uva 11754 Code Feat （中国剩余定理）

UVA 11754

　　一道中国剩余定理加上搜索的题目。分两种情况来考虑，当组合总数比较大的时候，就选择枚举的方式，组合总数的时候比较小时就选择搜索然后用中国剩余定理求出得数。

代码如下：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <set>

 using namespace std;

 typedef long long LL;

 const int N = ;
 const int UB = ;
 int X[N], C, S;
 LL ans[UB], R[N];
 set<int> Y[N];
 #define ITOR iterator

 template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a;}
 void gcd(LL a, LL b, LL &d, LL &x, LL & y) {
     if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
     else d = a, x = , y = ;
 }
 LL lcm(LL a, LL b) { return a / gcd(a, b) * b;}

 LL cal() {
     LL a = X[], r = R[] % a, d, x, y, tmp;
     for (int i = ; i < C; i++) {
         gcd(a, X[i], d, x, y);
         if ((R[i] - r) % d) return -;
         tmp = a * x * ((R[i] - r) / d) + r;
         a = lcm(a, X[i]);
         r = (tmp % a + a) % a;
     }
     return r ? r : a + r;
 }

 void dfs(int p) {
     if (p == C) {
         LL tmp = cal();
         if (~tmp) ans[++ans[]] = tmp;
         return ;
     }
     set<int>::ITOR si;
     for (si = Y[p].begin(); si != Y[p].end(); si++) {
         R[p] = *si;
         dfs(p + );
     }
 }

 void workdfs() {
     LL LCM = ;
     for (int i = ; i < C; i++) LCM = lcm(LCM, X[i]);
     ans[] = ;
     dfs();
     sort(ans + , ans + ans[] + );
     ans[] = (int) (unique(ans + , ans + ans[] + ) - ans - );
     int mk = ans[];
     while (ans[] < S) ans[ans[] + ] = ans[ans[] +  - mk] + LCM, ans[]++;
 }

 bool test(LL x) {
     for (int i = ; i < C; i++) {
         if (Y[i].find(x % X[i]) == Y[i].end()) return false;
     }
     return true;
 }

 void workenum(int mk) {
     set<int>::ITOR si;
     ans[] = ;
     for (int t = ; ans[] < S; t++) {
         for (si = Y[mk].begin(); si != Y[mk].end() && ans[] < S; si++) {
             LL cur = (LL) t * X[mk] + *si;
             if ( cur && test(cur)) ans[++ans[]] = cur;
         }
     }
 }

 int main() {
     while (~scanf("%d%d", &C, &S) && (C || S)) {
         int mk = , tt = , y, k;
         bool enm = false;
         for (int i = ; i < C; i++) {
             scanf("%d%d", X + i, &k);
             Y[i].clear();
             for (int j = ; j <= k; j++) {
                 scanf("%d", &y);
                 Y[i].insert(y);
             }
             if (k * X[i] < k * X[mk]) mk = i;
             tt *= k;
             if (tt > UB) enm = true;
         }
         if (enm) workenum(mk);
         else workdfs();
         ans[] = min(ans[], (LL) S);
         for (int i = ; i <= ans[]; i++) printf("%lld\n", ans[i]);
         puts("");
     }
     return ;
 }

——written by Lyon