题目链接：uva 11754 - Code Feat

题目大意：求一个数N，给出C和S，表示有C个条件，每个条件有X 和 k，然后是该个条件的k个yi，即NmodX=yj,输出满足的最小的S个N，要求正整数。

解题思路：total为所有的k的乘积，也就是可以作为一组完整限定条件的可能数，当个确定条件可以用中国剩余定理处理。但是如果total太大的话，处理的情况比较多。不过total数大的时候，可以通过枚举N来判断，找到一组k/x最小的最为枚举基准，然后判断即可。

#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         using namespace std; typedef long long ll; const int maxc = 15; const int maxk = 105; const int limit = 10000; ll total; int C, S, X[maxc], k[maxc]; int bestc, Y[maxc][maxk]; set
        
          vis[maxc]; void init () { total = 1; bestc = 0; for (int i = 0; i < C; i++) { scanf("%d%d", &X[i], &k[i]); total *= k[i]; for (int j = 0; j < k[i]; j++) scanf("%d", &Y[i][j]); sort(Y[i], Y[i] + k[i]); if (k[i] * X[bestc] < k[bestc] * X[i]) bestc = i; } } void solveEnum (int s) { for (int i = 0; i < C; i++) { if (i == s) continue; vis[i].clear(); for (int j = 0; j < k[i]; j++) vis[i].insert(Y[i][j]); } for (int t = 0; S; t++) { for (int i = 0; i < k[s]; i++) { ll n = (ll)X[s] * t + Y[s][i]; if (n == 0) continue; bool flag = true; for (int c = 0; c < C; c++) { if (c == s) continue; if (!vis[c].count(n%X[c])) { flag = false; break; } } if (flag) { printf("%lld\n", n); if (--S == 0) break; } } } } int a[maxc]; vector
         
           sol; 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); } } int china (int n, int* s, 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; } 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 solveChina () { sol.clear(); dfs(0); sort(sol.begin(), sol.end()); ll M = 1; for (int i = 0; i < C; i++) M *= X[i]; for (int i = 0; S; 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 () { while (scanf("%d%d", &C, &S) == 2 && C + S) { init(); if (total > limit) solveEnum(bestc); else solveChina(); printf("\n"); } return 0; }