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;
}

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