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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#239504#7688. Alea Iacta Estucup-team2303#0 1ms9764kbC++174.9kb2023-11-04 20:59:092023-11-04 20:59:11

Judging History

你现在查看的是最新测评结果

  • [2023-11-04 20:59:11]
  • 评测
  • 测评结果:0
  • 用时:1ms
  • 内存:9764kb
  • [2023-11-04 20:59:09]
  • 提交

answer

/*
60 + 0 + 100 + 64 = 224.
*/
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#include <bits/stdc++.h>
using namespace std;
//#define int long long
#define ll long long
#define L(i, j, k) for (int i = (j); i <= (k); i++)
#define R(i, j, k) for (int i = (j); i >= (k); i--)
#define pb push_back
#define pii pair<int, int>
namespace FastIO {
    char buf[1 << 22], buf2[1 << 22], a[20], *p1 = buf, *p2 = buf, hh = '\n';
    //一共占了4M 1M=2^20B,2个2M的char数组
    //1char占1B(字节)
    //a用来临时存每个整数,不会超过20位
    int p, p3 = -1;  //p临时指针
    void read() {}
    void print() {}      //为空的时候就会调用这玩意而不是下面那个print,于是就退出了循环
    inline int getc() {  //超快速读入的getchar
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    inline void flush() {
        fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
        // 从buf2写,每个写入的元素占1个字节,写入p3+1个,写到标准输出stdout里面
    }
    template <typename T, typename... T2>
    inline void read(T &x, T2 &...oth) {
        int f = 0;
        x = 0;
        char ch = getc();//更快的读入
        while (!isdigit(ch)) {
            if (ch == '-') f = 1;
            ch = getc();
        }
        while (isdigit(ch)) {
            x = x * 10 + ch - 48;
            ch = getc();
        }
        x = f ? -x : x;
        read(oth...);
    }
    // template <typename T, typename... T2>
    inline void print(int x, char o) {  //没有引用
        if (p3 > 1 << 21)                //接近超空间自动刷新
            flush();
        if (x < 0)
            buf2[++p3] = 45, x = -x;  //45ascii是负号
        do {
            a[++p] = x % 10 + 48;
        } while (x /= 10);
        do {
            buf2[++p3] = a[p];
        } while (--p);
        buf2[++p3] = o;  //换行
        // print(oth...);
    }
}
#define read FastIO::read
#define print FastIO::print
const int N = 7e7 + 9, mod = 998244353;
inline void add(int &x, int y) {x = (x + y) % mod;}
inline void del(int &x, int y) {x = (x - y + mod) % mod;}
int T, x, y, siz, a, b, ca, cb;
int G[2][N], F[109], tmp[N], id = 0;
inline void Write(int X)
{
    print(X, ' ');
    // if(X > 9) Write(X / 10);;
    // putchar(X % 10 + '0');
}
inline void dfs(int fx, int fy, int fa, int fb, int nw)
{
    if(nw == siz + 1) return ;
    int fac = F[nw];
    if(x % fac != 0 && y % fac != 0) {dfs(fx, fy, fa, fb, nw + 1); return ;}
    if(x % fac != 0) swap(x, y), swap(fx, fy);
    if(a % fac != 0) id ^= 1, swap(fa, fb), swap(a, b), swap(ca, cb);
    x /= fac, a /= fac;
    int val = 1, cn = 0;
    L(i, 1, fac)
    {
        L(j, 1, ca) tmp[++cn] = G[id][j] + val - 1;
        val += fx;
    }
    L(i, 1, cn) G[id][i] = tmp[i]; ca = cn;
    fx *= fac, fa *= fac;
//  puts("----------------------");
//  cout << ca << " " << cb << endl;
//  L(i, 1, ca) cout << G[id][i] << " "; cout << endl;
//  L(i, 1, cb) cout << G[id ^ 1][i] << " "; cout << endl;
//  puts("------------------------");
    dfs(fx, fy, fa, fb, nw); return ;
}
inline void work()
{
    ca = cb = 1;
    G[0][1] = 1, G[1][1] = 1; id = 0;
    dfs(1, 1, 1, 1, 1);
    Write(ca);
    L(i, 1, ca)
    {
        Write(G[id][i]);
    }
    FastIO::buf2[++FastIO::p3] = '\n';
    // cout << '\n';
    Write(cb);
    L(i, 1, cb)
    {
        Write(G[id ^ 1][i]);
    }
    FastIO::buf2[++FastIO::p3] = '\n';
    return ;
}
inline void get(int x, int y, int a, int b)
{
    if(x % a != 0 && x % b != 0) swap(x, y);
    if(x % a != 0) swap(a, b);
    Write(a);
    L(i, 1, a)
    {
        Write(i);
    }
    FastIO::buf2[++FastIO::p3] = '\n';

    Write(b);
    int nw = 1;
    L(i, 1, b / y)
    {
        L(j, 1, y)
        {
            Write(nw + j - 1);;
        }
        nw += a;
    }
        FastIO::buf2[++FastIO::p3] = '\n';
 return ;
}
inline void solve()
{
    read(x), read(y);
    ll t = 1ll * x * y, f = sqrt(t); siz = 0;
    ll ff = t;
    L(i, 2, f)
        if(ff % i == 0)
        {
            F[++siz] = i;
            while(ff % i == 0) ff /= i;
        }
    if(ff != 1)
    {
        if(ff == t) {print(0, '\n'), print(0, '\n'); return ;}
        F[++siz] = ff;
    }
    R(i, max(1ll, f), 1)
    {
        if(t % i != 0) continue;
        a = i, b = t / i;
        if(a == x || a == y) continue;
        if(x % a == 0 || y % a == 0 || x % b == 0 || y % b == 0) get(x, y, a, b);
        else work();
        return ;
    }
    print(0, '\n'), print(0, '\n');
    // puts("0"); puts("0");
    return ;
}
signed main()
{
    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);
    read(T);
    L(i, 1, T) solve();
    FastIO::flush();  //手动输出
    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 9764kb

input:

3
2 8
1 9
2 9

output:

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

result:

ok Correct. (3 test cases)

Test #2:

score: -100
Output Limit Exceeded

input:

1
40013 40013

output:

1 1 
1601040169 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98...

result: