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 = 11;
const int UB = 11111;
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 = 1, y = 0;
}
LL lcm(LL a, LL b) { return a / gcd(a, b) * b;}

LL cal() {
    LL a = X[0], r = R[0] % a, d, x, y, tmp;
    for (int i = 1; i < C; i++) {
        gcd(a, X[i], d, x, y);
        if ((R[i] - r) % d) return -1;
        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[0]] = tmp;
        return ;
    }
    set<int>::ITOR si;
    for (si = Y[p].begin(); si != Y[p].end(); si++) {
        R[p] = *si;
        dfs(p + 1);
    }
}

void workdfs() {
    LL LCM = 1;
    for (int i = 0; i < C; i++) LCM = lcm(LCM, X[i]);
    ans[0] = 0;
    dfs(0);
    sort(ans + 1, ans + ans[0] + 1);
    ans[0] = (int) (unique(ans + 1, ans + ans[0] + 1) - ans - 1);
    int mk = ans[0];
    while (ans[0] < S) ans[ans[0] + 1] = ans[ans[0] + 1 - mk] + LCM, ans[0]++;
}

bool test(LL x) {
    for (int i = 0; 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[0] = 0;
    for (int t = 0; ans[0] < S; t++) {
        for (si = Y[mk].begin(); si != Y[mk].end() && ans[0] < S; si++) {
            LL cur = (LL) t * X[mk] + *si;
            if ( cur && test(cur)) ans[++ans[0]] = cur;
        }
    }
}

int main() {
    while (~scanf("%d%d", &C, &S) && (C || S)) {
        int mk = 0, tt = 1, y, k; 
        bool enm = false;
        for (int i = 0; i < C; i++) {
            scanf("%d%d", X + i, &k);
            Y[i].clear();
            for (int j = 1; 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[0] = min(ans[0], (LL) S);
        for (int i = 1; i <= ans[0]; i++) printf("%lld
", ans[i]);
        puts("");
    }
    return 0;
}

——written by Lyon