UVA 11754 - Code Feat(数论)

UVA 11754 - Code Feat

题目链接

题意：给定一个c个x, y1,y2,y3..yk形式，前s小的答案满足s % x在集合y1, y2, y3 ... yk中

思路：LRJ大白例题，分两种情况讨论

1、全部x之积较小时候，暴力枚举每一个集合选哪个y。然后中国剩余定理求解

2、全部x之积较大时候，选定一个k/x尽可能小的序列，枚举x * t + y (t = 1, 2, 3...)去暴力求解。

代码：

#include <stdio.h>
#include <string.h>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

const int N = 15;
const int M = 105;

int c, s, x[N], k[N], y[N][M], now;
set<int> value[N];
vector<long long> ans;
long long a[N];

void solve_enum() {
	for (int i = 0; i < c; i++) {
		if (c == now) continue;
		value[i].clear();
		for (int j = 0; j < k[i]; j++)
			value[i].insert(y[i][j]);
 	}
 	for (int t = 0; ; t++) {
 		for (int i = 0; i < k[now]; i++) {
 			long long n = (long long)x[now] * t + y[now][i];
 			if (n == 0) continue;
 			bool ok = true;
 			for (int i = 0; i < c; i++) {
 				if (i == now) continue;
 				if (!value[i].count(n % x[i])) {ok = false; break;}
    		}
    		if (ok) {printf("%lld\n", n); if (--s == 0) return;}
   		}
  	}
}

long long exgcd(long long a, long long b, long long &x, long long &y) {
	if (!b) {x = 1; y = 0; return a;}
	long long d = exgcd(b, a % b, y, x);
	y -= a / b * x;
 	return d;
}

long long china() {
	long long M = 1, ans = 0;
 	for (int i = 0; i < c; i++)
 		M *= x[i];
	for (int i = 0; i < c; i++) {
		long long w = M / x[i];
		long long xx, yy;
		exgcd(x[i], w, xx, yy);
		ans = (ans + w * yy * a[i]) % M;
 	}
 	return (ans + M) % M;
}

void dfs(int d) {
	if (d == c) {
		ans.push_back(china());
		return;
 	}
 	for (int i = 0; i < k[d]; i++) {
 		a[d] = y[d][i];
 		dfs(d + 1);
	}
}

void solve_china() {
	ans.clear();
	dfs(0);
	sort(ans.begin(), ans.end());
	long long M = 1;
	for (int i = 0; i < c; i++) M *= x[i];
	for (int i = 0; ; i++) {
		for (int j = 0; j < ans.size(); j++) {
			long long n = M * i + ans[j];
			if (n > 0) {printf("%lld\n", n); if (--s == 0) return;}
  		}
 	}
}

int main() {
	while (~scanf("%d%d", &c, &s) && s || c) {
		now = 0;
		long long sum = 1;
  		for (int i = 0; i < c; i++) {
			scanf("%d%d", &x[i], &k[i]);
			sum *= k[i];
			if (k[i] * x[now] < k[now] * x[i])
				now = i;
			for (int j = 0; j < k[i]; j++)
				scanf("%d", &y[i][j]);
			sort(y[i], y[i] + k[i]);
  		}
  		if (sum > 10000) solve_enum();
  		else solve_china();
  		printf("\n");
	}
	return 0;
}