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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#858926#9922. Mah-jongGodwangTL 0ms3584kbC++235.9kb2025-01-17 10:06:372025-01-17 10:06:37

Judging History

This is the latest submission verdict.

  • [2025-01-17 10:06:37]
  • Judged
  • Verdict: TL
  • Time: 0ms
  • Memory: 3584kb
  • [2025-01-17 10:06:37]
  • Submitted

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 = 1e6 + 10, M = 1e6 + 10;

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

int tt;
int n;
int a[N];
map<ll,ll > ma[2];
int tempa[100],num[20];
int now;

///////////////////////////////////
ll count()
{
    ll ret=0;
    rep(i,1,8)
    {
        rep(j,1,num[i])
        {
            ret*=10ll;
            ret+=1ll*i;
        }
    }
    return ret;
}
void solve(ll zhi,int add,ll sum)
{
    fill(num+1,num+9,0);
    num[add]=1;
    while (zhi)
    {
        num[zhi%10]++;
        zhi/=10;
    }

    bool flag=1;
    rep(i,1,8)
    {
        if(num[i]>=3)
        {
            flag=0;
            num[i]-=3;
            ma[now][count()]=sum;
            num[i]+=3;
        }
    }
    rep(i,1,6)
    {
        if(num[i]&&num[i+1]&&num[i+2])
        {
            flag=0;
            num[i]--;
            num[i+1]--;
            num[i+2]--;
            ma[now][count()]=sum;
            num[i]++;
            num[i+1]++;
            num[i+2]++;
        }
    }
    if(flag)
    {
        ma[now][count()]=sum;
    }
    
}

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

void init()
{
    now=0;
    ma[0].clear();
    ma[1].clear();
}

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

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--)
    {
        ll ans=0;
        cin>>n;
        rep(i,1,n)
        {
            cin>>a[i];
        }
        init();
        rep(i,1,n)
        {
            ma[now][0]++;
            now^=1;
            ma[now].clear();
            for(auto j:ma[now^1])
            {
                solve(j.first,a[i],j.second);
            }   
            ans+=ma[now][0];
            //
          //  cout<<ans<<endl;
        }
        cout<<ans<<endl;
    }
    


    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3584kb

input:

5
4
1 1 1 1
6
1 2 3 1 2 3
7
6 5 8 7 6 3 2
8
1 2 1 2 1 2 1 3
9
2 2 4 4 1 1 1 3 3

output:

2
5
1
3
2

result:

ok 5 number(s): "2 5 1 3 2"

Test #2:

score: -100
Time Limit Exceeded

input:

100
992
8 1 8 1 2 3 6 6 1 3 1 8 7 7 4 7 7 1 6 6 4 8 3 7 3 5 1 4 4 7 5 7 5 7 4 3 7 5 2 8 7 1 6 3 6 2 4 3 2 3 1 6 3 1 3 2 6 6 7 4 6 1 1 4 6 4 7 7 8 5 6 4 1 5 4 8 2 4 4 2 1 3 5 7 6 8 3 7 6 6 5 6 4 2 5 4 3 7 3 5 5 3 3 2 7 8 2 7 2 4 4 3 4 1 1 3 5 5 4 6 3 3 3 2 6 1 2 6 4 8 8 6 6 8 7 3 1 1 8 8 7 2 5 6 3 5 ...

output:


result: