma)
        {
            maxx=max(maxx,i.second);
        }
        cout<<s.size()-maxx<<endl;
    }
    


    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3464kb

input:

2
abcb
xxx

output:

2
0

result:

ok 2 number(s): "2 0"

Test #2:

score: 0
Accepted
time: 19ms
memory: 3744kb

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