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Description
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.
Input
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).
Output
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.
Sample Input
4 5
2 3 5 12
Yes
2 7
2 3
No
提示：
以下是引用http://blog.csdn.net/aozil_yang/article/details/51812908
反证法，假设解不唯一，有x1，x2，那么x1，x2满足：
x1 % ci == x2 % ci 并且x1 % k != x2 % k,所以 (x1 - x2) % ci == 0 ，(x1-x2) % k != 0
并且lcm(c1,c2,,,,cn) % ci == 0
,所以lcm % （x1-x2） == 0.
且 (x1-x2) % k != 0;
所以lcm % k != 0
所以命题： 如果解不唯一，那么lcm % k != 0
逆否命题为： 若lcm % k == 0 那么 解唯一！
所以只需要判断lcm是否k 的整数倍，
也就是说这个问题可以转换为：
是否存在x使得 x是c1，c2，c3，，，cn，k的整数倍！
是的话就是yes，否则就是no！
代码：
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; ll gcd(ll a,ll b) { return !b ? a : gcd(b,a%b); } ll lcm(ll a,ll b) { return a*b/gcd(a,b); } int main() { // printf("%d\n",gcd(6,13)); int n; ll k; scanf("%d%I64d",&n,&k); ll ans = 1; bool ok = false; for (int i = 0; i < n; ++i) { ll x; scanf("%I64d",&x); if (ans) ans = lcm(ans,x) % k; if (ans == 0)ok=true; } printf("%s\n",ok? "Yes" : "No" ); return 0; }
codeforces 360 D - Remainders Game
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原文地址：http://www.cnblogs.com/shawn-ji/p/5671624.html
