Uva 11754 Code Feat

题意概述：

有一个正整数$N$满足$C$个条件，每个条件都形如“它除以$X$的余数在集合$\{Y_1, Y_2, ..., Y_k\}$中”，所有条件中的$X$两两互质，

你的任务是找出最小的S个解。

数据范围：

$1\leq C\leq9, 1 \leq S \leq 10, X \geq 2, 1 \leq k \leq 100, 0 \leq Y_i \leq X$

分析：

如果每个集合元素个数为1，那么我们直接使用中国剩余定理求解即可。

因此我们想到枚举余数，但是余数的组合最多会有$100^9$种可能，太多了。我们发现加入$k$的积不大的话，使用这种方法是可行的。比如

我们假设每个集合元素都不超过5。而当集合元素增加时，由于可行解是在模$M = \prod X_i$下的在枚举复杂度提高的同时也意味着可行解在

数轴上变得稠密，这意味着我们可以通过从小到大枚举答案并判定的方法解决。为了加快此过程，我们希望不可行的枚举尽快被否定，因此我们

希望首先使用$k$较小并且$X$较大的条件判定，因为这个条件是更强的，于是考虑按照$\frac{k}{X}$从小到大的顺序检验答案$i$是否满足当前

条件。由此问题便可解决。

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <iostream>
#include <assert.h>
#define pi acos(-1.)
using namespace std;
typedef long long ll;
const int int_inf = 0x3f3f3f3f;
const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
const int INT_INF = (int)((1ll << 31) - 1);
const int mod = 1e9 + 9;
const double double_inf = 1e30;
typedef unsigned long long ul;
#pragma comment(linker, "/STACK:102400000,102400000")
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define mp make_pair
#define st first
#define nd second
#define keyn (root->ch[1]->ch[0])
#define lson (u << 1)
#define rson (u << 1 | 1)
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pb push_back
#define type(x) __typeof(x.begin())
#define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
#define FOR(i, s, t) for(int i = (s); i <= (t); i++)
#define ROF(i, t, s) for(int i = (t); i >= (s); i--)
#define dbg(x) cout << x << endl
#define dbg2(x, y) cout << x << " " << y << endl
#define clr(x, i) memset(x, (i), sizeof(x))
#define maximize(x, y) x = max((x), (y))
#define minimize(x, y) x = min((x), (y))
#define low_bit(x) ((x) & (-x))

inline int readint(){
    int x;
    scanf("%d", &x);
    return x;
}

inline int readstr(char *s){
    scanf("%s", s);
    return strlen(s);
}

class cmpt{
public:
    bool operator () (const int &x, const int &y) const{
        return x > y;
    }
};

int Rand(int x, int o){
    //if o set, return [1, x], else return [0, x - 1]
    if(!x) return 0;
    int tem = (int)((double)rand() / RAND_MAX * x) % x;
    return o ? tem + 1 : tem;
}

void data_gen(){
    srand(time(0));
    freopen("in.txt", "w", stdout);
    int times = 100;
    printf("%d\n", times);
    while(times--){
        int r = Rand(1000, 1), a = Rand(1000, 1), c = Rand(1000, 1);
        int b = Rand(r, 1), d = Rand(r, 1);
        int m = Rand(100, 1), n = Rand(m, 1);
        printf("%d %d %d %d %d %d %d\n", n, m, a, b, c, d, r);
    }
}

struct cmpx{
    bool operator () (int x, int y) { return x > y; }
};
int debug = 1;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
//-------------------------------------------------------------------------
int C, S;
ll mt[15][110];
ll X[15], k[15];
ll ans[50];
ll buf[50];
ll w[15], _w[15];
ll M;
int id[15];
pair<pll, int> _buf[15];
const int lim = 1e5;

void egcd(ll a, ll b, ll &d, ll &x, ll &y){
    if(!b){
        d = a, x = 1, y = 0;
        return;
    }
    egcd(b, a % b, d, x, y);
    ll x1 = x, y1 = y;
    x = y1, y = x1 - a / b * y1;
}

void dfs(int pos){
    if(pos == C + 1){
        ll tem = 0;
        FOR(i, 1, C) tem += buf[i] * w[i] % M * _w[i] % M, tem %= M;
        if(!tem) tem = M;
        FOR(i, 0, S - 1) ans[i + S] = tem + M * i;
        sort(ans, ans + 2 * S);
        return;
    }
    FOR(i, 1, k[pos]){
        buf[pos] = mt[pos][i];
        dfs(1 + pos);
    }
}

bool cmp(pair<pll, int> x, pair<pll, int> y){
    return x.st.st * y.st.nd < x.st.nd * y.st.st;
}

void enu_solve(){
    FOR(i, 1, C) _buf[i] = mp(mp(k[i], X[i]), i);
    sort(_buf + 1, _buf + C + 1);
    FOR(i, 1, C) id[i] = _buf[i].nd;
    int cnt = 0;
    int bg = 1;
    while(cnt < S){
        int ok1 = 1;
        FOR(i, 1, C){
            int j = id[i];
            int tem = bg % X[j];
            int sz = k[j];
            int ok = 0;
            FOR(u, 1, sz) if(tem == mt[j][u]) { ok = 1; break; }
            if(!ok) { ok1 = 0; break; }
        }
        if(ok1) ans[cnt++] = bg;
        bg++;
    }
}

void crt_solve(){
    M = 1;
    FOR(i, 1, C) M *= X[i];
    FOR(i, 1, C) w[i] = M / X[i] % M;
    FOR(i, 1, C){
        ll d, x, y;
        egcd(w[i], X[i], d, x, y);
        _w[i] = (x % X[i] + X[i]) % X[i];
    }
    clr(ans, ll_inf);
    dfs(1);
}

void solve(){
    ll num = 1;
    int ok = 1;
    FOR(i, 1, C){
        num *= k[i];
        if(num > lim){
            ok = 0;
            break;
        }
    }
    if(ok) crt_solve();
    else enu_solve();
}

//-------------------------------------------------------------------------
int main(){
    //data_gen(); return 0;
    //C(); return 0;
    debug = 0;
    ///////////////////////////////////////////////////////////////////////////////////////////////////////////////
    if(debug) freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while(~scanf("%d%d", &C, &S) && C){
        FOR(i, 1, C){
            X[i] = readint(), k[i] = readint();
            FOR(j, 1, k[i]) mt[i][j] = readint();
        }
        solve();
        FOR(i, 0, S - 1) printf("%lld\n", ans[i]);
        printf("\n");
    }
    //////////////////////////////////////////////////////////////////////////////////////////////////////////////
    return 0;
}