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CRT Problem!!!
By Ismail_Alrifai, history, 10 months ago, In English

Hello codeforces!

I need help with this problem: UVa_11754

If all Ki = 1, we can use Chinese Remainder Theorem directly, but the number of possible states is (Π Ki = 100^9)

I believe there is a brute_force solution but i couldn't find it ):

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10 months ago, # |
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Auto comment: topic has been updated by Ismail_Alrifai (previous revision, new revision, compare).

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10 months ago, # |
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The number of possible states should actually be bounded by $$$2^{32}$$$ as the problem statement guarantees that the product of $$$X$$$s fit into a 32-bit integer. That said, I believe you should be able to partition the congruences into two disjoint sets (of roughly equal number of states, so upper bounded by  $$$2^{18}$$$) and generate all possible solutions in each set and use some kind of meet in the middle technique.

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    10 months ago, # ^ |
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    But the Ki <= 100, i meant in "possible state" as all combinations of Yij, try all Yij from each clue so the total 100 ^ 9.

    if i missed some thing can you please explain how (possible states should actually be bounded by 2^32)

    thank you :)

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      10 months ago, # ^ |
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      k is less than x and they say "the product of the X’s will fit into a 32-bit integer"

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10 months ago, # |
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it can be solved by two ways: enum the x and test whether it fits the conditions. dfs search all remainder and use chinese remainder.

luckily, there is a code by RuJia.

// UVa11754 Code Feat
// 刘汝佳
typedef long long LL;

// 即使a, b在int范围内，x和y有可能超出int范围
void gcd(LL a, LL b, LL& d, LL& x, LL& y) {
  if (!b) d = a, x = 1, y = 0;
  else gcd(b, a % b, d, y, x), y -= x * (a / b);
}

// n个方程：x=a[i](mod m[i]) (0<=i<n)
LL china(int n, int* a, int* m) {
  LL M = 1, d, y, x = 0;
  for (int i = 0; i < n; i++) M *= m[i];
  for (int i = 0; i < n; i++) {
    LL w = M / m[i];
    gcd(m[i], w, d, d, y);
    x = (x + y * w * a[i]) % M;
  }
  return (x + M) % M;
}

#include <algorithm>
#include <cstdio>
#include <set>
#include <vector>
using namespace std;

const int maxc = 9, maxk = 100, LIMIT = 10000;
set<int> values[maxc];
int C, X[maxc], k[maxc], Y[maxc][maxk];

void solve_enum(int S, int bc) {
  for (int c = 0; c < C; c++)
    if (c != bc) {
      values[c].clear();
      for (int i = 0; i < k[c]; i++) values[c].insert(Y[c][i]);
    }
  for (int t = 0; S != 0; t++) {
    for (int i = 0; i < k[bc]; i++) {
      LL n = (LL)X[bc] * t + Y[bc][i];
      if (n == 0) continue;  // 只输出正数解
      bool ok = true;
      for (int c = 0; c < C; c++)
        if (c != bc)
          if (!values[c].count(n % X[c])) {
            ok = false;
            break;
          }
      if (ok) {
        printf("%lld\n", n);
        if (--S == 0) break;
      }
    }
  }
}

int a[maxc];  // 搜索对象，用于中国剩余定理
vector<LL> sol;

void dfs(int dep) {
  if (dep == C)
    sol.push_back(china(C, a, X));
  else
    for (int i = 0; i < k[dep]; i++) a[dep] = Y[dep][i], dfs(dep + 1);
}

void solve_china(int S) {
  sol.clear();
  dfs(0);
  sort(sol.begin(), sol.end());

  LL M = 1;
  for (int i = 0; i < C; i++) M *= X[i];

  vector<LL> ans;
  for (int i = 0; S != 0; i++) {
    for (int j = 0; j < sol.size(); j++) {
      LL n = M * i + sol[j];
      if (n > 0) {
        printf("%lld\n", n);
        if (--S == 0) break;
      }
    }
  }
}

int main() {
  int S;
  while (scanf("%d%d", &C, &S) == 2 && C) {
    LL tot = 1;
    int bestc = 0;
    for (int c = 0; c < C; c++) {
      scanf("%d%d", &X[c], &k[c]);
      tot *= k[c];
      for (int i = 0; i < k[c]; i++) scanf("%d", &Y[c][i]);
      sort(Y[c], Y[c] + k[c]);
      if (k[c] * X[bestc] < k[bestc] * X[c]) bestc = c;
    }
    if (tot > LIMIT)
      solve_enum(S, bestc);
    else
      solve_china(S);
    printf("\n");
  }
  return 0;
}
/*
算法分析请参考: 《入门经典训练指南-升级版》2.3.2节 例题10
*/
// Accepted 10ms 2326 C++5.3.0 2020-12-09 17:27:16 25828703
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