Codeforces Round #360 (Div. 2) D. Remainders Game 中国剩余定理

题目链接：

题目

D. Remainders Game
time limit per test 1 second
memory limit per test 256 megabytes

问题描述

Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value . There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value for any positive integer x?

Note, that means the remainder of x after dividing it by y.

输入

The first line of the input contains two integers n and k (1 ≤ n,  k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).

输出

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

样例

input
4 5
2 3 5 12
output
Yes

input
2 7
2 3
output
No

题意

给你n个数和一个k，求x%k的值，没有告诉你x是多少，只告诉你能够计算x%ci的值。问能不能根据这n次测试唯一确定x%k。

题解

思路1：
结论：无法唯一确定x%k <==> k不能整除lcm(c1,...,cn)；
充分性：
无法唯一确定x%k --> 存在两个数x1,x2对于任意的ci取余的值都相等，对k取值的值却不等。

• x1,x2对任意的ci取余都相等 --> ci|(x1-x2) --> lcm(ci)|(x1-x2).
• x1,x2对k取余的值不等 --> k不整除(x1-x2) --> k不整除lcm(ci)

必要性：
我们令x1=2*lcm(ci)，x2=lcm(ci)，则易知有x1,x2对于任意的ci取余的值都相等，且因为k不能整除lcm(ci),所以x1，x2对k取余的值不等。 所以ci不能确定x%k的值。

思路2：
题目已经给我们n个线性同余方程：x%c[i]==a[i]%c[i]。
由于a数组是由我们任意选择的，所以我们可以构造所有的a相等，从而使得对于任意的i，j，有a[i]%(gcd(ci,cj))等于a[j]%(gcd(ci,cj))。这样，由中国剩余定理就可以知道x%(lcm(c1,...,cn))有唯一解了。 那么我们只要使得k能整除lcm(c1,...,cn)那么易知x%k的值也是固定的。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef __int64 LL;

LL n, k;

LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a%b); }
LL lcm(LL a, LL b) { return a*b / gcd(a, b); }

int main() {
	scanf("%I64d%I64d", &n, &k);
	LL lcm_ci = 1;
	bool su = 0;
	for (int i = 1; i <= n; i++) {
		LL x;
		scanf("%I64d", &x);
		lcm_ci = lcm(lcm_ci, x);
		lcm_ci = gcd(k, lcm_ci);
		if (lcm_ci == k) {
			su = 1; break;
		}
	}
	if (su) puts("Yes");
	else puts("No");
	return 0;
}